physics notes18

1773

Chapter 18 Heat and the First Law of Thermodynamics Conceptual Problems 1 • Object A has a mass that is twice the mass of object B and body A has a specific heat that is twice the specific heat of object B. If equal amounts of heat are transferred to these objects, how do the subsequent changes in their temperatures compare? (a) A B4Δ = ΔT T , (b) A B2Δ = ΔT T , (c) A BΔ = ΔT T ,

(d) A B12

Δ = ΔT T , (e) A B14

Δ = ΔT T

Picture the Problem We can use the relationship TmcQ Δ= to relate the temperature changes of objects A and B to their masses, specific heats, and the amount of heat supplied to each. Express the change in temperature of object A in terms of its mass, specific heat, and the amount of heat supplied to it:

AAA cm

QT =Δ

Express the change in temperature of object B in terms of its mass, specific heat, and the amount of heat supplied to it:

BBB cm

QT =Δ

Divide the second of these equations by the first to obtain:

BB

AA

A

B

cmcm

TT

=ΔΔ

Substitute and simplify to obtain:

( )( ) 422

BB

BB

A

B ==ΔΔ

cmcm

TT

⇒ B41

A ΔΔ TT =

and ( )e is correct.

2 • Object A has a mass that is twice the mass of object B. The temperature change of object A is equal to the temperature change of object B when the objects absorb equal amounts of heat. It follows that their specific heats are related by (a) A B2=c c , (b) A B2 =c c , (c) A B=c c , (d) none of the above Picture the Problem We can use the relationship TmcQ Δ= to relate the temperature changes of objects A and B to their masses, specific heats, and the amount of heat supplied to each.

Chapter 18

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Relate the temperature change of object A to its specific heat and mass:

AAAΔ

cmQT =

Relate the temperature change of object B to its specific heat and mass:

BBBΔ

cmQT =

Equate the temperature changes and simplify to obtain:

AABB

11cmcm

=

Solve for cA: B2

1B

B

BB

A

BA 2

ccm

mcmmc ===

and )( b is correct.

3 • [SSM] The specific heat of aluminum is more than twice the specific heat of copper. A block of copper and a block of aluminum have the same mass and temperature (20ºC). The blocks are simultaneously dropped into a single calorimeter containing water at 40ºC. Which statement is true when thermal equilibrium is reached? (a) The aluminum block is at a higher temperature than the copper block. (b) The aluminum block has absorbed less energy than the copper block. (c) The aluminum block has absorbed more energy than the copper block. (d) Both (a) and (c) are correct statements. Picture the Problem We can use the relationship TmcQ Δ= to relate the amount of energy absorbed by the aluminum and copper blocks to their masses, specific heats, and temperature changes. Express the energy absorbed by the aluminum block:

TcmQ Δ= AlAlAl

Express the energy absorbed by the copper block:

TcmQ Δ= CuCuCu


TcmTcm

QQ

ΔΔ

=AlAl

CuCu

Al

Cu

Because the block’s masses are the same and they experience the same change in temperature:

1Al

Cu

Al

Cu <=cc

QQ

or

AlCu QQ < and )( c is correct.

Heat and the First Law of Thermodynamics

1775

4 • A block of copper is in a pot of boiling water and has a temperature of 100°C. The block is removed from the boiling water and immediately placed in an insulated container which has a quantity of water that has a temperature of 20°C and has the same mass as the block of copper. (The heat capacity of the insulated container is negligible.) The final temperature will be closest to (a) 40°C, (b) 60°C, (c) 80°C. Determine the Concept We can use the relationship TmcQ Δ= to relate the temperature changes of the block of copper and the boiling water to their masses, specific heats, and the amount of heat supplied to or absorbed by each. Relate the heat supplied by the block of copper to its specific heat, mass, and temperature change as it cools to its equilibrium temperature:

( )C100ΔΔ

equilCuCu

CuCuCuCu

°−==

TcmTcmQ

Relate the heat absorbed by the water to its specific heat, mass, and temperature change as it warms to its equilibrium temperature:

( )C20

ΔΔ

equilOHOH

OHOHOHOH

22

2222

°−=

=

Tcm

TcmQ

Because thermal energy is conserved in this mixing process:

0ΔΔΔ OHCu 2=+= QQQ

or ( ) ( ) 0C20C100 equilOHOHequilCuCu 22

=°−+°− TcmTcm

Because the mass of the water is equal to the mass of the block of copper:

( ) ( ) 0C20C100 equilOHequilCu 2=°−+°− TcTc

Solving for Tequil yields:

( ) ( )OHCu

OHCuequil

2

2C20C100

cccc

T+

°+°=

Substitute numerical values (See Table 18-1 for the specific heats of copper and water.)and evaluate Tequil:

( )( ) ( )( ) C40KJ/mol 75.2KJ/mol 5.24

KJ/mol 75.2C20KJ/mol 5.24C100equil °=

⋅+⋅⋅°+⋅°

=T

and ( )b is correct.

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5 • You pour both a certain amount of water at 100°C and an equal amount of water at room temperature (20°C) into an insulated container. The final temperature of the mixture will be (a) 60°C, (b) less than 60°C, (c) greater than 60°C. Determine the Concept We can use the relationship TmcQ Δ= to relate the temperature changes of the room temperature water and the boiling water to their masses, specific heats, and the amount of heat supplied to or absorbed by each. Relate the heat supplied by the water to its specific heat, mass, and temperature change as it cools to its equilibrium temperature:

( )C100

ΔΔ

equil

OHhot

O2Hhot O2Hhot

O2Hhot O2H boilingO2Hhot 2

°−=

=

Tcm

TcmQ

Relate the heat absorbed by the water to its specific heat, mass, and temperature change as it warms to its equilibrium temperature:

( )C20

ΔΔ

equilOHOH

OHOHOHOH

22

2222

°−=

=

Tcm

TcmQ

Because thermal energy is conserved in this mixing process:

0ΔΔΔ OHOHhot 22=+= QQQ

or ( ) ( ) 0C20C100 equilOHOHequilOHhot OHhot 2222

=°−+°− TcmTcm

Because the mass of the boiling water is equal to the mass of the water that is initially at 20°C and the specific heat of water is independent of its temperature:

0C20C100 equilequil =°−+°− TT

Solving for Tequil yields:

C60equil °=T

and ( )a is correct.

6 • You pour some water at 100°C and some ice cubes at 0°C into an insulated container. The final temperature of the mixture will be (a) 50°C, (b) less than 50°C but larger than 0°C, (c) 0°C, (d) You cannot tell the final temperature from the data given. Determine the Concept We would need to know the mass of the boiling water and the mass of the melting ice in order to determine the final temperature of the mixture. ( )d is correct.


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7 • You pour water at 100oC and some ice cubes at 0°C into an insulated container. When thermal equilibrium is reached, you notice some ice remains and floats in liquid water. The final temperature of the mixture is (a) above 0°C, (b) less than 0°C, (c) 0°C, (d) You cannot tell the final temperature from the data given Determine the Concept Because some ice remains when thermal equilibrium is reached, the equilibrium temperature must be the temperature of the ice cubes at their melting point. ( )c is correct.

8 • Joule’s experiment establishing the mechanical equivalence of heat involved the conversion of mechanical energy into internal energy. Give some everyday examples in which some of the internal energy of a system is converted into mechanical energy. Determine the Concept One example from everyday life is that of a hot gas that expands and does work−as in the piston of an engine. Another example is the gas escaping from a can of spray paint. As it escapes, it moves air molecules and, in the process, does work against atmospheric pressure. 9 • Can a gas absorb heat while its internal energy does not change? If so, give an example. If not, explain why not. Determine the Concept Yes, if the absorbed heat is converted into mechanical work. Providing this condition is satisfied, no energy goes into changing the internal energy of the gas and its temperature will remain constant. 10 • The equation ΔEint = Q + W is the formal statement of the first law of thermodynamics. In this equation, the quantities Q and W, respectively, represent (a) the heat absorbed by the system and the work done by the system, (b) the heat absorbed by the system and the work done on the system, (c) the heat released by the system and the work done by the system, (d) the heat released by the system and the work done on the system. Determine the Concept According to the first law of thermodynamics, the change in the internal energy of the system is equal to the heat that enters the system plus the work done on the system. )( b is correct.

11 • [SSM] A real gas cools during a free expansion, while an ideal gas does not cool during a free expansion. Explain the reason for this difference. Determine the Concept Particles that attract each other have more potential energy the farther apart they are. In a real gas the molecules exert weak attractive forces on each other. These forces increase the internal potential energy during an

Chapter 18

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expansion. An increase in potential energy means a decrease in kinetic energy, and a decrease in kinetic energy means a decrease in translational kinetic energy. Thus, there is a decrease in temperature. 12 • An ideal gas that has a pressure of 1.0 atm and a temperature of 300 K is confined to half of an insulated container by a thin partition. The other half of the container is a vacuum. The partition is punctured and equilibrium is quickly reestablished. Which of the following is correct? (a) The gas pressure is 0.50 atm and the temperature of the gas is 150 K. (b) The gas pressure is 1.0 atm and the temperature of the gas is 150 K. (c) The gas pressure is 0.50 atm and the temperature of the gas is 300 K. (d) None of the above. Determine the Concept Because the container is insulated, no energy is exchanged with the surroundings during the expansion of the gas. Neither is any work done on or by the gas during this process. Hence, the internal energy of the gas does not change and we can conclude that the equilibrium temperature will be the same as the initial temperature. Applying the ideal-gas law for a fixed amount of gas we see that the pressure at equilibrium must be half an atmosphere. )( c

is correct. 13 • A gas consists of ions that repel each other. The gas undergoes a free expansion in which no heat is absorbed or released and no work is done. Does the temperature of the gas increase, decrease, or remain the same? Explain your answer. Determine the Concept Particles that repel each other have more potential energy the closer together they are. The repulsive forces decrease the internal potential energy during an expansion. A decrease in potential energy means an increase in kinetic energy, and an increase in kinetic energy means an increase in translational kinetic energy. Thus, there is an increase in temperature. 14 • Two gas-filled rubber balloons that have equal volumes are located at the bottom of a dark, cold lake. The temperature of the water decreases with increasing depth. One balloon rises rapidly and expands adiabatically as it rises. The other balloon rises more slowly and expands isothermally. The pressure in each balloon remains equal to the pressure in the water just next to the balloon. Which balloon has the larger volume when it reaches the surface of the lake? Explain your answer. Determine the Concept The balloon that expands isothermally is larger when it reaches the surface. The balloon that expands adiabatically will be at a lower temperature than the one that expands isothermally. Because each balloon has the


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same number of gas molecules and are at the same pressure, the one with the higher temperature will be bigger. An analytical argument that leads to the same conclusion is shown below. Letting the subscript ″a″ denote the adiabatic process and the subscript ″i″ denote the isothermal process, express the equation of state for the adiabatic balloon:

γγaf,f00 VPVP = ⇒

γ1

f

00af, ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

For the isothermal balloon: if,f00 VPVP = ⇒ ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

f

00if, P

PVV

Divide the second of these equations by the first and simplify to obtain:

λ

γ

11

f

01

f

00

f

00

af,

if,−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

=PP

PPV

PPV

VV

Because P0/Pf > 1 and γ > 1:

af,if, VV >

15 • A gas changes its state quasi-statically from A to C along the paths shown in Figure 18-21. The work done by the gas is (a) greatest for path A→B→C, (b) least for path A→C, (c) greatest for path A→D→C, (d) The same for all three paths. Determine the Concept The work done along each of these paths equals the area under its curve. The area is greatest for the path A→B→C and least for the path A→D→C. )( a is correct.

16 • When an ideal gas undergoes an adiabatic process, (a) no work is done by the system, (b) no heat is transferred to the system, (c) the internal energy of the system remains constant, (d) the amount of heat transferred into the system equals the amount of work done by the system. Determine the Concept An adiabatic process is, by definition, one for which no heat enters or leaves the system. )( b is correct.

Chapter 18

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17 • True or false: (a) When a system can go from state 1 to state 2 by several different processes,

the amount of heat absorbed by the system will be the same for all processes.

(b) When a system can go from state 1 to state 2 by several different processes, the work done on the system will be the same for all processes.

(c) When a system can go from state 1 to state 2 by several different processes, the change in the internal energy of the system will be the same for all processes.

(d) The internal energy of a given amount of an ideal gas depends only on its absolute temperature.

(e) A quasi-static process is one in which the system is never far from being in equilibrium.

(f) For any substance that expands when heated, its Cp is greater than its Cv. (a) False. The amount the internal energy of the system changes along any path depends only on the temperatures at state 1 and state 2, but because

oninintΔ WQE += , Qin depends on Won, which, in turn, depends on the path taken from state 1 to state 2. (b) False. The work done on the system depends on the path taken from state 1 to state 2. (c) True. The amount the internal energy of the system changes along any path depends only on the temperatures at state 1 and state 2. (d) True. For an ideal gas, nRTE 2

3int = . For a ″given amount″ of an ideal gas, n

is constant. (e) True. This is the definition of a quasi-static process. (f) True. All materials have values for Cp and Cv for which it is true that Cp is greater than Cv. For liquids and solids we generally ignore the very small difference between Cp and Cv . 18 • The volume of a sample of gas remains constant while its pressure increases. (a) The internal energy of the system is unchanged. (b) The system does work. (c) The system absorbs no heat. (d) The change in internal energy must equal the heat absorbed by the system. (e) None of the above. Determine the Concept For a constant-volume process, no work is done on or by the gas. Applying the first law of thermodynamics, we obtain Qin = ΔEint. Because the temperature must change during such a process, we can conclude that


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ΔEint ≠ 0 and hence Qin ≠ 0. )(d is correct.

19 • When an ideal gas undergoes an isothermal process, (a) no work is done by the system, (b) no heat is absorbed by the system, (c) the heat absorbed by the system equals the change in the system’s internal energy, (d) the heat absorbed by the system equals the work done by the system. Determine the Concept Because the temperature does not change during an isothermal process, the change in the internal energy of the gas is zero. Applying the first law of thermodynamics, we obtain Qin = −Won = Wby the system. Hence

)(d is correct.

20 •• Consider the following series of sequential quasi-static processes that a system undergoes: (1) an adiabatic expansion, (2) an isothermal expansion, (3) an adiabatic compression and (4) an isothermal compression which brings the system back to its original state. Sketch the series of processes on a PV diagram, and then sketch the series of processes on a VT diagram (in which volume is plotted as function of temperature). Determine the Concept In the graphs shown below, ″I″ denotes an isothermal process, ″A″ denotes an adiabatic process, ″E″ denotes an expansion, and ″C″ denotes a compression.

V

P

AE

IE

AC

IC

V

AE

IE

AC

IC

T 21 •• [SSM] An ideal gas in a cylinder is at pressure P and volume V. During a quasi-static adiabatic process, the gas is compressed until its volume has decreased to V/2. Then, in a quasi-static isothermal process, the gas is allowed to expand until its volume again has a value of V. What kind of process will return the system to its original state? Sketch the cycle on a graph.

Chapter 18

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Determine the Concept Adiabatic processes are steeper than isothermal processes. During an adiabatic process the pressure varies as γ−V , whereas during an isothermal processes the pressure varies as 1−V . Thus when a gas expands isothermally, the pressure doesn’t quite go down as far as the starting point. Thus the final process necessary is a constant-volume process dropping back to the initial pressure. Heat must be removed from the system in order for this to occur.

V

P

adiabatic

isothermal

iVi21

f VV =

constant volume

22 •• Metal A is denser than metal B. Which would you expect to have a higher heat capacity per unit mass, metal A or metal B? Why? Determine the Concept We can use the definition of heat capacity, the Dulong-Petit law, and the relationship between the mass of a substance and its molar mass to predict whether metal A or metal B has a higher heat capacity per unit mass. The specific heat of metal A is given by:

AA M

c'c =

From the Dulong-Petit law, the molar heat capacity of most metals is approximately:

Rc' 3=

Substituting for c′ yields: A

A3M

Rc =

Similarly, the specific heat of metal B is given by: B

B3M

Rc =

Dividing cA by cB and simplifying yields:

A

B

B

A

B

A

3

3

MM

MR

MR

cc

== ⇒ BA

BA c

MMc =

Because metal A is denser than metal B: BA cc >


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23 •• An ideal gas undergoes a process during which P V = constant and the volume of the gas decreases. Does its temperature increase, decrease, or remain the same during this process? Explain. Picture the Problem We can use the given dependence of the pressure on the volume and the ideal-gas law to show that if the volume decreases, so does the temperature. We’re given that: constant=VP

Because the gas is an ideal gas: ( )

nRTVVVPPV

=×== constant

Solving for T yields: ( ) V

nRVT ∝=

constant

Because T varies with the square root of V, if the volume decreases, the temperature decreases. Estimation and Approximation 24 • During the early stages of designing a modern electric generating plant, you are in charge of the team of environmental engineers. The new plant is to be located on the ocean and will use ocean water for cooling. The plant will produce electrical power at the rate of 1.00 GW. Because the plant will have an efficiency of one-third (typical of most modern plants), heat will be released to the cooling water at the rate of 2.00 GW. If environmental codes require that only water with a temperature increase of 15°F or less can be returned to the ocean, estimate the flow rate (in kg/s) of cooling water through the plant. Picture the Problem The rate at which thermal energy is delivered to the ocean by the power plant is given by tQP ΔΔ= where TmcQ Δ= Express the rate at which thermal energy is delivered to the ocean: ( ) Tc

tmTmc

ttQP Δ

ΔΔΔ

ΔΔ

ΔΔ

===

Our interest is in the flow rate of the cooling water. Solving for tm ΔΔ yields:

TcP

tm

ΔΔΔ

=

Chapter 18

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Substitute numerical values and evaluate tm ΔΔ :

kg/s 107.5

F 9C 5F 15

CkgkJ 4.184

J/s 1000.2ΔΔ

4

9

×=

⎟⎠⎞

⎜⎝⎛

°°

×°⎟⎟⎠

⎞⎜⎜⎝

⎛°⋅

×=

tm

25 •• [SSM] A ″typical″ microwave oven has a power consumption of about 1200 W. Estimate how long it should take to boil a cup of water in the microwave assuming that 50% of the electrical power consumption goes into heating the water. How does this estimate correspond to everyday experience? Picture the Problem Assume that the water is initially at 30°C and that the cup contains 200 g of water. We can use the definition of power to express the required time to bring the water to a boil in terms of its mass, heat capacity, change in temperature, and the rate at which energy is supplied to the water by the microwave oven. Use the definition of power to relate the energy needed to warm the water to the elapsed time:

tTmc

tWP

ΔΔ

=ΔΔ

= ⇒P

Tmct Δ=Δ

Substitute numerical values and evaluate Δt:

( ) ( )min1.6s63.97

W600

C30C100Kkg

kJ184.4kg200.0Δ ≈=

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=t ,

an elapsed time that seems to be consistent with experience. 26 •• A demonstration of the heating of a gas under adiabatic compression involves putting a small strip of paper into a large glass test tube, which is then sealed off with a piston. If the piston compresses the trapped air very rapidly, the paper will catch fire. Assuming that the burning point of paper is 451ºF, estimate the factor by which the volume of the air trapped by the piston must be reduced for this demonstration to work. Picture the Problem The adiabatic compression from an initial volume V1 to a final volume V2 between the isotherms at temperatures T1 and T2 is shown to the right. We’ll assume a room temperature of 300 K and apply the equation for a quasi-static adiabatic compression to 451ºF (506 K) with γair = 1.4 to solve for the ratio of the initial to the final volume of the air.


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P

V

K 3001 =T

K 5062 =T

2V 1V Express constant1 =−γTV in terms of the initial and final values of T and V:

122

111

−− = γγ VTVT ⇒1

1

1

2

2

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γ

TT

VV

Substitute numerical values and evaluate V1/V2:

69.3K300K506 14.1

1

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

VV

27 •• A small change in the volume of a liquid occurs when heating the liquid at constant pressure. Use the following data to estimate the contribution this fractional change makes to the heat capacity of water between 4ºC and 100ºC. The density of water at 4.00ºC and 1.00 atm pressure is 1.000 g/cm3. The density of liquid water at 100ºC and 1.00 atm pressure is 0.9584 g/cm3. Picture the Problem The heat capacity Cp of a sample of water is equal to the mass m of the sample times the specific heat capacity cp of water, where cp = 4186 J/kg·K. When the sample is heated, its temperature and its volume increases. The increase in temperature reflects the increase in the internal energy of the sample, and the increase in volume reflects the amount of work Won done by the sample on its surroundings. That is, Won = PΔV, where P is the pressure and ΔV is the change in volume of the sample. We are looking for the ratio of the work done by the sample to the heat absorbed by the sample. The heat that enters the water during the constant pressure process is given by:

TmcQ Δ= pin

The work that is done by the sample during this process is given by:

( )C4.00C100by °° −=Δ= VVPVPW

Chapter 18

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The volumes of the sample at these temperatures is related to its densities at these temperatures:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

°°

°°

C4.00C100

C4.00C100by

11ρρ

ρρ

Pm

mmPW

where m is the mass of the sample.

Taking the ratio of Wby to Qin yields:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ=

°°

°°

C00.4C100p

C00.4C100pin

by

11

11

ρρ

ρρ

TcP

TmcPm

QW

Substitute numerical values and evaluate Wby/Qin:

( )

3by 3

in3 3

101.325 10 Pa 1 1 1.1 10 %kg kgJ 954.3 10004186 96 K m mkg K

−

⎛ ⎞⎜ ⎟×

= − = ×⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⋅⎝ ⎠

WQ

Heat Capacity, Specific Heat, Latent Heat 28 • You designed a solar home that contains 1.00 × 105 kg of concrete (specific heat = 1.00 kJ/kg⋅K). How much heat is released by the concrete at night when it cools from 25.0ºC to 20.0ºC? Picture the Problem We can use the relationship TmcQ Δ= to calculate the amount of heat given off by the concrete as it cools from 25.0°C to 20.0°C. Relate the heat given off by the concrete to its mass, specific heat, and change in temperature:

TmcQ Δ=

Substitute numerical values and evaluate Q:

( ) ( ) MJ500C0.02C0.52Kkg

kJ1.00kg101.00 5 =°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

×=Q

29 • [SSM] How much heat must be absorbed by 60.0 g of ice at –10.0ºC to transform it into 60.0 g of water at 40.0ºC? Picture the Problem We can find the amount of heat that must be absorbed by adding the heat required to warm the ice from −10.0°C to 0°C, the heat required to


1787

melt the ice, and the heat required to warm the water formed from the ice to 40.0°C. Express the total heat required: waterwarmicemelticewarm QQQQ ++=

Substitute for each term to obtain: ( )waterwaterficeice

waterwaterficeice

TcLTcmTmcmLTmcQ

Δ++Δ=Δ++Δ=

Substitute numerical values (See Tables 18-1 and 18-2) and evaluate Q:

( ) ( )( )

( )

kJ3.31

C0C0.04Kkg

kJ184.4

kgkJ5.333C0.01C0

KkgkJ05.2kg0.0600

=

⎥⎦

⎤°−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

+

+⎢⎣

⎡°−−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

=Q

30 •• How much heat must be released by 0.100 kg of steam at 150ºC to transform it into 0.100 kg of ice at 0ºC? Picture the Problem We can find the amount of heat that must be removed by adding the heat that must be removed to cool the steam from 150°C to 100°C, the heat that must be removed to condense the steam to water, the heat that must be removed to cool the water from 100°C to 0°C, and the heat that must be removed to freeze the water. Express the total heat that must be removed:

waterfreezewatercool

steamcondensesteamcool

QQ

QQQ

++

+=

Substitute for each term and simplify to obtain:

( )fwaterwatervsteamsteam

fwaterwatervsteamsteam

ΔΔΔΔ

LTcLTcmmLTmcmLTmcQ

+++=+++=

Chapter 18

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Substitute numerical values (See Tables 18-1 and 18-2) and evaluate Q:

( ) ( )

( )

kJ311

kgkJ5.333C0C100

KkgkJ184.4

kgMJ26.2C100C150

KkgkJ02.2kg0.100

=

⎥⎦

⎤+°−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

+

+⎢⎣

⎡°−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

=Q

31 •• A 50.0-g piece of aluminum at 20ºC is cooled to –196ºC by placing it in a large container of liquid nitrogen at that temperature. How much nitrogen is vaporized? (Assume that the specific heat of aluminum is constant over this temperature range.) Picture the Problem We can find the amount of nitrogen vaporized by equating the heat gained by the liquid nitrogen and the heat lost by the piece of aluminum. Express the heat gained by the liquid nitrogen as it cools the piece of aluminum:

Nv,NN LmQ =

Express the heat lost by the piece of aluminum as it cools:

AlAlAlAl TcmQ Δ=

Equate these two expressions and solve for mN:

AlAlAlNv,N ΔTcmLm =

and

Nv,

AlAlAlN

ΔL

Tcmm =

Substitute numerical values (see Table 18-1 for the specific heat of aluminum) and evaluate mN:

( ) ( )( )g48.8

kgkJ199

C196C20Kkg

kJ0.900kg0.0500

N =°−−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

=m

32 •• You are supervising the creation of some lead castings for use in the construction industry. Each casting involves one of your workers pouring 0.500 kg of molten lead that has a temperature of 327ºC into a cavity in a large block of ice at 0ºC. How much liquid water should you plan on draining per hour if there


1789

are 100 workers who are able to each average one casting every 10.0 min? Picture the Problem Because the heat lost by the lead as it cools is gained by the block of ice (we’re assuming no heat is lost to the surroundings), we can apply the conservation of energy to determine how much ice melts. Apply the conservation of energy to this process:

0ΔΔΔ WPb =+= QQQ or

( ) 0Δ wf,casting 1 w,PbPbPbf,Pb =++− LmTcLm

Solving for mw yields: ( )

Wf,

PbPbPbf,Pbcasting 1 W,

ΔL

TcLmm

+=

Because there are N workers each turning out n castings per hour: ( )

Wf,

PbPbPbf,Pb

casting 1 W.

ΔL

TcLnNm

nNmm

+=

=

Substitute numerical values and evaluate mw:

( )( )( ) ( )kg60

kgkJ333.5

C0C327Kkg

kJ0.128kgkJ7.24kg500.06100

≈⎟⎟⎠

⎞⎜⎜⎝

⎛°−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

+

=m

Calorimetry 33 • [SSM] While spending the summer on your uncle’s horse farm, you spend a week apprenticing with his farrier (a person who makes and fits horseshoes). You observe the way he cools a shoe after pounding the hot, pliable shoe into the correct size and shape. Suppose a 750-g iron horseshoe is taken from the farrier’s fire, shaped, and at a temperature of 650°C, dropped into a 25.0-L bucket of water at 10.0°C. What is the final temperature of the water after the horseshoe and water arrive at equilibrium? Neglect any heating of the bucket and assume the specific heat of iron is K)J/(kg 460 ⋅ . Picture the Problem During this process the water will gain energy at the expense of the horseshoe. We can use conservation of energy to find the equilibrium temperature. See Table 18-1 for the specific heat of water.

Chapter 18

1790

Apply conservation of energy to obtain:

0horseshoe thecool water thewarmi

i =+=∑ QQQ

or ( ) ( ) 0C650C0.10 fFeFefwaterwater =°−+°− tcmtcm

Solve for tf to obtain:

( ) ( )FeFewaterwater

FeFewaterwaterf

C650C0.10cmcmcmcmt

+°+°

=

Substitute numerical values and evaluate tf:

( ) ( ) ( ) ( )

( ) ( )

C1.12

KkgkJ 460.0kg 750.0

KkgkJ 184.4kg 0.25

C650Kkg

kJ 460.0kg 750.0C0.10Kkg

kJ 184.4kg 0.25

f

°=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=t

34 • The specific heat of a certain metal can be determined by measuring the temperature change that occurs when a piece of the metal is heated and then placed in an insulated container that is made of the same material and contains water. Suppose the piece of metal has a mass of 100 g and is initially at 100ºC. The container has a mass of 200 g and contains 500 g of water at an initial temperature of 20.0ºC. The final temperature is 21.4ºC. What is the specific heat of the metal? Picture the Problem During this process the water and the container will gain thermal energy at the expense of the piece of metal. Applying conservation of energy will lead to an expression you can solve for the specific heat of the metal. Apply conservation of energy to obtain:

0sample metal thecoolcontainer thewarm water thewarmi

i =++=∑ QQQQ

or 0ΔΔΔ metalmetalmetalwmetalcontainerwww =++ TcmTcmTcm

Solving for cmetal yields:

wcontainermetalmetal

wwwmetal ΔΔ

ΔTmTm

Tcmc−

=


1791

Substitute numerical values and evaluate cmetal :

( ) ( )

( )( ) ( )( )

KkgkJ 39.0

C0.20C4.21kg0.200C4.21C100kg0.100

C0.20C4.21Kkg

kJ184.4kg0.500

metal

⋅=

°−°−°−°

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=c

Remarks: Consulting Table 18-1, we note that the metal is copper. 35 •• During his many appearances at the Tour de France, champion bicyclist Lance Armstrong typically expended an average power of 400 W, 5.0 hours a day for 20 days. What quantity of water, initially at 24ºC, could be brought to a boil if you could harness all of that energy? Picture the Problem We can use TmcQ Δ= to express the mass m of water that can be heated through a temperature interval ΔT by an amount of heat energy Q. We can then find the amount of heat energy expended by Armstrong from the definition of power. Express the amount of heat energy Q required to raise the temperature of a mass m of water by ΔT:

TmcQ Δ= ⇒Tc

QmΔ

=

Use the definition of power to relate the heat energy expended by Armstrong to the rate at which he expended the energy:

tQPΔ

= ⇒ tPQ Δ=

Substitute for Q to obtain: Tc

tPmΔΔ

=

Substitute numerical values and evaluate m:

( )

kg105.4

C24C100Kkg

kJ184.4

d 20dh0.5

hs 3600

sJ400

2×=

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛ ×⎟

⎠⎞

⎜⎝⎛ ×

=m

36 •• A 25.0-g glass tumbler contains 200 mL of water at 24.0ºC. If two 15.0-g ice cubes, each at a temperature of –3.00ºC, are dropped into the tumbler, what is the final temperature of the drink? Neglect any heat transfer between the tumbler and the room.

Chapter 18

1792

Picture the Problem First you need to convince yourself that there is enough energy in the water to melt all the ice. Once you’ve done that you can use conservation of energy to find the final equilibrium temperature. See Tables 18-1 and 18-2 for specific heats and the latent heat of fusion of water. First, determine the energy required to warm and melt all the ice:

( ) ( ) ( )( )

kJ 10.19kJ 01.10kJ 1845.0

kJ/kg 5.333kg 0300.0C 00.3Kkg

kJ2.05kg 0300.0

Δ ficeice warmiceiceicemelt ice warmneeded

=+=

+°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=

+=+= LmTcmQQQ

The maximum amount of energy available from the water and the tumbler is:

( ) ( )

( )kJ 20.59 kJ 08.20kJ 5040.0

C0.24C0Kkg

kJ 184.4L

m 10L 10200mkg 1000.1

C0.24C0Kkg

kJ .8400kg 0250.0

ΔΔ

ΔΔ

333

33

watermax,waterwaterwater tumblermax,glasstumbler

watermax,waterwater tumblermax,glasstumblermax available,

−=−−=°−°×

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛××⎟

⎠⎞

⎜⎝⎛ ×+

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=

+=

+=

−−

TcVTcm

TcmTcmQ

ρ

Because max available,Q > neededQ , all the ice will melt and the final temperature will be

greater than 0°C. Apply conservation of energy to obtain:

0 waterice thewarm

ice theallmelt ice thewarm water warm thecool tumbler thecooli

i

=+

+++=∑Q

QQQQQ

or

0Δ

ΔΔΔ

watericewatericeice f,ice

icewarn iceicewaterwaterwatertumblerglasstumbler

=++

++

TcmLm

TcmTcmTcm


1793

Substituting numerical values yields:

( ) ( ) ( ) ( )

( ) ( ) 0C0Kkg

kJ184.4kg 0300.0kJ 01.10 kJ 1845.0

C0.24 Kkg

kJ184.4kg 200.0C0.24Kkg

kJ840.0kg 0250.0

f

ff

=°−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+++

°−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+°−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

t

tt

Solving for tf yields:

C6.10f °=t

37 •• [SSM] A 200-g piece of ice at 0ºC is placed in 500 g of water at 20ºC. This system is in a container of negligible heat capacity and is insulated from its surroundings. (a) What is the final equilibrium temperature of the system? (b) How much of the ice melts? Picture the Problem Because we can not tell, without performing a couple of calculations, whether there is enough heat available in the 500 g of water to melt all of the ice, we’ll need to resolve this question first. See Tables 18-1 and 18-2 for specific heats and the latent heat of fusion of water. (a) Determine the energy required to melt 200 g of ice: ( )

kJ70.66kgkJ5.333kg0.200ficeicemelt

=

⎟⎟⎠

⎞⎜⎜⎝

⎛== LmQ

The energy available from 500 g of water at 20ºC is:

( ) ( )

kJ84.14

C02C0Kkg

kJ184.4kg0.500Δ waterwaterwatermax available,

−=

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

== TcmQ

Because max available,Q < icemeltQ : C.0 is re temperatufinal The °

(b) Equate the energy available from the water max available,Q to miceLf and

solve for mice to obtain:

f

max available,ice L

Qm =

Substitute numerical values and evaluate mice:

g125

kgkJ5.333

kJ84.14ice ==m

Chapter 18

1794

38 •• A 3.5-kg block of copper at a temperature of 80ºC is dropped into a bucket containing a mixture of ice and water whose total mass is 1.2 kg. When thermal equilibrium is reached, the temperature of the water is 8.0ºC. How much ice was in the bucket before the copper block was placed in it? (Assume that the heat capacity of the bucket is negligible.) Picture the Problem Because the bucket contains a mixture of ice and water initially, we know that its temperature must be 0°C. We can apply conservation of energy to find the amount of ice initially in the bucket. See Tables 18-1 and 18-2 for specific heats and the heats of fusion and vaporization of water. Apply conservation of energy to obtain:

0blockcopper thecool

waterice thewarm

ice the allmelt

ii =++=∑ QQQQ

Substitute for

ice the allmelt Q ,

waterice thewarmQ , and

blockcopper thecoolQ to obtain:

0ΔΔ CuCuCuwatericewaterwatericefice =++ TcmTcmLm

Solving for mice yields:

f

watericewaterwatericeCuCuCuice L

TcmTcmm

Δ−Δ=

Substitute numerical values and evaluate mice:

( ) ( )

( ) ( )

kg17.0

kgkJ5.333

C0C0.8Kkg

kJ184.4kg2.1

kgkJ5.333

C0.8C80Kkg

kJ386.0kg5.3

ice

=

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

−

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=m

39 •• A well-insulated bucket of negligible heat capacity contains 150 g of ice at 0ºC. (a) If 20 g of steam at 100ºC is injected into the bucket, what is the final equilibrium temperature of the system? (b) Is any ice left after the system reaches equilibrium?


1795

Picture the Problem First you need to convince yourself that there is enough energy in the steam to melt all the ice. Once you’ve done that you can use conservation of energy to find the final equilibrium temperature. See Tables 18-1 and 18-2 for specific heats and the heats of fusion and vaporization of water. (a) First, determine the energy required to melt all the ice:

( )( )kJ 03.50

kJ/kg 5.333kg 150.0ficeicemelt

=== LmQ

Find the maximum amount of energy available from the steam is:

( ) ( ) ( )

kJ 51.53kJ 8.37 kJ14.45

C001C0Kkg

kJ 184.4kg 020.0kgkJ 2257kg 020.0

Δ maxwater watersteamvsteammax steam,

−=−−=

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

+= TcmLmQ

Because max steam,Q > icemelt Q , all the ice will melt and the final temperature will be


0 waterice

thewarmice the

allmelt hot water

thecoolsteam

condensei

i =+++=∑ QQQQQ

or ( ) ( ) 0C0 kJ 03.50C100 kJ 14.45 fwatericefwaterhot water =°−++°−+− tcmtcm


( ) ( )

( ) ( ) 0C0Kkg

kJ184.4kg 150.0

kJ 03.50C100 Kkg

kJ18.4kg 020.0kJ 14.45

f

f

=°−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+

+°−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+−

t

t

Solving for tf yields:

C9.4f °=t

(b) Because the final temperature is greater than 0°C, no ice is left. 40 •• A calorimeter of negligible heat capacity contains 1.00 kg of water at 303 K and 50.0 g of ice at 273 K. (a) Find the final temperature T. (b) Find the final temperature T if the mass of ice is 500 g.

Chapter 18

1796

Picture the Problem First you need to convince yourself that there is enough energy in the water to melt all the ice. Once you’ve done that you can use conservation of energy to find the final equilibrium temperature. See Tables 18-1 and 18-2 for specific heats and the heat of fusion of water. (a) Find the heat available to melt the ice:

( ) ( )

kJ5.251

K033K732Kkg

kJ184.4kg1.00Δ waterwaterwatermax available,

−=

−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

== TcmQ

Find the heat required to melt all of the ice:

( ) kJ68.61kgkJ5.333kg0.0500ficeicemelt =⎟⎟

⎠

⎞⎜⎜⎝

⎛== LmQ

Because icemeltmax available, QQ > , we know that the final temperature will be

greater than 273 K. Apply conservation of energy to obtain:

0 waterice

thewarmice the

allmelt waterwarm

thecooli

i =++=∑ QQQQ

or

( ) ( )

( ) ( ) 0K 273Kkg

kJ184.4kg 0500.0

kJ 68.16K 303 Kkg

kJ184.4kg 00.1

=−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+

+−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

T

T

Solving for Tf yields: K298f =T

(b) Find the heat required to melt 500 g of ice: ( )( )

kJ671kJ/kg5.333kg0.500

ficeicemelt

≈=

= LmQ

Because the heat required to melt 500 g of ice is greater than the heat available, the final temperature will be 0°C. 41 •• A 200-g aluminum calorimeter contains 600 g of water at 20.0ºC. A 100-g piece of ice cooled to –20.0ºC is placed in the calorimeter. (a) Find the final temperature of the system, assuming no heat is transferred to or from the system.


1797

(b) A 200-g piece of ice at –20.0ºC is added. How much ice remains in the system after the system reaches equilibrium? (c) Would the answer for Part (b) change if both pieces of ice were added at the same time? Picture the Problem First you need to convince yourself that there is enough energy in the water and the calorimeter to melt all the ice. Once you’ve done that you can use conservation of energy to find the final equilibrium temperature in (a). In Part (b) you can find the energy required to raise the temperature of the 200 g of ice to 0°C and then, noting that there are now 700 g of water in the calorimeter, find the energy available from cooling the calorimeter and water from their equilibrium temperature to 0°C. Finally, you can find the amount of ice that will melt in terms of the difference between the energy available and the energy required to warm the ice. See Tables 18-1 and 18-2 for specific heats and the latent heat of fusion of water. Find the energy available to melt the ice:

( )

( ) ( ) ( )

kJ81.53

C0.20C0Kkg

kJ900.0kg200.0Kkg

kJ184.4kg600.0

ΔΔΔ watercalcalwaterwaterwatercalcalwaterwaterwatermax available,

=

°−°⎥⎦

⎤⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=

+=+= TcmcmTcmTcmQ

Find the energy required to warm and melt all of the ice:

( ) ( )( ) ( )

kJ45.37kgkJ5.333kg0.100C0.20C0

KkgkJ2.05kg0.100

Δ ficeiceiceiceice melt the

andwarm

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+°−−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

=

+= LmTcmQ

(a) Because icemeltmax available, QQ > , we know that the final temperature will be


0 waterice

thewarmice melt the and warm

waterwarm thecool

rcalorimete thecool

ii =+++=∑ QQQQQ

or, using the results from our preliminary calculations,

( ) ( )( ) 0C0

kJ 45.37C0.20C0.20

fwaterice

fwaterwaterfCurcalorimete

=°−++°−+°−

tcmtcmtcm

Chapter 18

1798


( ) ( ) ( ) ( )

( ) 0Kkg

kJ184.4kg100.0kJ 37.45

C0.20Kkg

kJ4.184kg600.0C0.20Kkg

kJ0.900kg200.0

f

ff

=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

++

°−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+°−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

t

tt

Solve for tf to obtain: C26.5C262.5f °=°=t

(b) The mass of ice remaining in the system when equilibrium is reached is the difference between the initial mass of ice and the mass of the ice that has melted:

f

C0 toice warmavail

initiallyice

meltedice

initiallyice

remainingice L

QQmmmm °

−−=−= (1)

Find the energy required to raise the temperature of the 200 g of ice to 0°C:

( ) ( )( ) kJ080.8C0.20C0Kkg

kJ2.02kg0.200Δ iceiceiceC0 toicewarm =°−−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

==°

TcmQ

Noting that there are now 700 g of water in the calorimeter, find the energy available from cooling the calorimeter and water from 5.262°C to 0°C:

( )

( ) ( ) ( )

kJ359.16

C0C262.5Kkg

kJ900.0kg200.0Kkg

kJ184.4kg700.0

ΔΔΔ watercalcalwaterwaterwatercalcalwaterwaterwateravail

=

°−°⎥⎦

⎤⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=

+=+= TcmcmTcmTcmQ

Substitute numerical values in equation (1) and evaluate

remainingicem :

g 002kJ/kg333.5

kJ8.080kJ359.61g 200remainingice =

−−=m

(c) No. Because the initial and final conditions are the same, the answer would be the same.


1799

42 •• The specific heat of a 100-g block of a substance is to be determined. The block is placed in a 25-g copper calorimeter holding 60 g of water initially at 20ºC. Then, 120 mL of water at 80ºC are added to the calorimeter. When thermal equilibrium is reached, the temperature of the system is 54ºC. Determine the specific heat of the block. Picture the Problem Let the subscript B denote the block, w1 the water initially in the calorimeter, and w2 the 120 mL of water that is added to the calorimeter vessel. We can use conservation of energy to find the specific heat of the block. See Table 18-1 for specific heats. Apply conservation of energy to obtain:

021 wcool wwarm

rcalorimete thewarm

block the warm

ii =+++=∑ QQQQQ

or 0ΔΔΔΔ

22211 wwwwwCuCuBB =+++ TcmTcmTcmTcm

where ΔT is the common temperature change of the calorimeter, block, and water initially in the calorimeter. Substitute numerical values to obtain:

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) 0C80C54Kkg

kJ4.184kg0.120

C02C54Kkg

kJ4.184kg0.060

C02C54Kkg

kJ0.386kg025.0C02C54kg100.0 B

=°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+°−°c

Solving for cB yields: KkJ/kg2.1B ⋅=c 43 •• [SSM] A 100-g piece of copper is heated in a furnace to a temperature tC. The copper is then inserted into a 150-g copper calorimeter containing 200 g of water. The initial temperature of the water and calorimeter is 16.0ºC, and the temperature after equilibrium is established is 38.0ºC. When the calorimeter and its contents are weighed, 1.20 g of water are found to have evaporated. What was the temperature tC? Picture the Problem We can find the temperature t by applying conservation of energy to this calorimetry problem. See Tables 18-1 and 18-2 for specific heats and the heat of vaporization of water.

Chapter 18

1800

Use conservation of energy to obtain:

0sampleCu

thecoolrcalorimete

thewarm waterthe

warmwater

vaporizei

i =+++=∑ QQQQQ

or 0ΔΔΔ CuCuCuwcalcalOHOHOHwf,vaporizedO,H 2222

=+++ TcmTcmTcmLm


( ) ( ) ( )

( ) ( ) ( ) ( ) 0C0.38Kkg

kJ386.0g100C0.16C0.38Kkg

kJ386.0g150

C0.16C0.38Kkg

kJ184.4g200Kkg

kJ2257g1.20

C =−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

t

Solving for tC yields: C618C °=t 44 •• A 200-g aluminum calorimeter contains 500 g of water at 20.0ºC. Aluminum shot with a mass equal to 300 g is heated to 100.0ºC and is then placed in the calorimeter. Find the final temperature of the system, assuming that there is no heat is transfer to the surroundings. Picture the Problem We can find the final temperature by applying conservation of energy to the calorimeter, the water in the calorimeter, and to the cooling aluminum shot. See Table 18-1 for the specific heats of aluminum and water. Use conservation of energy to obtain:

0shot Al

thecoolrcalorimete

thewarmwater

thewarmi

i =++=∑ QQQQ

or 0ΔΔΔ AlAlshotAlcalww =++ TcmTcmTcm

where ΔT is the common temperature change of the water and calorimeter.


( ) ( ) ( )

( ) ( ) 0C100Kkg

kJ0.900g300

C0.20Kkg

kJ386.0g200Kkg

kJ184.4g500

f

f

=°−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+

°−⎥⎦

⎤⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

t

t


1801

Solve for tf to obtain: C9.28f °=t

First Law of Thermodynamics 45 • A diatomic gas does 300 J of work and also absorbs 2.50 kJ of heat. What is the change in internal energy of the gas? Picture the Problem We can apply the first law of thermodynamics to find the change in internal energy of the gas during this process. Apply the first law of thermodynamics to express the change in internal energy of the gas in terms of the heat added to the system and the work done on the gas:

oninint WQE +=Δ

The work done by the gas equals the negative of the work done on the gas. Substitute numerical values and evaluate ΔEint:

kJ2.20J300kJ 50.2Δ int =−=E

46 • If a gas absorbs 1.67 MJ of heat while doing 800 kJ of work, what is the change in the internal energy of the gas? Picture the Problem We can apply the first law of thermodynamics to find the change in internal energy of the gas during this process. Apply the first law of thermodynamics to express the change in internal energy of the gas in terms of the heat added to the system and the work done on the gas:

oninint WQE +=Δ

The work done by the gas is the negative of the work done on the gas. Substitute numerical values and evaluate ΔEint:

MJ70.8kJ800MJ 67.1Δ int =−=E

47 • If a gas absorbs 84 J while doing 30 J of work, what is the change in the internal energy of the gas?

Chapter 18

1802

Picture the Problem We can apply the first law of thermodynamics to find the change in internal energy of the gas during this process. Apply the first law of thermodynamics to express the change in internal energy of the gas in terms of the heat added to the system and the work done on the gas:

oninint WQE +=Δ

The work done by the gas is the negative of the work done on the gas. Substitute numerical values and evaluate ΔEint:

J54J03J 84Δ int =−=E

48 •• A lead bullet initially at 30ºC just melts upon striking a target. Assuming that all of the initial kinetic energy of the bullet goes into the internal energy of the bullet, calculate the impact speed of the bullet. Picture the Problem We can use the definition of kinetic energy to express the speed of the bullet upon impact in terms of its kinetic energy. The heat absorbed by the bullet is the sum of the heat required to warm the bullet from 303 K to its melting temperature of 600 K and the heat required to melt it. We can use the first law of thermodynamics to relate the impact speed of the bullet to the change in its internal energy. See Table 18-1 for the specific heat and melting temperature of lead. Using the first law of thermodynamics, relate the change in the internal energy of the bullet to the work done on it by the target:

oninint WQE +=Δ or, because Qin = 0,

( )iKKKWE −−=Δ==Δ fonint

Substitute for ΔEint, Kf, and Ki to obtain:

( ) 2212

21

Pbf,PbPb 0 mvmvmLTmc =−−=+Δ

or ( ) 2

21

Pbf,iMPPb mvmLTTmc =+−

Solving for v yields:

( )[ ]Pbf,iMPPb2 LTTcv +−=

Substitute numerical values and evaluate v:

( ) m/s354kgkJ24.7K303K600

KkgkJ0.1282 =

⎩⎨⎧

⎭⎬⎫

+−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=v


1803

49 •• During a cold day, you can warm your hands by rubbing them together. Assume the coefficient of kinetic friction between your hands is 0.500, the normal force between your hands is 35.0 N, and that you rub them together at an average relative speed of 35.0 cm/s. (a) What is the rate at which mechanical energy is dissipated? (b) Assume further that the mass of each of your hands is 350 g, the specific heat of your hands is 4.00 kJ/kg⋅K, and that all the dissipated mechanical energy goes into increasing the temperature of your hands. How long must you rub your hands together to produce a 5.00°C increase in their temperature? Picture the Problem We can find the rate at which heat is generated when you rub your hands together using the definition of power and the rubbing time to produce a 5.00 °C increase in temperature from ( ) tdtdQQ Δ=Δ and Q = mcΔT. (a) The rate at which heat is generated as a function of the friction force and the average relative speed of your hands is given by:

vFvfPdtdQ

nk μ===

Substitute numerical values and evaluate dQ/dt:

( )( )( )

W6.13

m/s0.350N35.0.5000

=

=dtdQ

(b) Relate the heat required to raise the temperature of your hands 5.00°C to the rate at which it is being generated:

TmctdtdQQ Δ=Δ=Δ ⇒

dtdQTmct Δ

=Δ

Substitute numerical values and evaluate Δt: ( ) ( )

min1.38s60

min1s2862

W6.13

C5.00Kkg

kJ4.00kg0.3502Δ

=×=

°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=t

Work and the PV Diagram for a Gas 50 • The gas is allowed to expand at constant pressure until it reaches its final volume. It is then cooled at constant volume until it reaches its final pressure. (a) Illustrate this process on a PV diagram and calculate the work done by the gas. (b) Find the heat absorbed by the gas during this process.

Chapter 18

1804

Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isobaric expansion. We can then use the first law of thermodynamics to find the heat added to the system during this process. (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram. 2

0

1

0

P, atm

3.00

2.00

1.00

1.00 2.00 3.00V, L

The work done by the gas equals the area under the shaded curve:

( )( )

J608

Lm10L2.00

atmkPa101.325atm3.00L2.00atm3.00Δ

33

gasby

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎠⎞

⎜⎝⎛ ×===

−

VPW

(b) The work done by the gas is the negative of the work done on the gas. Apply the first law of thermodynamics to the system to obtain:

( ) ( )( ) gasby int,1int,2

gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−Δ=

Substitute numerical values and evaluate Qin:

( )kJ1.06

J608J456J912in

=

+−=Q

51 • [SSM] The gas is first cooled at constant volume until it reaches its final pressure. It is then allowed to expand at constant pressure until it reaches its final volume. (a) Illustrate this process on a PV diagram and calculate the work done by the gas. (b) Find the heat absorbed by the gas during this process. Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isobaric expansion. We can then use the first law of thermodynamics to find the heat absorbed by the gas during this process


1805

(a) The path from the initial state (1) to the final state (2) is shown on the PV diagram. 2

0

1

0

P, atm

3.00

2.00

1.00

1.00 2.00 3.00V, L

The work done by the gas equals the area under the curve:

( )( )

J054

Lm10L2.00

atmkPa101.325atm2.00L2.00atm2.00Δ

33

gasby

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎠⎞

⎜⎝⎛ ×===

−

VPW



gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−Δ=


( ) J618J054J456J912in =+−=Q

52 •• The gas is allowed to expand isothermally until it reaches its final volume and its pressure is 1.00 atm. It is then heated at constant volume until it reaches its final pressure. (a) Illustrate this process on a PV diagram and calculate the work done by the gas. (b) Find the heat absorbed by the gas during this process. Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isothermal expansion. We can then use the first law of thermodynamics to find the heat absorbed by the gas during this process.

Chapter 18

1806

(a) The path from the initial state (1) to the final state (2) is shown on the PV diagram. 2

0

1

0

P, atm

3.00

3.00

2.00

2.00

1.00

1.00V, L


[ ]

3ln

ln

11

L3.00L00.111

L00.3

L00.111

L3

L11gasby

2

1

VP

VVPVdVVP

VdVnRTdVPW

V

V

=

==

==

∫

∫∫

Substitute numerical values and evaluate Wby gas:

J3343lnL

m10L1.00atm

kPa101.325atm3.0033

gasby =⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎠⎞

⎜⎝⎛ ×=

−

W



gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−Δ=


( ) J790J334J456J912in =+−=Q

53 •• The gas is heated and is allowed to expand such that it follows a single straight-line path on a PV diagram from its initial state to its final state. (a) Illustrate this process on a PV diagram and calculate the work done by the gas. (b) Find the heat absorbed by the gas during this process. Picture the Problem We can find the work done by the gas during this process from the area under the curve. We can then use the first law of thermodynamics to find the heat absorbed by the gas during this process.


1807

(a) The path from the initial state (1) to the final state (2) is shown on the PV diagram:

00

1

2

V,L

P, atm

3.00

3.002.00

2.00

1.00

1.00


( )( )

J507Latm

J101.325Latm00.5

L2.00atm2.00atm3.0021

trapezoidgasby

=

⋅×⋅=

+=

= AW



gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−Δ=


( ) J639J507J456J912in =+−=Q

Remarks: You could use the linearity of the path connecting the initial and final states and the coordinates of the endpoints to express P as a function of V. You could then integrate this function between 1.00 and 3.00 L to find the work done by the gas as it goes from its initial to its final state. 54 •• In this problem, 1.00 mol of a dilute gas initially has a pressure equal to 1.00 atm and a volume equal to 25.0 L. As the gas is slowly heated, the plot of its state on a PV diagram moves in a straight line to the final state. The gas now has a pressure equal to 3.00 atm and a volume equal to 75.0 L. Find the work done and the heat absorbed by the gas. Picture the Problem We can find the work done by the gas during this process from the area under the curve and the heat absorbed by the gas from the 1st law of thermodynamics.

Chapter 18

1808

The path from the initial state i to the final state f is shown on the PV diagram:

00

P, atm

i

f

V,L

3.00

2.00

1.00

75.050.025.0


( )( )

kJ1.10Latm

J101.325Latm100

L 25.0L0.75atm3.00atm1.002

1

trapezoidgasby

=⋅

×⋅=

−×+=

= AW

The work done by the gas is the negative of the work done on the gas. Apply the first law of thermodynamics to the system to obtain:


gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−Δ=


( ) kJ6.10kJ1.10J456J912in =+−=Q

Remarks: You could use the linearity of the path connecting the initial and final states and the coordinates of the endpoints to express P as a function of V. You could then integrate this function between 25.0 and 75.0 L to find the work done by the gas as it goes from its initial to its final state. 55 •• In this problem, 1.00 mol of the ideal gas is heated while its volume changes, so that T = AP2, where A is a constant. The temperature changes from T0 to 4T0. Find the work done by the gas. Picture the Problem We can find the work done by the gas from the area under the PV curve provided we can find the pressure and volume coordinates of the initial and final states. We can find these coordinates by using the ideal gas law and the condition .2APT =


1809

Apply the ideal-gas law with n = 1.00 mol and 2APT = to obtain:

2RAPPV = ⇒ RAPV = (1) This result tells us that the volume varies linearly with the pressure.

Solve T = AP2 for the pressure of the gas:

ATP 0

0 =

Find the pressure when the temperature is 4T0:

000 224 P

AT

ATP ===

Using equation (1), express the coordinates of the final state:

( )00 2,2 PV

The PV diagram for the process is shown to the right:

P

2P0

P0

V0

T0

4T0

2V0V

i

f


( )( )

0023

000021

trapezoidgasby 22

VP

VVPPAW

=

−+==

56 •• A sealed, almost-empty spray paint can still contains a residual amount of the propellant: 0.020 mol of nitrogen gas. The can’s warning label clearly states: ″Do Not Dispose by Incineration.″ (a) Explain this warning and draw the PV diagram for the gas if, in fact, the can is subject to a high temperature. (b) You are in charge of testing the can. The manufacturer claims it can withstand an internal gas pressure of 6.00 atm before it breaks. The can is initially at room-temperature and standard pressure in your testing laboratory. You begin to heat it uniformly using a heating element that has a power output of 200 W. The can and element are in an insulating oven, and you can assume 1.0% of the heat released by the heating element is absorbed by the gas in the can. How long should you expect the heating element to remain on before the can bursts? Picture the Problem This is a constant-volume process and so no work is done on the system. The heat that is added to the gas by the heating element increases the internal energy (and, hence, the temperature) of the gas. We can use the definition of power to relate the minimum time-to-bursting to the internal energy required to raise the temperature of the gas in the can and the rate at which energy is supplied by the heating element. The energy required to burst the can is related

Chapter 18

1810

to the increase in the temperature of the can through the definition of the heat capacity at constant volume Cv. (a) Because the volume of the can is fixed (constant), the heat absorbed by the gas in the can all goes into increasing the internal energy of the gas and, thus, its temperature. At constant volume, the pressure rise is the maximum possible, threatening the structural integrity of the can and possibly leading to a grenade-like shrapnel explosion.

f

i

fP

fT

iT

iP

P

V

(b) The heating time required to burst the container is related to the rate at which the heating element delivers energy to it:

PQt ΔΔ =

The heat that the heating element must supply to the gas is given by:

TnRTCQ ΔΔΔ 25

v ==

Substituting for ΔQ yields: P

TnRP

TnRt

2Δ5Δ

Δ 25

== (1)

The temperature difference that corresponds to the bursting threshold is given by:

ifΔ TTT −= (2)

Apply the ideal-gas law to the gas in the can to obtain:

f

ff

i

ii

TVP

TVP

=

or, because Vi = Vf,

f

f

i

i

TP

TP=

Solving for the ratio of the final temperature to the initial temperature yields:

6i

f

i

f ==PP

TT

⇒ if 6TT =

Substitute for Tf in equation (2) and evaluate ΔT:

iii 56Δ TTTT =−=


1811

Substituting for ΔT in equation (1) yields:

PnRTt2

25Δ i=

Substitute numerical values and evaluate Δt:

( ) ( )

( )( ) min .15s 60

min 1s 053 W200010.02

K 293Kmol

J314.8mol 020.025Δ =⋅=

⎟⎠⎞

⎜⎝⎛

⋅=t

57 •• [SSM] An ideal gas initially at 20ºC and 200 kPa has a volume of 4.00 L. It undergoes a quasi-static, isothermal expansion until its pressure is reduced to 100 kPa. Find (a) the work done by the gas, and (b) the heat absorbed by the gas during the expansion. Picture the Problem The PV diagram shows the isothermal expansion of the ideal gas from its initial state 1 to its final state 2. We can use the ideal-gas law for a fixed amount of gas to find V2 and then evaluate ∫ PdV for an

isothermal process to find the work done by the gas. In Part (b) of the problem we can apply the first law of thermodynamics to find the heat added to the gas during the expansion.

200

100 2

1

V2

P, kPa

293 K

4.00V, L

(a) Express the work done by a gas during an isothermal process:

∫∫∫ ===2

1

2

1

2

1

11gasby

V

V

V

V

V

V VdVVP

VdVnRTdVPW

Apply the ideal-gas law for a fixed amount of gas undergoing an isothermal process:

2211 VPVP = ⇒ 12

12 V

PPV =

Substitute numerical values and evaluate V2:

( ) L00.8L4.00kPa100kPa200

2 ==V

Chapter 18

1812

Substitute numerical values and evaluate W:

( )( ) ( )[ ]

( )

J 555

Lm 10LkPa 5.554

L 4.00L 00.8lnLkPa 800

lnLkPa 800L 00.4kPa 200

33

L 8.00L 00.4

L 8.00

L 00.4gasby

=

×⋅=⎟⎠⎞

⎜⎝⎛⋅=

⋅==

−

∫ VVdVW

(b) Apply the first law of thermodynamics to the system to obtain:

onintin WEQ −Δ= or, because ΔEint = 0 for an isothermal process,

onin WQ −=

Because the work done by the gas is the negative of the work done on the gas:

( ) J555gasby gasby in ==−−= WWQ

Remarks: in an isothermal expansion the heat added to the gas is always equal to the work done by the gas (ΔEint = 0). Heat Capacities of Gases and the Equipartition Theorem 58 • The heat capacity at constant volume of a certain amount of a monatomic gas is 49.8 J/K. (a) Find the number of moles of the gas. (b) What is the internal energy of the gas at T = 300 K? (c) What is the heat capacity at constant pressure of the gas? Picture the Problem We can find the number of moles of the gas from its heat capacity at constant volume using nRC 2

3V = . We can find the internal energy of

the gas from TCE Vint = and the heat capacity at constant pressure using nRCC += VP .

(a) Express CV in terms of the number of moles in the monatomic gas:

nRC 23

V = ⇒R

Cn3

2 V=


1813

Substitute numerical values and evaluate n:

mol 99.3

mol 993.3

KmolJ314.83

KJ8.492

=

=⎟⎠⎞

⎜⎝⎛

⋅

⎟⎠⎞

⎜⎝⎛

=n

(b) The internal energy of the gas is related to its temperature:

TCE Vint =

Substitute numerical values and evaluate Eint:

( ) kJ 9.14K 300KJ8.49int =⎟⎠⎞

⎜⎝⎛=E

(c) Relate the heat capacity at constant pressure to the heat capacity at constant volume:

nRnRnRnRCC 25

23

VP =+=+=

Substitute numerical values and evaluate CP:

( )

J/K 0.83

KmolJ314.8mol 993.32

5P

=

⎟⎠⎞

⎜⎝⎛

⋅=C

59 •• [SSM] The heat capacity at constant pressure of a certain amount of a diatomic gas is 14.4 J/K. (a) Find the number of moles of the gas. (b) What is the internal energy of the gas at T = 300 K? (c) What is the molar heat capacity of this gas at constant volume? (d) What is the heat capacity of this gas at constant volume? Picture the Problem (a) The number of moles of the gas is related to its heat capacity at constant pressure and its molar heat capacity at constant pressure according to PP nc'C = . For a diatomic gas, the molar heat capacity at constant pressure is Rc' 2

7P = . (b) The internal energy of a gas depends on its number of

degrees of freedom and, for a diatomic gas, is given by nRTE 25

int = . (c) The molar heat capacity of this gas at constant volume is related to its molar heat capacity at constant pressure according to Rc'c' −= PV . (d) The heat capacity of this gas at constant volume is the product of the number of moles in the gas and its molar heat capacity at constant volume. (a) The number of moles of the gas is the ratio of its heat capacity at constant pressure to its molar heat capacity at constant pressure:

P

P

c'Cn =

Chapter 18

1814

For a diatomic gas, the molar heat capacity is given by: Kmol

J1.2927

P ⋅== Rc'


mol 495.0

mol 4948.0

KmolJ1.29

KJ4.14

=

=

⋅

=n

(b) With 5 degrees of freedom at this temperature:

nRTE 25

int =

Substitute numerical values and evaluate Eint:

( ) ( ) kJ 09.3K 300Kmol

J314.8mol 4948.025

int =⎟⎠⎞

⎜⎝⎛

⋅=E

(c) The molar heat capacity of this gas at constant volume is the difference between the molar heat capacity at constant pressure and the gas constant R:

Rc'c' −= PV

Because Rc' 27

P = for a diatomic gas:

RRRc' 25

27

V =−=

Substitute the numerical value of R to obtain:

KmolJ8.20

KmolJ79.20

KmolJ314.82

5V

⋅=

⋅=⎟

⎠⎞

⎜⎝⎛

⋅=c'

(d) The heat capacity of this gas at constant volume is given by:

VV nc'C' =

Substitute numerical values and evaluate VC' : ( )

KJ3.10

KmolJ79.20mol 4948.0V

=

⎟⎠⎞

⎜⎝⎛

⋅=C'

60 •• (a) Calculate the heat capacity per unit mass of air at constant volume and the heat capacity per unit mass of air at constant pressure. Assume that air has a temperature of 300 K and a pressure of 1.00 × 105 N/m2. Also assume that air is


1815

composed of 74.0% N2 molecules (molecular weight 28.0 g/mol) and 26.0% O2 molecules (molar mass of 32.0 g/mol) and that both components are ideal gases. (b) Compare your answer for the specific heat at constant pressure to the value listed in the Handbook of Chemistry and Physics of 1.032 kJ/kg⋅K. Picture the Problem The specific heats of air at constant volume and constant pressure are given by cV = CV/m and cP = CP/m and the heat capacities at constant volume and constant pressure are given by nRC 2

5V = and nRC 2

7P = ,

respectively. (a) Express the heat capacity per unit mass of air at constant volume and constant pressure:

mCc V

V = (1)

and

mCc P

P = (2)

Express the heat capacities of a diatomic gas in terms of the gas constant R, the number of moles n, and the number of degrees of freedom:

nRC 25

V = and

nRC 27

P =

The mass of 1.00 mol of air is given by:

22 ON 26.074.0 MMm +=

Substitute for CV and m in equation (1) to obtain:

( )22 ON

V 26.074.025

MMnRc+

=

Substitute numerical values and evaluate cV:

( )

( )( ) ( )( )[ ] KkgJ716

kg100.32260.0kg100.28740.02Kmol

J314.8mol00.15

33V ⋅=

×+×

⎟⎠⎞

⎜⎝⎛

⋅= −−c

Substitute for CP and m in equation (2) to obtain:

( )22 ON

P 26.074.027

MMnRc+

=

Chapter 18

1816

Substitute numerical values and evaluate cP:

( )

( )( ) ( )( )[ ] KkgkJ00.1

kg100.32260.0kg100.28740.02Kmol

J314.8mol00.17

33P ⋅=

×+×

⎟⎠⎞

⎜⎝⎛

⋅= −−c

(b) The percent difference between the value from the Handbook of Chemistry and Physics and the calculated value is:

%3KJ/kg1.032

KkJ/kg1.00KkJ/kg1.032≈

⋅⋅−⋅

61 •• In this problem, 1.00 mol of an ideal diatomic gas is heated at constant volume from 300 to 600 K. (a) Find the increase in the internal energy of the gas , the work done by the gas, and the heat absorbed by the gas. (b) Find the same quantities if this gas is heated from 300 to 600 K at constant pressure. Use the first law of thermodynamics and your results for (a) to calculate the work done by the gas. (c) Again calculate the work done in Part (b). This time calculate it by integrating the equation dW = P dV. Picture the Problem (a) We know that, during a constant-volume process, no work is done and that we can calculate the heat added during this expansion from the heat capacity at constant volume and the change in the absolute temperature. We can then use the first law of thermodynamics to find the change in the internal energy of the gas. In Part (b), we can proceed similarly; using the heat capacity at constant pressure rather than constant volume. (a) The increase in the internal energy of a fixed amount of an ideal diatomic gas depends only on its change in temperature as it goes from one state to another:

TnRE Δ=Δ 25

int

Substitute numerical values and evaluate ΔEint:

( ) ( ) kJ24.6kJ 236.6K300Kmol

J314.8mol00.1Δ 25

int ==⎟⎠⎞

⎜⎝⎛

⋅=E

For a constant-volume process: 0on =W


1817

From the 1st law of thermodynamics we have:

kJ6.24

0kJ24.6Δ onintin

=

−=−= WEQ

(b) Because ΔEint depends only on the temperature difference:

kJ24.6int =ΔE

Relate the heat added to the gas to its heat capacity at constant pressure and the change in its temperature:

( ) TnRTnRnRTCQ Δ=Δ+=Δ= 27

25

Pin


( ) ( ) kJ73.8kJ 730.8K300Kmol

J8.314mol1.0027

in ==⎟⎠⎞

⎜⎝⎛

⋅=Q

Apply the first law of thermodynamics to find W: kJ2.49

kJ6.236kJ730.8Δ ininton

=

−=−= QEW

(c) Integrate dWon = P dV to obtain: ( ) ( )ififon

f

i

TTnRVVPPdVWV

V

−=−== ∫

Substitute numerical values and evaluate Won: ( ) ( )

kJ2.49

K300Kmol

J8.314mol1.00on

=

⎟⎠⎞

⎜⎝⎛

⋅=W

62 •• A diatomic gas is confined to a closed container of constant volume V0 and at a pressure P0. The gas is heated until its pressure triples. What amount of heat had to be absorbed by the gas in order to triple the pressure? Picture the Problem Because this is a constant-volume process, we can use

TCQ Δ= V to express Q in terms of the temperature change and the ideal-gas law for a fixed amount of gas to find ΔT. Express the amount of heat Q that must be absorbed by the gas if its pressure is to triple:

( )0f25

V

TTnRTCQ

−=

Δ=

Chapter 18

1818

Using the ideal-gas law for a fixed amount of gas, relate the initial and final temperatures, pressures and volumes:

f

0

0

0 3T

VPTVP= ⇒ 0f 3TT =

Substitute for Tf and simplify to obtain:

( ) ( ) VPnRTTTnRQ 000025 553 ==−=

63 •• In this problem, 1.00 mol of air is confined in a cylinder with a piston. The confined air is maintained at a constant pressure of 1.00 atm. The air is initially at 0ºC and has volume V0. Find the volume after 13,200 J of heat are absorbed by the trapped air. Picture the Problem Let the subscripts i and f refer to the initial and final states of the gas, respectively. We can use the ideal-gas law for a fixed amount of gas to express V′ in terms of V and the change in temperature of the gas when 13,200 J of heat are transferred to it. We can find this change in temperature using

TCQ Δ= P . Using the ideal-gas law for a fixed amount of gas, relate the initial and final temperatures, volumes, and pressures:

f

f

i

i

TV'P

TVP=

Noting that the process is isobaric, solve for V′ to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛ Δ+=

Δ+==

ii

i

i

f 1TTV

TTTV

TTVV'

Relate the heat transferred to the gas to the change in its temperature:

TnRTCQ Δ=Δ= 27

P ⇒nRQT

72

=Δ

Substitute for ΔT to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

i721nRT

QVV'

One mol of gas at standard temperature and pressure occupies 22.4 L. Substitute numerical values and evaluate V′:

( ) ( )( ) ( )

L6.59K273

KmolJ8.314mol1.007

kJ13.221m1022.4 33 =⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛

⋅

+×= −V'


1819

64 •• The heat capacity at constant pressure of a sample of gas is greater than the heat capacity at constant volume by 29.1 J/K. (a) How many moles of the gas are present? (b) If the gas is monatomic, what are Cv and Cp? (c) What are the values of Cv and Cp at normal room temperatures? Picture the Problem We can use the relationship between CP and CV ( nRCC += VP ) to find the number of moles of this particular gas. In Parts (b) and (c) we can use the number of degrees of freedom associated with monatomic and diatomic gases, respectively, to find CP and CV. (a) Express the heat capacity of the gas at constant pressure to its heat capacity at constant volume:

nRCC += VP ⇒R

CCn VP −=


mol3.50

mol 500.3

KmolJ8.314

KJ29.1

=

=

⋅

=n

(b) CV for a monatomic gas is given by:

nRC 23

V =

Substitute numerical values and evaluate CV:

( )

J/K43.6J/K 65.43

KmolJ8.314mol3.5002

3V

==

⎟⎠⎞

⎜⎝⎛

⋅=C

Express CP for a monatomic gas:

nRC 25

P =

Substitute numerical values and evaluate CP:

( )

J/K7.27

KmolJ8.314mol3.5002

5P

=

⎟⎠⎞

⎜⎝⎛

⋅=C

(c) At normal room temperature, if the diatomic molecules rotate but do not vibrate they have 5 degrees of freedom. Hence:

nRC 25

V = and

nRRCC 27

VP =+=

Chapter 18

1820

Substitute numerical values and evaluate CV and CP:

( )

J/K7.27

KmolJ8.314mol3.5002

5V

=

⎟⎠⎞

⎜⎝⎛

⋅=C

and ( )( )

J/K102

KJ/mol8.314mol3.50027

P

=

⋅=C

65 •• [SSM] Carbon dioxide (CO2) at a pressure of 1.00 atm and a temperature of –78.5ºC sublimates directly from a solid to a gaseous state without going through a liquid phase. What is the change in the heat capacity at constant pressure per mole of CO2 when it undergoes sublimation? (Assume that the gas molecules can rotate but do not vibrate.) Is the change in the heat capacity positive or negative during sublimation? The CO2 molecule is pictured in Figure 18-22. Picture the Problem We can find the change in the heat capacity at constant pressure as CO2 undergoes sublimation from the energy per molecule of CO2 in the solid and gaseous states. Express the change in the heat capacity (at constant pressure) per mole as the CO2 undergoes sublimation:

solidP,gasP,P CCC −=Δ (1)

Express Cp,gas in terms of the number of degrees of freedom per molecule:

( ) NkNkfC 25

21

gasP, == because each molecule has three translational and two rotational degrees of freedom in the gaseous state.

We know, from the Dulong-Petit Law, that the molar specific heat of most solids is 3R = 3Nk. This result is essentially a per-atom result as it was obtained for a monatomic solid with six degrees of freedom. Use this result and the fact CO2 is triatomic to express CP,solid:

NkNkC 9atoms3atom3

solidP, =×=

Substitute in equation (1) to obtain: NkNkNkC 213

218

25

PΔ −=−= 66 •• In this problem, 1.00 mol of a monatomic ideal gas is initially at 273 K and 1.00 atm. (a) What is the initial internal energy of the gas? (b) Find the work done by the gas when 500 J of heat are absorbed by the gas at constant pressure.


1821

What is the final internal energy of the gas? (c) Find the work done by the gas when 500 J of heat are absorbed by the gas at constant volume. What is the final internal energy of the gas? Picture the Problem We can find the initial internal energy of the gas from

nRTU 23

i = and the final internal energy from the change in internal energy resulting from the addition of 500 J of heat. The work done during a constant-volume process is zero and the work done during the constant-pressure process can be found from the first law of thermodynamics. (a) Express the initial internal energy of the gas in terms of its temperature:

nRTE 23

iint, =

Substitute numerical values and evaluate Eint,i:

( ) ( ) kJ3.40kJ 405.3K273Kmol

J8.314mol1.0023

iint, ==⎟⎠⎞

⎜⎝⎛

⋅=E

(b) Relate the final internal energy of the gas to its initial internal energy:

TCEEEE Δ+=Δ+= Viint,intiint,fint,

Express the change in temperature of the gas resulting from the addition of heat:

P

in

CQT =Δ

Substitute to obtain: in

P

Viint,fint, Q

CCEE +=

Substitute numerical values and evaluate Eint,f:

( )

kJ71.3kJ 705.3

J500kJ405.32523

fint,

==

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

nRnR

E

The work done by the gas is the negative of the work done on the gas and is related to the change in internal energy and the heat added to the gas through the first law of thermodynamics:

( )iniint,fint,

iniint,fint,

inintongasby Δ

QEEQEE

QEWW

++−=

+−−=

+−=−=

Chapter 18

1822


kJ 19.0

J 500kJ 40.3kJ 71.3gasby

=

++−=W

(c) Relate the final internal energy of the gas to its initial internal energy:

intiint,fint, EEE Δ+=

Apply the first law of thermodynamics to the constant-volume process:

oninint WQE +=Δ or, because Won = 0,

J500inint ==Δ QE

Substitute numerical values and evaluate Eint,f:

kJ3.91J500kJ.4053fint, =+=E

and, because Won = 0 = −Wby gas, 0gasby =W

67 •• List all of the degrees of freedom possible for a water molecule and estimate the heat capacity of water at a temperature very far above its boiling point. (Ignore the fact the molecule might dissociate at high temperatures.) Think carefully about all of the different ways in which a water molecule can vibrate. Picture the Problem We can use ( )NkfC 2

1waterV, = to express CV,water and then

count the number of degrees of freedom associated with a water molecule to determine f. Express CV,water in terms of the number of degrees of freedom per molecule:

( )NkfC 21

waterV, = where f is the number of degrees of freedom associated with a water molecule.

There are 3 translational degrees of freedom and three rotational degrees of freedom. In addition, each of the hydrogen atoms can vibrate against the oxygen atom, resulting in an additional 4 degrees of freedom (2 per atom). Substitute for f to obtain: ( )1

210 5 5V,waterC Nk Nk nR= = =

Heat Capacities of Solids and the Dulong-Petit Law 68 • The Dulong–Petit law was originally used to determine the molar mass of a substance from its measured heat capacity. The specific heat of a certain solid substance is measured to be 0.447 kJ/kg⋅K. (a) Find the molar mass of the substance. (b) What element has this specific heat value?


1823

Picture the Problem The Dulong-Petit law gives the molar specific heat of a solid, c′. The specific heat is defined as c = c′/M where M is the molar mass. Hence we can use this definition to find M and a periodic table to identify the element. (a) Apply the Dulong-Petit law: Rc' 3= ⇒

MRc 3

= ⇒cRM 3

=

Substitute numerical values and evaluate M:

g/mol7.55

KkgkJ0.447

KmolJ24.9

=

⋅

⋅=M

(b) Consulting the periodic table of elements we see that the element is most likely iron . Quasi-Static Adiabatic Expansion of a Gas 69 •• [SSM] A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is (a) isothermal and (b) adiabatic. Picture the Problem We can use the ideal-gas law to find the initial volume of the gas. In Part (a) we can apply the ideal-gas law for a fixed amount of gas to find the final volume and the expression for the work done in an isothermal process. Application of the first law of thermodynamics will allow us to find the heat absorbed by the gas during this process. In Part (b) we can use the relationship between the pressures and volumes for a quasi-static adiabatic process to find the final volume of the gas. We can apply the ideal-gas law to find the final temperature and, as in (a), apply the first law of thermodynamics, this time to find the work done by the gas. Use the ideal-gas law to express the initial volume of the gas:

i

ii P

nRTV =

Substitute numerical values and evaluate Vi:

( ) ( )33

i m103.118kPa400

K300Kmol

J8.314mol0.500−×=

⎟⎠⎞

⎜⎝⎛

⋅=V

Chapter 18

1824

(a) Because the process is isothermal:

K300if == TT

Use the ideal-gas law for a fixed amount of gas to express Vf: f

ff

i

ii

TVP

TVP

=

or, because T = constant,

f

iif P

PVV =

Substitute numerical values and evaluate Vf:

( )

L7.80

L 795.7kPa160kPa400L3.118f

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Express the work done by the gas during the isothermal expansion:

i

fgasby ln

VVnRTW =


( )

( )

kJ14.1

L3.118L7.795lnK300

KmolJ8.314mol0.500gasby

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎠⎞

⎜⎝⎛

⋅=W

Noting that the work done by the gas during the process equals the negative of the work done on the gas, apply the first law of thermodynamics to find the heat absorbed by the gas:

( )kJ1.14

kJ1.140onintin

=

−−=−Δ= WEQ

(b) Using γ = 5/3 and the relationship between the pressures and volumes for a quasi-static adiabatic process, express Vf:

γγffii VPVP = ⇒

γ1

f

iif ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Substitute numerical values and evaluate Vf: ( )

L5.40

L 403.5kPa160kPa400L118.3

53

f

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛=V


1825

Apply the ideal-gas law to find the final temperature of the gas: nR

VPT fff =

Substitute numerical values and evaluate Tf:

( )( )( )

K208

KmolJ8.314mol0.500

m105.403kPa160 33

f

=

⎟⎠⎞

⎜⎝⎛

⋅

×=

−

T

For an adiabatic process: 0in =Q

Apply the first law of thermodynamics to express the work done on the gas during the adiabatic process:

TnRTCQEW Δ=−Δ=−Δ= 23

Vininton 0

Substitute numerical values and evaluate Won:

( )( )( )

J745K300K208

KJ/mol8.314mol0.50023

on

−=−×

⋅=W

Because the work done by the gas equals the negative of the work done on the gas:

( ) J745J574gasby =−−=W

70 •• A 0.500-mol sample of an ideal diatomic gas at 400 kPa and 300 K expands until the pressure decreases to 160 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is (a) isothermal and (b) adiabatic. Picture the Problem We can use the ideal-gas law to find the initial volume of the gas. In Part (a) we can apply the ideal-gas law for a fixed amount of gas to find the final volume and the expression for the work done in an isothermal process. Application of the first law of thermodynamics will allow us to find the heat absorbed by the gas during this process. In Part (b) we can use the relationship between the pressures and volumes for a quasi-static adiabatic process to find the final volume of the gas. We can apply the ideal-gas law to find the final temperature and, as in (a), apply the first law of thermodynamics, this time to find the work done by the gas. Use the ideal-gas law to express the initial volume of the gas:

i

ii P

nRTV =

Chapter 18

1826


( ) ( )

L3.118kPa400

K300Kmol

J8.314mol0.500i

=

⎟⎠⎞

⎜⎝⎛

⋅=V

(a) Because the process is isothermal:

K300if == TT

Use the ideal-gas law for a fixed amount of gas to express Vf: f

ff

i

ii

TVP

TVP

=

or, because T = constant,

f

iif P

PVV =


( )

L7.80

L 795.7kPa160kPa400L3.118f

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Express the work done by the gas during the isothermal expansion:

i

fgasby ln

VVnRTW =


( )

( )

kJ14.1

L3.118L7.795lnK300

KmolJ8.314mol0.500gasby

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎠⎞

⎜⎝⎛

⋅=W

Noting that the work done by the gas during the isothermal expansion equals the negative of the work done on the gas, apply the first law of thermodynamics to find the heat absorbed by the gas:

( )kJ1.14

kJ1.140onintin

=

−−=−Δ= WEQ

(b) Using γ = 1.4 and the relationship between the pressures and volumes for a quasi-static adiabatic process, express Vf:

γγffii VPVP = ⇒

γ1

f

iif ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV


1827

Substitute numerical values and evaluate Vf: ( )

L00.6

L 000.6kPa160kPa400L118.3

1.41

f

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Apply the ideal-gas law to express the final temperature of the gas:

nRVPT ff

f =


( )( )( )( )

K231

KJ/mol8.314mol0.500m106.000kPa160 33

f

=

⋅×

=−

T

For an adiabatic process: 0in =Q

Apply the first law of thermodynamics to express the work done on the gas during the adiabatic expansion:

TnRTCQEW Δ=−Δ=−Δ= 25

Vininton 0


( )( )( )

J717K300K312

KJ/mol8.314mol0.50025

on

−=−×

⋅=W

Noting that the work done by the gas during the adiabatic expansion is the negative of the work done on the gas, we have:

( ) J717J717gasby =−−=W

71 •• A 0.500-mol sample of helium gas expands adiabatically and quasi-statically from an initial pressure of 5.00 atm and temperature of 500 K to a final pressure of 1.00 atm. Find (a) the final temperature of the gas, (b) the final volume of the gas, (c) the work done by the gas, and (d) the change in the internal energy of the gas. Picture the Problem We can eliminate the volumes from the equations relating the temperatures and volumes and the pressures and volumes for a quasi-static adiabatic process to obtain a relationship between the temperatures and pressures. We can find the initial volume of the gas using the ideal-gas law and the final volume using the pressure-volume relationship. In Parts (d) and (c) we can find the change in the internal energy of the gas from the change in its temperature and use the first law of thermodynamics to find the work done by the gas during its expansion.

Chapter 18

1828

(a) Express the relationship between temperatures and volumes for a quasi-static adiabatic process:

1ff

1ii

−− = γγ VTVT

Express the relationship between pressures and volumes for a quasi-static adiabatic process:

γγffii VPVP = (1)

Eliminate the volume between these two equations to obtain:

γ11

i

fif

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

PPTT

Substitute numerical values and evaluate Tf: ( ) K263

atm5.00atm1.00K500

3511

f =⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

T

(b) Solve equation (1) for Vf: γ

1

f

iif ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Apply the ideal-gas law to express Vi:

i

ii P

nRTV =


( ) ( )

L4.103atm

kPa101.3255.00atm

K500Kmol

J8.314mol0.500i

=

×

⎟⎠⎞

⎜⎝⎛

⋅=V

Substitute for Vi and evaluate Vf:

( ) L8.10atm1.00atm5.00L103.4

53

f =⎟⎟⎠

⎞⎜⎜⎝

⎛=V

(c) The work done by the gas during an adiabatic expansion is the negative of the work done on the gas during the expansion:

( )( )

int

int

inintongas by the

Δ0Δ

Δ

EE

QEWW

−=−−=

−−=−=

Because TnRTCE Δ=Δ=Δ 23

Vint :

TnRW Δ23

gas by the −=


1829

Substitute numerical values and evaluate Wby the gas:

( ) ( ) kJ 48.1K 500K 263Kmol

J8.314mol 500.023

gas by the =−⎟⎠⎞

⎜⎝⎛

⋅−=W

(d) Apply the first law of thermodynamics for an adiabatic process to obtain:

( )gas by the

gas by theoninint 0Δ

W

WWQE

−=

−+=+=

Substitute Wby the gas from (c): ( ) kJ48.1kJ 48.1Δ int −=−=E Cyclic Processes 72 •• A 1.00-mol sample of N2 gas at 20.0ºC and 5.00 atm is allowed to expand adiabatically and quasi-statically until its pressure equals 1.00 atm. It is then heated at constant pressure until its temperature is again 20.0ºC. After it reaches a temperature of 20.0ºC, it is heated at constant volume until its pressure is again 5.00 atm. It is then compressed at constant pressure until it is back to its original state. (a) Construct a PV diagram showing each process in the cycle. (b) From your graph, determine the work done by the gas during the complete cycle. (c) How much heat is absorbed (or released) by the gas during the complete cycle? Picture the Problem To construct the PV diagram we’ll need to determine the volume occupied by the gas at the beginning and ending points for each process. Let these points be A, B, C, and D. We can apply the ideal-gas law to the starting point (A) to find VA. To find the volume at point B, we can use the relationship between pressure and volume for a quasi-static adiabatic process. We can use the ideal-gas law to find the volume at point C and, because they are equal, the volume at point D. We can apply the first law of thermodynamics to find the amount of heat added to or subtracted from the gas during the complete cycle. (a) Using the ideal-gas law, express the volume of the gas at the starting point A of the cycle:

A

AA P

nRTV =

Substitute numerical values and evaluate VA:

( ) ( )

L4.808m10 4.808atm

kPa101.325atm5.00

K293Kmol

J8.314mol1.00

33

A

=×=

×

⎟⎠⎞

⎜⎝⎛

⋅=

−

V

Chapter 18

1830

Use the relationship between pressure and volume for a quasi-static adiabatic process to express the volume of the gas at point B; the end point of the adiabatic expansion:

γ1

B

AAB ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Substitute numerical values and evaluate VB: ( ) L18.15

atm1.00atm5.00L808.4

4.11

B =⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Using the ideal-gas law for a fixed amount of gas, express the volume occupied by the gas at points C and D:

C

CDC P

nRTVV ==

Substitute numerical values and evaluate VC:

( ) ( )

L04.24m 1004.24atm

kPa101.325atm1.00

K293Kmol

J8.314mol1.00

33

C

=×=

×

⎟⎠⎞

⎜⎝⎛

⋅=

−

V

The complete cycle is shown in the diagram.

0 50

10 15 20 25

A 0

CB1.00

2.00

3.00

4.00

5.00

atm ,P

L ,V

(b) Note that for the paths A→B and B→C, Wby gas, the work done by the gas, is positive. For the path D→A, Wby gas is negative, and greater in magnitude than WA→C. Therefore the total work done by the gas is negative. Find the area enclosed by the cycle by noting that each rectangle of dotted lines equals 5 atm⋅L and counting the rectangles:

( ) ( )

kJ6.6

J 586.6Latm

J101.325Latm65rectangle

Latm5.00rectangles13gasby

−=

−=⎟⎠⎞

⎜⎝⎛

⋅⋅−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅−≈W


1831

(c) The work done on the gas equals the negative of the work done by the gas. Apply the first law of thermodynamics to find the amount of heat added to or subtracted from the gas during the complete cycle:

( )kJ6.6

kJ6.590Δ onintin

=

−−=−= WEQ

because ΔEint = 0 for the complete cycle.

73 •• [SSM] A 1.00-mol sample of an ideal diatomic gas is allowed to expand. This expansion is represented by the straight line from 1 to 2 in the PV diagram (Figure 18-23). The gas is then compressed isothermally. This compression is represented by the straight line from 2 to 1 in the PV diagram. Calculate the work per cycle done by the gas. Picture the Problem The total work done as the gas is taken through this cycle is the area bounded by the two processes. Because the process from 1→2 is linear, we can use the formula for the area of a trapezoid to find the work done during this expansion. We can use ( )ifprocess isothermal ln VVnRTW = to find the work done

on the gas during the process 2→1. The net work done during this cycle is then the sum of these two terms. Express the net work done per cycle:

1221

gas on thegas by thenet

→→ +=

+=

WW

WWW (1)

Work is done by the gas during its expansion from 1 to 2 and hence is equal to the negative of the area of the trapezoid defined by this path and the vertical lines at V1 = 11.5 L and V2 = 23 L. Use the formula for the area of a trapezoid to express W1→2:

( )( )

atmL3.17atm1.0atm2.0

L11.5L2321

trap21

⋅−=+×

−−=

−=→ AW

Work is done on the gas during the isothermal compression from V2 to V1 and hence is equal to the area under the curve representing this process. Use the expression for the work done during an isothermal process to express W2→1:

⎟⎟⎠

⎞⎜⎜⎝

⎛=→

i

f12 ln

VVnRTW

Chapter 18

1832

Apply the ideal-gas law at point 1 to find the temperature along the isotherm 2→1:

( )( )( )( ) K280

Katm/molL10206.8mol1.00L5.11atm0.2

2 =⋅⋅×

== −nRPVT

Substitute numerical values and evaluate W2→1:

( )( )( ) atmL9.15L23L5.11lnK280Katm/molL10206.8mol00.1 2

12 ⋅=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅×= −

→W

Substitute in equation (1) and evaluate Wnet:

kJ14.0atmL

J101.325atmL40.1

atmL15.9atmL3.17net

−=⋅

×⋅−=

⋅+⋅−=W

74 •• A 2.00-mol sample of an ideal monatomic gas have an initial pressure of 2.00 atm and an initial volume of 2.00 L. The gas is taken through the following quasi-static cycle: It is expanded isothermally until it has a volume of 4.00 L. It is next heated at constant volume until it has a pressure of 2.00 atm. It is then cooled at constant pressure until it is back to its initial state. (a) Show this cycle on a PV diagram. (b) Find the temperatures at each end of each part of the cycle. (c) Calculate the heat absorbed and the work done by the gas during each part of the cycle. Picture the Problem Denote the initial and final points of each part of the cycle by the numerals 1, 2 and 3. We can apply the ideal-gas law to find the temperatures T1, T2, and T3. We can use the appropriate work and heat equations to calculate the heat added and the work done by the gas for the isothermal process (1→2), the constant-volume process (2→3), and the isobaric process (3→1). (a) The cycle is shown in the PV diagram to the right:

00

31

2

2.00

1.00

2.001.00 3.00 4.00

atm ,P

L ,V


1833

(b) Use the ideal-gas law to find T1:

( )( )( )

K4.42K 37.24

KmolatmL108.206mol2.00

L2.00atm2.002

111

==

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

Because the process 1→2 is isothermal:

K4.422 =T

Use the ideal-gas law to find T3:

( )( )( )

K7.84K 74.48


L4.00atm2.002

333

==

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

(c) Because the process 1→2 is isothermal, Qin,1→2 = Wby gas,1→2: ⎟⎟

⎠

⎞⎜⎜⎝

⎛== →→

1

221gas,by 21 in, ln

VVnRTWQ

Substitute numerical values and evaluate Qin,1→2:

( ) ( ) J281L2.00L4.00lnK24.37

KmolJ8.314mol2.0021 in, =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅=→Q

Because process 2→3 takes place at constant volume:

032 =→W

Because process 2→3 takes place at constant volume, Won,2→3 = 0, and:

( )2323

V3int,23in,2 ΔΔ

TTnR

TCEQ

−=

== →→

Substitute numerical values and evaluate Qin,2→3:

( ) ( ) J608K37.42K48.74Kmol

J8.314mol2.0023

3in,2 =−⎟⎠⎞

⎜⎝⎛

⋅=→Q

Because process 3→1 is isobaric: )( 312

5P13 TTnRTCQ −=Δ=→

Chapter 18

1834

Substitute numerical values and evaluate Q3→1:

( ) ( ) kJ01.1K74.84K37.24Kmol

J8.314mol2.0025

13 −=−⎟⎠⎞

⎜⎝⎛

⋅=→Q

The work done by the gas from 3 to 1 equals the negative of the work done on the gas:

( )313,13,11gas,3by VVPVPW −=Δ−=→

Substitute numerical values and evaluate Wby gas,3→2:

( )( ) ( )

J405

LatmJ101.325Latm4.00L4.00L2.00atm2.001gas,3by

=

⎟⎠⎞

⎜⎝⎛

⋅⋅−−=−−=→W

75 ••• [SSM] At point D in Figure 18-24 the pressure and temperature of 2.00 mol of an ideal monatomic gas are 2.00 atm and 360 K, respectively. The volume of the gas at point B on the PV diagram is three times that at point D and its pressure is twice that at point C. Paths AB and CD represent isothermal processes. The gas is carried through a complete cycle along the path DABCD. Determine the total amount of work done by the gas and the heat absorbed by the gas along each portion of the cycle. Picture the Problem We can find the temperatures, pressures, and volumes at all points for this ideal monatomic gas (3 degrees of freedom) using the ideal-gas law and the work for each process by finding the areas under each curve. We can find the heat exchanged for each process from the heat capacities and the initial and final temperatures for each process. Express the total work done by the gas per cycle:

DCCBBAADtotgas,by →→→→ +++= WWWWW

1. Use the ideal-gas law to find the volume of the gas at point D:

( ) ( )

( )

L29.54atmkPa101.325atm2.00

K360Kmol

J8.314mol2.00

D

DD

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅=

=P

nRTV


1835

2. We’re given that the volume of the gas at point B is three times that at point D:

( )L62.88

L 54.2933 DCB

==== VVV

Use the ideal-gas law to find the pressure of the gas at point C:

( ) ( )atm6667.0

L62.88

K360Kmol

atmL10206.8mol2.00 2

C

CC =

⎟⎠⎞

⎜⎝⎛

⋅⋅

×==

−

VnRTP

We’re given that the pressure at point B is twice that at point C:

( ) atm333.1atm6667.022 CB === PP

3. Because path DC represents an isothermal process:

K360CD == TT

Use the ideal-gas law to find the temperatures at points B and A: ( )( )

( )

K719.8Kmol

atmL108.206mol2.00

L88.62atm1.3332

BBBA

=

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

==

−

nRVPTT

Because the temperature at point A is twice that at D and the volumes are the same, we can conclude that:

atm00.42 DA == PP

The pressure, volume, and temperature at points A, B, C, and D are summarized in the table to the right.

Point P V T

(atm) (L) (K) A 4.00 29.5 720 B 1.33 88.6 720 C 0.667 88.6 360 D 2.00 29.5 360

4. For the path D→A, 0AD =→W and: ( )DA2

3

AD23

AD int,AD

TTnRTnREQ

−=

Δ=Δ= →→→

Chapter 18

1836

Substitute numerical values and evaluate QD→A:

( ) ( ) kJ979.8K360K720Kmol

J8.314mol2.0023

AD =−⎟⎠⎞

⎜⎝⎛

⋅=→Q

For the path A→B:

⎟⎟⎠

⎞⎜⎜⎝

⎛== →→

A

BBA,BABA ln

VVnRTQW

Substitute numerical values and evaluate WA→B:

( ) ( ) kJ15.13L29.54L88.62lnK720

KmolJ8.314mol2.00BA =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅=→W

and, because process A→B is isothermal, 0BA int, =Δ →E

For the path B→C, 0CB =→W , and:

( )BC23

VCBCB ΔΔTTnR

TCUQ−=

== →→

Substitute numerical values and evaluate QB→C:

( )( )( ) kJ979.8K720K360KJ/mol8.314mol2.0023

CB −=−⋅=→Q

For the path C→D: ⎟⎟⎠

⎞⎜⎜⎝

⎛=→

C

DDC,DC ln

VVnRTW

Substitute numerical values and evaluate WC→D:

( ) ( ) kJ576.6L62.88L54.92lnK603

KmolJ8.314mol2.00DC −=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅=→W

Also, because process A→B is isothermal, 0BAint, =Δ →E , and kJ58.6DCDC −== →→ WQ

Qin, Won, and ΔEint are summarized for each of the processes in the table to the right.

Process Qin Won ΔEint

(kJ) (kJ) (kJ) D→A 98.8 0 8.98

A→B 2.13 −13.2 0

B→C 98.8− 0 −8.98


1837

C→D 58.6− 6.58 0

Referring to the table, find the total work done by the gas per cycle:

kJ6.6

kJ6.580kJ13.20DCCBBAAD totgas,by

=

−++=

+++= →→→→ WWWWW

Remarks: Note that, as it should be, ΔEint is zero for the complete cycle. 76 ••• At point D in Figure 18-24 the pressure and temperature of 2.00 mol of an ideal diatomic gas are 2.00 atm and 360 K, respectively. The volume of the gas at point B on the PV diagram is three times that at point D and its pressure is twice that at point C. Paths AB and CD represent isothermal processes. The gas is carried through a complete cycle along the path DABCD. Determine the total amount of work done by the gas and the heat absorbed by the gas along each portion of the cycle. Picture the Problem We can find the temperatures, pressures, and volumes at all points for this ideal diatomic gas (5 degrees of freedom) using the ideal-gas law and the work for each process by finding the areas under each curve. We can find the heat exchanged for each process from the heat capacities and the initial and final temperatures for each process. Express the total work done by the gas per cycle:

DCCBBAADtotgas,by →→→→ +++= WWWWW

1. Use the ideal-gas law to find the volume of the gas at point D:

( ) ( )

( )

L29.54atmkPa101.325atm2.00

K360Kmol

J8.314mol2.00

D

DD

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅=

=P

nRTV

2. We’re given that the volume of the gas at point B is three times that at point D:

( )L62.88

L 54.2933 DCB

==== VVV

Chapter 18

1838

Use the ideal-gas law to find the pressure of the gas at point C:

( ) ( )atm6667.0

L62.88

K360Kmol

atmL10206.8mol2.00 2

C

CC =

⎟⎠⎞

⎜⎝⎛

⋅⋅

×==

−

VnRTP

We’re given that the pressure at point B is twice that at point C:

( ) atm333.1atm6667.022 CB === PP

3. Because path DC represents an isothermal process:

K360CD == TT

Use the ideal-gas law to find the temperatures at points B and A: ( )( )

( )

K719.8Kmol

atmL108.206mol2.00

L88.62atm1.3332

BBBA

=

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

==

−

nRVPTT

Because the temperature at point A is twice that at D and the volumes are the same, we can conclude that:

atm00.42 DA == PP

The pressure, volume, and temperature at points A, B, C, and D are summarized in the table to the right.

Point P V T (atm) (L) (K)

A 4.00 29.5 720 B 1.33 88.6 720 C 0.667 88.6 360 D 2.00 29.5 360

4. For the path D→A, 0AD =→W and:

( )

( ) ( )

kJ97.14

K360K720Kmol

J8.314mol2.00

ΔΔ

25

DA25

AD25

ADAD

=

−⎟⎠⎞

⎜⎝⎛

⋅=

−=== →→→ TTnRTnRUQ


1839

For the path A→B:

( ) ( )

kJ15.13L29.54L88.62lnK720

KmolJ8.314mol2.00ln

A

BBA,BABA

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛== →→ V

VnRTQW

and, because process A→B is isothermal, 0BAint, =Δ →E

For the path B→C, 0CB =→W and:

( )

( ) ( )

kJ97.14

K720K360Kmol

J8.314mol2.00

ΔΔ

25

BC25

VCBCB

−=

−⎟⎠⎞

⎜⎝⎛

⋅=

−=== →→ TTnRTCUQ

For the path C→D:

( ) ( ) kJ576.6L62.88L54.92lnK603

KmolJ8.314mol2.00ln

C

DDC,DC −=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=→ V

VnRTW

Also, because process A→B is isothermal, 0BAint, =Δ →E and kJ576.6DCDC −== →→ WQ

Qin, Won, and ΔEint are summarized for each of the processes in the table to the right.

Process Qin Won ΔEint

(kJ) (kJ) (kJ) D→A 0.15 0 15.0

A→B 2.13 −13.2 0

B→C 0.15− 0 −15.0

C→D 58.6− 6.58 0

Referring to the table and noting that the work done by the gas equals the negative of the work done on the gas, find the total work done by the gas per cycle:

kJ6.6kJ6.580kJ13.20DCCBBAADtotgas,by =−++=+++= →→→→ WWWWW

Chapter 18

1840

Remarks: Note that ΔEint for the complete cycle is zero and that the total work done is the same for the diatomic gas of this problem and the monatomic gas of problem 89. 77 ••• A sample consisting of n moles of an ideal gas is initially at pressure P1, volume V1, and temperature Th. It expands isothermally until its pressure and volume are P2 and V2. It then expands adiabatically until its temperature is Tc and its pressure and volume are P3 and V3. It is then compressed isothermally until it is at a pressure P4 and a volume V4, which is related to its initial volume V1 by 1 1

c 4 h 1γ γ− −=T V T V . The gas is then compressed adiabatically until it is back in its

original state. (a) Assuming that each process is quasi-static, plot this cycle on a PV diagram. (This cycle is known as the Carnot cycle for an ideal gas.) (b) Show that the heat Qh absorbed during the isothermal expansion at Th is Qh = nRTh ln(V2/V1). (c) Show that the heat Qc released by the gas during the isothermal compression at Tc is Qc = nRTc ln (V3/V4). (d) Using the result that TVγ–1 is constant for a quasi-static adiabatic expansion, show that V2/V1 = V3/V4. (e) The efficiency of a Carnot cycle is defined as the net work done by the gas, divided by the heat absorbed Qh by the gas. Using the first law of thermodynamics, show that the efficiency is 1 – Qc/Qh. (f) Using your results from the previous parts of this problem, show that Qc/Qh = Tc/Th. Picture the Problem We can use the equations of state for adiabatic and isothermal processes to express the work done on or by the system, the heat entering or leaving the system, and the change in internal energy for each of the four processes making up the Carnot cycle. We can use the first law of thermodynamics and the definition of the efficiency of a Carnot cycle to show that the efficiency is 1 – Qc / Qh. (a) The cycle is shown on the PV diagram to the right: 1

2

3

4

Th

TcP3

P4

P2

P1

V

P

1V 2V 3V4V

(b) Because the process 1→2 is isothermal:

021int, =Δ →E


1841

Apply the first law of thermodynamics to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛=== →→

1

2h2121h ln

VVnRTWQQ

(c) Because the process 3→4 is isothermal:

043 =Δ →U


⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=== →→

4

3c

3

4c4343c

ln

ln

VVnRT

VVnRTWQQ

where the minus sign tells us that heat is given off by the gas during this process.

(d) Apply the equation for a quasi-static adiabatic process at points 4 and 1 to obtain:

11h

14c


1

h

c

4

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γ

TT

VV (1)

Apply the equation for a quasi-static adiabatic process at points 2 and 3 to obtain:

13c

12h


1

h

c

3

2−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γ

TT

VV (2)

Equate equations (1) and (2) and rearrange to obtain:

1

2

4

3

VV

VV

=

(e) Express the efficiency of the Carnot cycle:

hQW

=ε


( ) ( )chch

incycle int,on

0 QQQQ

QEW

−−=−−=

−Δ=

because Eint is a state function and 0cycle int, =ΔE .

Substitute to obtain:

h

c

h

ch

h

on

h

gas by the

1QQ

QQQ

QW

QW

−=

−=

−==ε

Chapter 18

1842

(f) In Part (b) we established that: ⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2hh ln

VVnRTQ

In Part (c) we established that the heat leaving the system along the path 3→4 is given by:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

4

3cc ln

VVnRTQ


h

c

1

2h

4

3c

h

c

ln

ln

TT

VVnRT

VVnRT

QQ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

because1

2

4

3

VV

VV

= .

Remarks: This last result establishes that the efficiency of a Carnot cycle is

also given byh

cC T

Tε −= 1 .

General Problems 78 • During the process of quasi-statically compressing an ideal diatomic gas to one-fifth of its initial volume, 180 kJ of work are done on the gas. (a) If this compression is accomplished isothermally at room temperature (293 K), how much heat released by the gas? (b) How many moles of gas are in this sample? Picture the Problem (a) We can use the first law of thermodynamics to relate the heat removed from the gas to the work done on the gas. (b) We can find the number of moles of the gas from the expression for the work done on or by a gas during an isothermal process. (a) Apply the first law of thermodynamics to this process:

ononintin WWEQ −=−Δ= because ΔEint = 0 for an isothermal process.

Substitute numerical values to obtain:

kJ180in −=Q

Because Qremoved = −Qin: kJ180removed =Q


1843

(b) The work done on the gas during the isothermal process is given by:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

i

flnVVnRTW ⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛=

i

flnVVRT

Wn

Substitute numerical values and evaluate n: ( )

mol45.9

51lnK293

KmolJ8.314

kJ180

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅

−=n

79 • [SSM] The PV diagram in Figure 18-25 represents 3.00 mol of an ideal monatomic gas. The gas is initially at point A. The paths AD and BC represent isothermal changes. If the system is brought to point C along the path AEC, find (a) the initial and final temperatures of the gas, (b) the work done by the gas, and (c) the heat absorbed by the gas. Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process EDC is isobaric, we can find the area under this line geometrically and then use the 1st law of thermodynamics to find QAEC. (a) Using the ideal-gas law, find the temperature at point A:

( )( )( )

K65K65.2


L4.01atm4.02

AAA

==

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

Using the ideal-gas law, find the temperature at point C: ( )( )

( )

K81K81.2


L0.02atm1.02

CCC

==

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

(b) Express the work done by the gas along the path AEC:

( )( )

kJ1.6kJ1.62atmL

J101.325atmL15.99

L4.01L20.0atm1.0Δ0 ECECECAEAEC

==

⋅×⋅=

−=+=+= VPWWW

Chapter 18

1844

(c) Apply the first law of thermodynamics to express QAEC: ( )ATTnRW

TnRWTCWEWQ

−+=Δ+=

Δ+=Δ+=

C23

AEC

23

AEC

VAECintAECAEC

Substitute numerical values and evaluate QAEC:

( )( )( ) kJ2.2K65.2K81.2KJ/mol8.314mol3.00kJ1.62 23

AEC =−⋅+=Q

Remarks The difference between WAEC and QAEC is the change in the internal energy ΔEint,AEC during this process. 80 •• The PV diagram in Figure 18-25 represents 3.00 mol of an ideal monatomic gas. The gas is initially at point A. The paths AD and BC represent isothermal changes. If the system is brought to point C along the path ABC, find (a) the initial and final temperatures of the gas, (b) the work done by the gas, and (c) the heat absorbed by the gas. Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process AB is isobaric, we can find the area under this line geometrically. We can use the expression for the work done during an isothermal expansion to find the work done between B and C and the first law of thermodynamics to find QABC. (a) Using the ideal-gas law, find the temperature at point A:

( )( )( )

K65K65.2


L4.01atm4.02

AAA

==

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

Use the ideal-gas law to find the temperature at point C: ( )( )

( )

K81K81.2


L0.02atm1.02

CCC

==

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT


1845

(b) Express the work done by the gas along the path ABC: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+Δ=

+=

B

CBABAB

BCABABC

lnVVnRTVP

WWW

Use the ideal-gas law to find the volume of the gas at point B:

( ) ( )L5.00

atm4.0

K81.2Kmol

atmL108.206mol3.00 2

B

BB =

⎟⎠⎞

⎜⎝⎛

⋅⋅

×==

−

PnRTV

Substitute numerical values and evaluate WABC:

( )( ) ( )

( )

kJ2.3kJ21.3atm

J101.325atmL1.73

L5.00L20.0lnK81.2

KmolatmL108.206mol3.00L4.01L5.00atm4.0 2

ABC

==×⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎠⎞

⎜⎝⎛

⋅⋅

×+−= −W

(c) Apply the first law of thermodynamics to obtain: ( )ATTnRW

TnRWTCWEWQ

−+=Δ+=

Δ+=Δ+=

C23

AEC

23

AEC

VABCintABCABC

Substitute numerical values and evaluate QABC:

( ) ( ) kJ8.3K65.2K81.2Kmol

J8.314mol3.00kJ21.3 23

ABC =−⎟⎠⎞

⎜⎝⎛

⋅+=Q

Remarks: The difference between WABC and QABC is the change in the internal energy ΔEint,ABC during this process. 81 •• The PV diagram in Figure 18-25 represents 3.00 mol of an ideal monatomic gas. The gas is initially at point A. The paths AD and BC represent isothermal changes. If the system is brought to point C along the path ADC, find (a) the initial and final temperatures of the gas, (b) the work done by the gas, and (c) the heat absorbed by the gas. Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process DC is isobaric, we can find the area under this line geometrically. We can use the expression for the work done during an isothermal

Chapter 18

1846

expansion to find the work done between A and D and the first law of thermodynamics to find QADC. (a) Using the ideal-gas law, find the temperature at point A:

( )( )( )

K65K65.2


L4.01atm4.02

AAA

==

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

Use the ideal-gas law to find the temperature at point C: ( )( )

( )

K81


L02atm1.02

CCC

=

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

(b) Express the work done by the gas along the path ADC: DCDC

A

DA

DCADADC

ln VPVV

nRT

WWW

Δ+⎟⎟⎠

⎞⎜⎜⎝

⎛=

+=

Use the ideal-gas law to find the volume of the gas at point D:

( ) ( )L051.61

atm1.0

K65.2Kmol

atmL108.206mol3.00 2

D

DD =

⎟⎠⎞

⎜⎝⎛

⋅⋅

×==

−

PnRTV

Substitute numerical values and evaluate WADC:

( ) ( )

( )( )

kJ7.2kJ 65.2atmL

J101.325atmL2.26

L1.61L0.02atm1.0L01.4L1.61lnK65.2

KmolatmL108.206mol3.00 2

ADC

==⋅

×⋅=

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅⋅

×= −W


1847

(c) Apply the first law of thermodynamics to obtain: ( )ATTnRW

TnRWTCWEWQ

−+=Δ+=

Δ+=Δ+=

C23

ADC

23

ADC

VADCintADCADC

Substitute numerical values and evaluate QADC:

( ) ( ) kJ3.3K65.2K81.2Kmol

J8.314mol3.00kJ65.2 23

ADC =−⎟⎠⎞

⎜⎝⎛

⋅+=Q

82 •• Suppose that the paths AD and BC in Figure 18-25 represent adiabatic processes. What are the work done by the gas and the heat absorbed by the gas in following the path ABC? Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process AB is isobaric, we can find the area under this line geometrically. We can find the work done during the adiabatic expansion between B and C using BCVBC TCW Δ−= and the first law of thermodynamics to find QABC. The work done by the gas along path ABC is given by:

BC23

ABAB

BCVABAB

BCABABC

TnRVPTCVP

WWW

Δ−Δ=

Δ−Δ=+=

because, with Qin = 0, WBC = −ΔEint,BC.

Use the ideal-gas law to find TA:

( )( )( )

K65.2Kmol

atmL108.206mol3.00

L4.01atm4.02

AAA

=

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

Use the ideal-gas law to express TB: nR

VPT BBB =

Substitute numerical values and evaluate TB:

( )( )( )

K142Kmol

atmL108.206mol3.00

L8.71atm4.02

B

=

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

−T

Use the ideal-gas law to express TC: nR

VPT CCC =

Chapter 18

1848

Substitute numerical values and evaluate TC:

( )( )( )

K81.2Kmol

atmL108.206mol3.00

L0.02atm1.02

C

=

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

−T

Apply the pressure-volume relationship for a quasi-static adiabatic process to the gas at points B and C to find the volume of the gas at point B:

γγCCBB VPVP =

and

( )

L71.8

L20.0atm4.0atm1.0 5

31

CB

CB

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= VPPV

γ

Substitute numerical values and evaluate WABC:

( )( ) ( )

( )

kJ2.4kJ18.4atm

J101.325atmL3.41

K142K81.2Kmol

atmL108.206mol3.00L4.01L71.8atm4.0 223

ABC

==×⋅=

−×

⎟⎠⎞

⎜⎝⎛

⋅⋅

×−−= −W

Apply the 1st law of thermodynamics to obtain: ( )ATTnRW

TnRWTCWEWQ

−+=Δ+=

Δ+=Δ+=

C23

ABC

23

ABC

VABCintABCABC

Substitute numerical values and evaluate QABC:

( ) ( ) kJ8.4K65.2K81.2Kmol

J8.314mol3.00kJ18.4 23

ABC =−⎟⎠⎞

⎜⎝⎛

⋅+=Q

83 •• [SSM] As part of a laboratory experiment, you test the calorie content of various foods. Assume that when you eat these foods, 100% of the energy released by the foods is absorbed by your body. Suppose you burn a 2.50-g potato chip, and the resulting flame warms a small aluminum can of water. After burning the potato chip, you measure its mass to be 2.20 g. The mass of the can is 25.0 g, and the volume of water contained in the can is 15.0 ml. If the temperature increase in the water is 12.5°C, how many kilocalories (1 kcal = 1 dietary calorie) per 150-g serving of these potato chips would you estimate there are? Assume the can of water captures 50.0 percent of the heat released during the burning of the potato chip. Note: Although the joule is the SI unit of choice in most thermodynamic situations, the food industry in the United States currently expresses the energy released during metabolism in terms of the ″dietary calorie,″ which is our kilocalorie.


1849

Picture the Problem The ratio of the energy in a 150-g serving to the energy in 0.30 g of potato chip is the same as the ratio of the masses of the serving and the amount of the chip burned while heating the aluminum can and the water in it. The ratio of the energy in a 150-g serving to the energy in 0.30 g of potato chip is the same as the ratio of the masses of the serving and the amount of the chip burned while heating the aluminum can and the water in it:

500g 0.30g 150

g 0.30

serving g-150 ==Q

Q

or g 0.30serving g 150 500QQ =

Letting f represent the fraction of the heat captured by the can of water, express the energy transferred to the aluminum can and the water in it during the burning of the potato chip:

( ) Tcmcm

TcmTcm

QQfQ

Δ

ΔΔ

OHOHAlAl

OHOHAlAl

OH Alg 0.30

22

22

2

+=

+=

+=

where ΔT is the common temperature change of the aluminum cup and the water it contains.

Substituting for g 0.30Q yields and solving for serving g-150Q yields:

( )f

TcmcmQ

Δ500 OHOHAlAlserving g-150

22+

=

Substitute numerical values and evaluate serving g-150Q :

( ) ( ) ( )

kcal 652cal 10256J 4.184

cal 1J 1007.1

500.0

C 5.12Kkg

kJ184.4kg 0150.0Kkg

kJ900.0kg 0250.0500

36

serving g-150

≈×=××=

°⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=Q

84 •• Diesel engines operate without spark plugs, unlike gasoline engines. The cycle that diesel engines undergo involves adiabatically compressing the air in a cylinder, and then fuel is injected. When the fuel is injected, if the air temperature inside the cylinder is above the fuel’s flashpoint, the fuel-air mixture will ignite. Most diesel engines have compression ratios in the range from 14:1 to 25:1. For this range of compression ratios (which are the ratio of maximum to minimum volume), what is the range of maximum temperatures of the air in the cylinder, assuming the air is taken into the cylinder at 35°C? Most modern gasoline engines typically have compression ratios on the order of 8:1. Explain why you expect the diesel engine to require a better (more efficient) cooling system than its gasoline counterpart.

Chapter 18

1850

Picture the Problem You can use the equation-of-state for an adiabatic process to find the range of maximum temperatures of the air in the cylinder. For an adiabatic process from state i to state f:

1ff

1ii

−− = γγ VTVT where, for a diatomic gas, γ = 1.4.

Solving for Tf yields:

1

f

ii1

f

1i

if

−

−

−

⎟⎟⎠

⎞⎜⎜⎝

⎛==

γ

γ

γ

VVT

VVTT

Substitute numerical values and evaluate Tf for an engine with a compression ratio of 14:1:

( )

C612K8851

14C 35K 27314.1

f

°==

⎟⎠⎞

⎜⎝⎛°+=

−

T

Substitute numerical values and evaluate Tf for an engine with a compression ratio of 25:1:

( )

C438K1116125C 35K 273

14.1

f

°==

⎟⎠⎞

⎜⎝⎛°+=

−

T

The range of maximum temperatures of the air in the cylinder, assuming it is taken into the cylinder at 35°C, is .C843C612 °→°

Gasoline engines, with much lower compression ratios, would clearly have lower maximum operating temperatures and thus be subject to overall, average, lower temperatures and not require as extensive cooling systems. 85 •• At very low temperatures, the specific heat of a metal is given by c = aT + bT3. For copper, a = 0.0108 J/kg⋅K2 and b = 7.62 × 10–4 J/kg⋅K4. (a) What is the specific heat of copper at 4.00 K? (b) How much heat is required to heat copper from 1.00 to 3.00 K? Picture the Problem We can find c at T = 4.00 K by direct substitution. Because c is a function of T, we can integrate dQ over the given temperature interval in order to find the heat required to heat copper from 1.00 to 3.00 K. (a) Substitute for a and b to obtain:

34

42 Kkg

J1062.7Kkg

J0.0108 TTc ⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

×+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

= −


1851

Evaluate c at T = 4.00 K:

( ) ( ) ( )

KkgJ1020.9

K00.4Kkg

J1062.7K00.4Kkg

J0.0108K00.4

2

34

42

⋅×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

×+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=

−

−c

(b) The heat required to heat copper from 1.00 to 3.00 K is given by:

( )∫=K 00.3

K 00.1

dTTcQ

Substituting for c(T) yields:

∫∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

×+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

= −K00.3

K00.1

34

4K00.3

K00.12 Kkg

J1062.7Kkg

J0108.0 dTTTdTQ

Evaluate this integral to obtain:

J/kg0584.04Kkg

J1062.72Kkg

J0108.0K3.00

K00.1

4

44

K3.00

K00.1

2

2 =⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

×+⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

= − TTQ

86 •• How much work must be done on 30.0 g of carbon monoxide (CO) at standard temperature and pressure to compress it to one-fifth of its initial volume if the process is (a) isothermal, (b) adiabatic? Picture the Problem Let the subscripts 1 and 2 refer to the initial and final state respectively. Because the gas is initially at STP, we know that V1 = 22.4 L/mol, P1 = 1 atm, and T1 = 273 K. We can use ( )12ln VVnRTW −= to find the work done on the gas during an isothermal compression. We can relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial and

final points on the path using 1

2211

−−

=γ

VPVPW and find P1 by eliminating P2 using

,2211γγ VPVP = where, for a diatomic gas, γ = 1.4. See Appendix C for the molar

mass of CO. (a) Express the work done on the gas in compressing it isothermally:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

2on ln

VVnRTW

Because the number of moles of CO is the mass m of CO divided by its molar mass:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

2on ln

VVRT

MmW

Chapter 18

1852


( ) kJ3.9151lnK273

KmolJ8.314

g/mol28.01g30.0

on =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛−=W

(b) Express the work done on the gas in compressing it adiabatically:

1

1

21

211

2211on

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−−

−=

γ

γ

VPPVP

VPVPW

Using the equation for a quasi-static adiabatic process, relate the initial and final pressures and volumes:

γγ2211 VPVP = ⇒

γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

1

2

VV

PP

Substitute for P2/P1and simplify to obtain:

1

2.01

1

5

12

111

1

2

1112

2

111

on −

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=γγγ

γγγ

VVVPV

VVVPV

VVVP

W


( )( ) ( )( )kJ49.5

14.1

50.21molL22.4mol1.071kPa101.325 1.4

=−

−⎟⎠⎞

⎜⎝⎛

−=W

87 •• How much work must be done on 30.0 g of carbon dioxide (CO2) at standard temperature and pressure to compress it to one-fifth of its initial volume if the process is (a) isothermal, (b) adiabatic? Picture the Problem Let the subscripts 1 and 2 refer to the initial and final state respectively. Because the gas is initially at STP, we know that V1 = 22.4 L/mol, P1 = 1 atm, and T1 = 273 K. We can use ( )12ln VVnRTW −= to find the work done on the gas during an isothermal compression. We can relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial

and final points on the path using 1

2211

−−

=γ

VPVPW and find P1 by eliminating P2

using γγ2211 VPVP = . We can find γ using the data in Table 18-3. See Appendix C

for the molar mass of CO2.


1853

(a) Express the work done on the gas in compressing it isothermally:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

2on ln

VVnRTW

Because the number of moles of CO2 is the mass m of CO2 divided by its molar mass:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

2on ln

VVRT

MmW


( ) kJ49.251lnK273

KmolJ8.314

g/mol01.44g30.0

on =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛−=W


1

1

21

211

2211on

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−−

−=

γ

γ

VPPVP

VPVPW


γγ2211 VPVP = ⇒

γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

1

2

VV

PP

Substitute for P2/P1 and simplify to obtain:

1

2.01

1

5

12

111

1

2

1112

2

111

on −

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=γγγ

γγγ

VVVPV

VVVPV

VVVP

W

From Table 18-3 we have:

Rc 39.3V = and

( ) RRc 41.402.139.3P =+=

Evaluate γ: 30.139.341.4

V

P ===RR

ccγ

Chapter 18

1854


( )( ) ( )( )kJ20.3

13.1

50.21molL22.4mol6817.0kPa101.325 1.30

on =−

−⎟⎠⎞

⎜⎝⎛

−=W

88 •• How much work must be done on 30.0 g of argon (Ar) at standard temperature and pressure to compress it to one-fifth of its initial volume if the process is (a) isothermal, (b) adiabatic? Picture the Problem Let the subscripts 1 and 2 refer to the initial and final states respectively. Because the gas is initially at STP, we know that V1 = 22.4 L/mol, P1 = 1 atm, and T1 = 273 K. We can use ( )12ln VVnRTW −= to find the work done on the gas during an isothermal compression. We can relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial

and final points on the path using 1

2211

−−

=γ

VPVPW and find P1 by eliminating P2

using ,2211γγ VPVP = where, for a monatomic gas, γ = 1.67. See Appendix C for the

molar mass of Ar. (a) Express the work done on the gas in compressing it isothermally:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

2on ln

VVnRTW

Because the number of moles of Ar is the mass m of Ar divided by its molar mass:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

2on ln

VVRT

MmW


( ) kJ74.251lnK273

KmolJ8.314

g/mol948.39g30.0

on =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛−=W


1

1

21

211

2211on

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−−

−=

γ

γ

VPPVP

VPVPW


γγ2211 VPVP = ⇒

γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

1

2

VV

PP


1855

Substitute for P2/P1 and simplify to obtain:

( )

1

2.01

1

5

12

111

1

2

1112

2

111

on −

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=γγγ

γγγ

VVVPV

VVVPV

VVVP

W


( )( ) ( )( )( )kJ93.4

167.1

50.21molL22.4mol75098.0kPa101.325 1.67

on =−

−⎟⎠⎞

⎜⎝⎛

−=W

89 •• [SSM] A thermally insulated system consists of 1.00 mol of a diatomic gas at 100 K and 2.00 mol of a solid at 200 K that are separated by a rigid insulating wall. Find the equilibrium temperature of the system after the insulating wall is removed, assuming that the gas obeys the ideal-gas law and that the solid obeys the Dulong–Petit law. Picture the Problem We can use conservation of energy to relate the equilibrium temperature to the heat capacities of the gas and the solid. We can apply the Dulong-Petit law to find the heat capacity of the solid at constant volume and use the fact that the gas is diatomic to find its heat capacity at constant volume. Apply conservation of energy to this process:

0Δ solidgas =+= QQQ

Use TCQ ΔV= to substitute for Qgas and Qsolid:

( ) ( ) 0K200K100 equilsolidV,equilgasV, =−+− TCTC

Solving for Tequil yields: ( )( ) ( )( )

solidV,gasV,

solidV,gasV,equil

K200K100CC

CCT

+

+=

Using the Dulong-Petit law, determine the heat capacity of the solid at constant volume:

RnC solidsolidV, 3=

The heat capacity of the gas at constant volume is given by:

RnC gas25

gasV, =

Chapter 18

1856

Substitute for CV,solid and CV,gas and simplify to obtain:

( )( ) ( )( ) ( )( ) ( )( )solidgas2

5

solidgas25

solidgas25

solidgas25

equil 33K200K100

33K200K100

nnnn

RnRnRnRn

T+

+=

+

+=

Substitute numerical values for ngas and nsolid and evaluate Tequil:

( )( )( ) ( )( )( )( ) ( ) K 171

mol 00.23mol 00.1mol 00.23K200mol 00.1K100

25

25

equil =++

=T

90 •• When an ideal gas undergoes a temperature change at constant volume, its internal energy change is given by the formula ΔEint = CvΔT. However, this formula correctly gives the change in internal energy whether the volume remains constant or not. (a) Explain why this formula gives correct results for an ideal gas even when the volume changes. (b) Using this formula, along with the first law of thermodynamics, show that for an ideal gas Cp = Cv + nR. Picture the Problem (a) TCE Δ=Δ Vint correctly gives the change in internal energy when the temperature changes and the volume is not constant because ΔEint is the same for all gas processes that have the same ΔT. For an ideal gas, the internal energy is the sum of the kinetic energies of the gas molecules and this sum is proportional to kT. Any two processes that change the thermal energy of the gas by ΔEint will cause the same temperature change ΔT. Consequently, ΔEint is independent of the process that takes the gas from one state to another. (b) Use the first law of thermodynamics to relate the work done on the gas, the heat entering the gas, and the change in the internal energy of the gas:

oninint WQE +=Δ

Substituting for ΔEint yields:

oninV WQTC +=Δ

The work done on the gas during a constant-pressure process is given by:

( )ifon VVPVPW −−=Δ−=


1857

Use the ideal gas law to substitute for Vf and Vi and then simplify to obtain:

( )

TnR

TTnRP

nRTP

nRTPW

Δ−=

−−=⎟⎠⎞

⎜⎝⎛ −−= if

ifon

Substituting for onW gives: TnRQTC Δ−=Δ inV

Solve for Qin and simplify to obtain:

( )TC

TnRCTnRTCQΔ=

Δ+=Δ+Δ=

P

VVin

where nRCC += VP 91 •• An insulated cylinder is fitted with an insulated movable piston to maintain constant pressure. The cylinder initially contains 100 g of ice at –10ºC. Heat is transferred to the ice at a constant rate by a 100-W heater. Make a graph showing the temperature of the ice/water/ steam as a function of time starting at ti when the temperature is –10ºC and ending at tf when the temperature is 110ºC. Picture the Problem Knowing the rate at which energy is supplied, we can obtain the data we need to plot this graph by finding the time required to warm the ice to 0°C, melt the ice, warm the water formed from the ice to 100°C, vaporize the water, and warm the water to 110°C. Find the time required to warm the ice to 0°C:

( ) ( )( )min 0.3s20.0

sJ100

C10C0Kkg

kJ2.0kg0.100ΔΔ ice

1 ≈=°−−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

==P

Tmct

Find the time required to melt the ice: ( )

min 5.6s5.333sJ100

kgkJ333.5kg0.100

Δ f2

≈=

⎟⎟⎠

⎞⎜⎜⎝

⎛

==P

mLt

Find the time required to heat the water to 100°C:

( ) ( )

min 7.0s418sJ100

C100Kkg

kJ18.4kg0.100

ΔΔ w3

≈=

°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=

=P

Tmct

Chapter 18

1858

Find the time required to vaporize the water: ( )

min 6.37s2257sJ100

kgkJ2572kg0.100

Δ V4

≈=

⎟⎟⎠

⎞⎜⎜⎝

⎛

==P

mLt

Find the time required to heat the vapor to 110°C:

( ) ( )

min 0.3s0.20sJ100

C10Kkg

kJ0.2kg0.100

ΔΔ steam5

≈=

°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=

=P

Tmct

A graph of T as a function of time follows:

50.512.95.90.3min ,t

110100

0−10

C , °T

8.50 92 •• (a) In this problem, 2.00 mol of a diatomic ideal gas expand adiabatically and quasi-statically. The initial temperature of the gas is 300 K. The work done by the gas during the expansion is 3.50 kJ. What is the final temperature of the gas? (b) Compare your result to the result you would get if the gas were monatomic. Picture the Problem We know that, for an adiabatic process, Qin = 0. Hence the work done by the expanding gas equals the change in its internal energy. Because we’re given the work done by the gas during the expansion, we can express the change in the temperature of the gas in terms of this work and CV.


1859

(a) Express the final temperature of the gas as a result of its expansion:

TTT Δ+= if

Apply the equation for adiabatic work and solve for ΔT:

TCW Δ−= Vadiabatic and

nRW

CWT

25adiabatic

V

adiabatic −=−=Δ

Substitute for ΔT to obtain:

nRWTT

25adiabatic

if −=

Substitute numerical values and evaluate Tf for a diatomic gas: ( )

K216

KmolJ8.314mol2.00

kJ3.50K30025

f

=

⎟⎠⎞

⎜⎝⎛

⋅

−=T

(b) Because a monatomic gas has only 3 degrees of freedom:

nRC 23

V = and

nRWTT

23adiabatic

if −=

Substitute numerical values and evaluate Tf for a monatomic gas: ( )

K160

KmolJ8.314mol2.00

kJ3.50K30023

f

=

⎟⎠⎞

⎜⎝⎛

⋅

−=T

93 •• A vertical insulated cylinder is divided into two parts by a movable piston of mass m. The piston is initially held at rest. The top part is evacuated and the bottom part is filled with 1.00 mol of diatomic ideal gas at temperature 300 K. After the piston is released and the system comes to equilibrium, the volume occupied by gas is halved. Find the final temperature of the gas. Picture the Problem Let the subscripts 1 and 2 refer to the initial and final states in this adiabatic expansion. We can use an equation describing a quasi-static adiabatic process to express the final temperature as a function of the initial temperature and the initial and final volumes.

Chapter 18

1860

Using the equation for a quasi-static adiabatic process, relate the initial and final volumes and temperatures:

111

122

−− = γγ VTVT

Solve for T2 and simplify to obtain: ( ) 1

1

1

121

11

1

2

112 2 −

−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= γ

γγ

TV

VTVVTT

Substitute the numerical value for T1 and evaluate T2:

( )( ) K3962K300 14.12 == −T

94 ••• According to the Einstein model of a crystalline solid, the internal

energy per mole is given by E

A E/

31

=−T T

N kTUe

where TE is a characteristic temperature

called the Einstein temperature, and T is the temperature of the solid in kelvins. Use this expression to show that a crystalline solid’s molar heat capacity at

constant volume is given by ( )22

V1

3E

E

−⎟⎠⎞

⎜⎝⎛=

TT

TTE

ee

TTRc' .

Picture the Problem The molar heat capacity at constant volume is related to the

internal energy per mole according todTdU

nc'v

1= . We can differentiate U with

respect to temperature and use nR = Nk or R = NAk to establish the result given in the problem statement. From the Einstein model of a crystalline solid, the internal energy per mole is given by:

13

E

EA

−= TTe

kTNU

Relate the molar heat capacity at constant volume to the internal energy per mol:

dTdU

nc' 1

V =

Use dTdU

nc' 1

V = to express Vc' :

( )( )

( ) ( )22

22E

2EEEA

V

13

113

11

131

131

31

E

EE

E

E

EEE

−⎟⎠⎞

⎜⎝⎛=⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−

−=

−⎥⎥⎦

⎤

⎢⎢⎣

⎡

−

−=⎥⎦

⎤⎢⎣⎡

−=⎥⎦

⎤⎢⎣⎡

−=

TT

TTEETT

TT

TT

TTTTTT

ee

TTR

TTe

eRT

edTd

eRT

edTdRT

ekTN

dTd

nc'


1861

95 ••• [SSM] (a) Use the results of Problem 94 to show that in the limit that T >> TE, the Einstein model gives the same expression for specific heat that the Dulong–Petit law does. (b) For diamond, TE is approximately 1060 K. Integrate numerically dEint = v′c dT to find the increase in the internal energy if 1.00 mol of diamond is heated from 300 to 600 K. Picture the Problem (a) We can rewrite our expression for 'cV by dividing its numerator and denominator by TTe E and then using the power series for ex to show that, for T > TE, Rc' 3V ≈ . In part (b), we can use the result of Problem 108 to obtain values for 'cV every 100 K between 300 K and 600 K and use this data to find ΔU numerically. (a) From Problem 94 we have:

( )2

2E

V1

3E

E

−⎟⎠⎞

⎜⎝⎛=

TT

TT

ee

TTRc'

Divide the numerator and denominator by TTe E to obtain:

TTTT

TT

TTTT

eeTTR

eeeT

TRc'

EE

E

EE

213

1213

2E

2

2E

V

−+−⎟⎠⎞

⎜⎝⎛=

+−⎟⎠⎞

⎜⎝⎛=

Express the exponential terms in their power series to obtain:

E

2E

2EE

2EE

for

...2112...

2112 EE

TTTT

TT

TT

TT

TTee TTTT

>>⎟⎠⎞

⎜⎝⎛≈

+⎟⎠⎞

⎜⎝⎛+−+−+⎟

⎠⎞

⎜⎝⎛++=+− −

Substitute for TTTT ee EE 2 −+− to obtain:

R

TTT

TRc' 313 2E

2E

V =

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛≈

(b) Use the result of Problem 94 to verify the following table:

T (K) 300 400 500 600 cV (J/mol⋅K) 9.65 14.33 17.38 19.35

Chapter 18

1862

The following graph of specific heat as a function of temperature was plotted using a spreadsheet program:

5

7

9

11

13

15

17

19

21

300 350 400 450 500 550 600

T (K)

CV (J

/mol

-K)

Integrate numerically, using the formula for the area of a trapezoid, to obtain:

( )( )( )

( )( )( )

( )( )( )

kJ62.4Kmol

J35.1938.17K100mol 00.1

KmolJ38.1733.14K100mol 00.1

KmolJ33.1465.9K100mol 00.1Δ

21

21

21

=⋅

++

⋅++

⋅+=U

96 ••• Use the results of the Einstein model in Problem 94 to determine the molar internal energy of diamond (TE = 1060 K) at 300 K and 600 K, and thereby the increase in internal energy as diamond is heated from 300 K to 600 K. Compare your result to that of Problem 95. Picture the Problem We can simplify our calculations by relating Avogadro’s number NA, Boltzmann’s constant k, the number of moles n, and the number of molecules N in the gas and solving for NAk. We can then calculate U300 K, U600 K, and their difference.


1863

Express the increase in internal energy per mole resulting from the heating of diamond:

K300K600 UUU −=Δ (1)

Express the relationship between Avogadro’s number NA, Boltzmann’s constant k, the number of moles n, and the number of molecules N in the gas:

NknR = ⇒ kNknNR A==

Substitute in the given equation to obtain:

13

E

E

−= TTe

RTU

Determine U300 K: ( )

J795J 4.7951

K1060Kmol

J314.83

K300K1060K300

==

−

⎟⎠⎞

⎜⎝⎛

⋅=e

U

Determine U600 K: ( )

kJ45.5kJ 4498.51

K1060Kmol

J314.83

K600K1060K600

==

−

⎟⎠⎞

⎜⎝⎛

⋅=e

U

Substituting in equation (1) yields:

kJ4.65J795.4kJ5.4498

Δ K300K600

=−=

−= UUU

This result agrees with the result of Problem 95 to within 1%.

97 ••• During an isothermal expansion, an ideal gas at an initial pressure P0 expands until its volume is twice its initial volume V0. (a) Find its pressure after the expansion. (b) The gas is then compressed adiabatically and quasi-statically until its volume is V0 and its pressure is 1.32P0. Is the gas monatomic, diatomic, or polyatomic? (c) How does the translational kinetic energy of the gas change in each stage of this process? Picture the Problem The isothermal expansion followed by an adiabatic compression is shown on the PV diagram. The path 1→2 is isothermal and the path 2→3 is adiabatic. We can apply the ideal-gas law for a fixed amount of gas and an isothermal process to find the pressure at point 2 and the pressure-volume relationship for a quasi-static adiabatic process to determineγ.

Chapter 18

1864

(a) Relate the initial and final pressures and volumes for the isothermal expansion and solve for and evaluate the final pressure:

2211 VPVP = and

021

1

10

2

112 2

PVVP

VVPP ===

(b) Relate the initial and final pressures and volumes for the adiabatic compression:

γγ3322 VPVP =

or ( ) γγ

000021 32.12 VPVP = which simplifies to 64.22 =γ

Take the natural logarithm of both sides of this equation and solve for and evaluate γ :

64.2ln2ln =γ ⇒ 40.12ln64.2ln

==γ

and the gas is diatomic .

(c) During the isothermal process, T is constant and the translational kinetic energy of the gas is unchanged. During the adiabatic process, T3 = 1.32T0, and the translational kinetic energy of the gas increased by a factor of 1.32. 98 ••• If a hole is punctured in a tire, the gas inside will gradually leak out. Assume the following: the area of the hole is A; the tire volume is V; and the time, τ, it takes for most of the air to leak out of the tire can be expressed in terms of the ratio A/V, the temperature T, the Boltzmann constant k, and the initial mass m of the gas molecules inside the tire. (a) Based on these assumptions, use dimensional analysis to find an estimate for τ. (b) Use the result of Part (a) to estimate the time it takes for a car tire with a nail hole punched in it to go flat. Picture the Problem In (a) we’ll assume that τ = f (A/V, T, k, m) with the factors dependent on constants a, b, c, and d that we’ll find using dimensional analysis. In (b) we’ll use our result from (a) and assume that the diameter of the puncture is


1865

about 2 mm, that the tire volume is 0.1 m3, and that the air temperature is 20°C. (a) Express τ = f (A/V, T, k, m):

( ) ( ) ( ) dcba

mkTVA⎟⎠⎞

⎜⎝⎛=τ (1)

Rewrite this equation in terms of the dimensions of the physical quantities to obtain:

( ) ( ) ( ) dc

ba MKT

MLKLT 2

21

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

where K represents the dimension of temperature.

Simplify this dimensional equation to obtain:

dcccba MTKLMKLT -2c21 −−= or

2c21 TMKLT −+−−= dccbac

Equate exponents to obtain: 12:T =− c , 02:L =− ac ,

0:K =− cb , and

0:M =+ dc

Solve these equations simultaneously to obtain:

1−=a , 21−=b , 2

1−=c , and 21=d

Substituting in equation (1) yields: ( ) ( ) ( )

kTm

AVmkT

VA

=⎟⎠⎞

⎜⎝⎛= −−

−

21

21

21

1

τ

(b) Substitute numerical values and evaluate τ:

( )( )( )

( )( ) min4s232K293KJ/mol314.8

m1.0kg/m293.1

m1024

m1.0 33

23

3

≈=⋅×

=−πτ

Chapter 18

1866

physics notes18

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