1
HEAT AND THERMODYNAMICS
HEAT Heat is that form of energy. Zeroth Law of Thermodynamics And Temperature
If a system A is in thermal equilibrium with system B and the system B is in thermal equilibrium with system C, then systems A and C are in thermal equilibrium with each other.
Temperature scalesRelation between Celsius, Kelvin and Fahrenheit Scale:
Example 1: The electrical resistance of pure platinum increases linearly
with increasing temperature. This property is used in a
Platinum resistance thermometer. The relation between R(
(Resistance at (K) and R0 (Resistance at (0K) is given as
where ( = temperature coefficient of resistance. Now, if a
Platinum resistance thermometer reads 00 Celsius when its resistance is 80( and 100o when its resistance is 90(, find the temperature at which its resistance is 86(.
Solution: Using the given relationship,
we have
. . . (i)
. . . (ii)
Where
is the desired temperature.
Taking the ratio of (i) & (ii)
CALORIMETRY
Principle of Calorimetry
When two objects having different temperatures are brought in contact, heat flows from the hot object to the cold object. If the system is sufficiently thermally isolated from its surrounding, the heat lost by the hot object = the heat gained by the cold object.
Specific heat capacity
The amount of heat needed to raise the temperature of unit mass of a substance by 1( is known as its specific heat capacity. If Q amount of heat raises the temperature of m mass of a material by
, then its specific heat capacity is given as:
Also the amount of heat supplied per unit increase in temperature for any body is known as its heat capacity,
.
Specific Latent Heat
In order to change the state of a substance (from solid to liquid or from liquid to gas) heat has to be supplied to it. During this process temperature remains constant. The amount of heat supplied per unit mass for such a process is known as the Specific Latent Heat of that substance for that process.
Example 2:5 gm of water at 30oC and 5 gm of ice at - 20oC are mixed together in a calorimeter. Find the final temperature of mixture and also the final masses of ice and water. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal/gmoC and latent heat of ice = 80 cal/gm.
Solution:In this case heat is given by water and taken by ice
Heat available with water to cool from 30oC to 0oC
= ms( =.
Heat required by 5 gm ice to increase its temperature up to 0oC
ms(
Out of 150 cal heat available, 50 cal is used for increasing
temperature of ice from - 20oC to 0oC. The remaining heat 100 cal is used for melting the ice.
if mass of ice melted is m gm then
(
Thus 1.25 gm ice out of 5 gm melts and the mixture of ice and
water is at 0oC.
Mechanical Equivalent of Heat
1 calorie is the quantity of heat required to raise the temperature of pure water at 1 atm pressure from 14.5(C to 15.5(C.
When work is completely converted to heat, the quantity of heat produced (Q) is found to be proportional to the quantity of work (W) that was converted into heat
W ( Q
Or,W = JQ, where J is known as the Mechanical Equivalent of Heat.
J = 4.2 ( 107 erg cal-1 = 4.2 J/cal
Example 3: A bullet splinter of mass of 10 gm moving with a speed of 400 m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets stuck in it. How much ice will melt if 50% of the lost Kinetic energy goes to ice? (Temperature of ice block = 0oC).
Solution: Velocity of bullet + ice block, V = m/s
V =4 m/s
Loss of K.E. =
= [0.01((400)2(1((4)2] = [1600 ( 16]= (1584/2) J
Heat generated = = 95 Cal
Mass of ice melted = = 1.2 gm.
Thermal Expansion
An increase in the temperature of a body is generally accompanied by an increase in its size. This is known as Thermal Expansion.
The value of linear coefficient of thermal expansion at temperature T can be found by taking limit
.
or
For most of the solids the value of
is small and nearly independent of T. In such cases the linear dimension of an object at a different temperature is given by:
(Here
is the initial length and
is the change in temperature).
Example 4:A clock with an iron pendulum is made so as to keep correct time at 10oC. Given . How fast or slow does the clock move per day if the temperature rises to 250 C? Given (iron = 12 ( 10(6 per 0C.
Solution: When the pendulum keeps correct time, its period of vibration is 2
sec and so it makes
Vibration /day
If length of pendulum at 10oC is
and at 25oC is
EMBED Equation.2
as
i.e. T (((
i.e. n is no. of vibrations per sec.
EMBED Equation.2
EMBED Equation.2 = 43200[1 ( 0.00009] =
That is the clock makes (43200 ( 43196.12) = 3.88 vibration loss
per day. That is clock losses 3.88 ( 2 = 7.76 sec per day.
Area and Volume expansion
If the temperature of a two-dimensional object (lamina) is changed, its area changes. If the coefficient of linear expansion of the material of lamina is small and constant, then its final area is given by
, where
is the initial area,
is the change in temperature and
is the area coefficient of thermal expansion. It can be shown that
. Similar relation also holds for the volume of a three-dimensional object
(Here
is known as the coefficient of volume expansion). For most solids, ( ~ 10(6/(C
Thermal Expansion in liquids:The experimental measurement of ( for a liquid becomes slightly difficult due to expansion of the container, when a liquid is heated in a container. The initial level of the liquid falls due to expansion of the container. But afterwards it rises due to faster expansion of the liquid.
The actual increase in the = volume of the liquid the apparent increase in the volume of liquid+The increase in the volume of the container.
(liquid = (apparent + (container
(, for liquids, is of the order of 10(4/(C
Anomalous expansion of water: If the temperature of water is increased from 0oC, it contracts until the temperature reaches 4oC and expands thereafter. In other words the density of water is highest at 4oC.
Example 5:A glass vessel of volume
is completely filled with a liquid
and its temperature is raised by
. What volume of the liquid
will overflow? Coefficient of linear expansion of glass =
and
coefficient of volume expansion of the liquid = ((.
Solution: Volume of the liquid over flown
= Increase in the volume of the liquid ( increase in the volume of
the container
=
= =
Thermal Stress
When a rod of length L is held between two rigid supports and the temperature of rod is increased by
, the rigid supports prevent the rod from expanding. This causes compressive stress in the rod. As the length of the rod remains unchanged,
Thermal expansion = Mechanical compression
Thermal stress
, which is compressive.
KINETIC THEORY OF GASES
Fluids consist of molecules which are in random motion, which collide with each other and with the walls of the container. The intermolecular separation in a gas is larger by an order of magnitude than the intermolecular separation in a liquid. The intermolecular separation in a gas increases as the temperature increases, and decreases with increasing density.
it is experimentally observed that most gases follow a universal equation of state
pV = n RT
Gas molecules are relatively free:the interactions between them are small. This means that the total energy of the gas molecules is mostly kinetic.
A simple model of a gas, with point particles representing molecules and their collisions with the walls of the confining vessel causing the pressure exerted by the gas leads to some important conclusions:(a)The pressure exerted by an ideal gas is given by
p = ( mc, where ( = number of molecules per unit volume,
m = mass of each molecule, crms = the rms speed of each molecule.
(b)The total internal energy of an ideal gs is
U = Nf where N = number of molecules in the gas, f = number of degree of freedom, T = absolute temperature of the gas, kB = = Boltzmanns constant.
(C)The rms speed is given by
crms=, where M represents the molar mass of the gas (i.e. mass of 1 mole of the gas)
(d)The pressure of an ideal gas is given by
p =
EMBED Equation.3 , where Utran is the translational energy of the molecules of the gas.
Example6:Find r.m.s speed of Hydrogen molecules at room temperature
(=300 k).
Solution:Mass of 1 mole of Hydrogen gas= 2 gm = 2 (10-3 kg
( Vrms =
=
= 1.93 ( 103 m/s.
2.When the container contains more than one gas, total pressure exerted by all the gases on the wall is sum of pressures exerted by each gas as it would while filling the container alone. In a way, each gas behaves independent of each other. Thus we have
P = P1+P2+P3+, where P1, P2, P3 are the partial pressures of gases 1, 2 & 3 respectively
This is known as Daltons Law of partial pressures.
3.One mole of any gas occupies a volume of 22.4 litre at standard temperature and Pressure which are 273.15 (=0C) and 1.013(105 Pa (=1atm) respectively,
Example 7: 4 gm Hydrogen is mixed with 11.2 litre of He at S.T.P. in a
container of volume 20 litre. If the final temperature is 300 K find
the pressure.
Solution:4 gm Hydrogen = 2 moles Hydrogen
11.2 ( He at S.T.P. = 1/2 mole of He
P = PH + PHe
= (nH+nHe) = (2+)
= 3.12( 105 N/m2.
Internal Energy
Internal energy, of any body is sum total of kinetic energies and potential energies of its constituents (at molecular level). In case of an ideal gas, as there are no intermolecular forces, except during collision the possibility of potential energy is ruled out, so it is only kinetic energy. The kinetic energy of the molecules can be of three types.
(i)Translational
(ii)Rotational
(iii)Vibrational
In a way, it means that the energy of molecules is shared in various modes. These independent modes of motions are called degrees of freedom. The table given below gives the number of degrees of freedom for various types of molecules at normal temperature.
Nature of motion
AtomicityDegree of Freedom (f)
TranslationalRotationalVibrational*Total
(1) Monoatomic3003
(2) Diatomic3205
PolyLinear3205
Non-linear3306
*At room temperature the energy associated with vibrational motion is negligibly small in comparison to translational and rotational K.E.
Equipartition of Energy
According to the Law of equipartition of energy the average K.E. of a molecule is equally shared among different degrees of freedom. The average energy per degree of freedom of a molecule is kT, where k is the Boltzmanns constant and T is the absolute temperature. Thus, for a monoatomic ideal gas:U (the internal energy) = kT
Also for one mole U = RT
Example 8:Find the average kinetic energy per molecule at temperature T for an equimolar mixture of two ideal gases A and B where A is monoatomic and B is diatomic.
Solution:No. of degrees of freedom per molecule for A = 3
No. of degrees of freedom per molecule for B = 5
Since the mixture is equimolar, the average kinetic energy per
molecule will be given by the average of the two values i.e.
kT = 4kT
where k is Boltzmanns constant.
Work Done In Different Processes
Work done by an enclosed gas on its surroundings is given by
W = (p dV
For different types of processes, we have got different relations between p (Pressure) and V (Volume) and accordingly we have different expressions for work.
(a)Isochoric Process
Here, the volume is constant throughout and therefore the work done by the gas, irrespective of the manner in which pressure varies, is zero
Wisochoric = 0
(b)Isobaric Process
In this case, pressure of the gas remains constant throughout the process. Hence,
( pdV = p (V
= n R(T(n = number of moles)
(( T= change in absolute temperature)
(C)Isothermal process
The temperature remains constant throughout the process.
Using ideal gas equation, we get,
p=
Hence,
( pdV = n RT = n RT
(d) Adiabatic Process:
pV( = Constant = C (say)
(p = CV-(
(W = ( pdV =
= = (nCv(T2 ( T1)
The work done by a gas can also be evaluated from the p-V diagram of the process.
Area enclosed by the curve in a p-V diagram = work done by the gas
First Law of Thermodynamics
First law of thermodynamics is simply a re-statement of the principle of conservation of energy for a thermally isolated system.
If (Q, (U & (W represent the heat given to the system, change in its internal energy and the work done by the system respectively, the first law of thermodynamics states that,
(Q = (U + (W
The heat transferred to the system ((Q) is either utilised to do work ((W) or increase the internal energy of the system ((U).
Example 9:3000 J of heat is given to a gas at constant pressure of 2 ( 105 N/m2. If its volume increases by 10 litres during the process find the change in the internal energy of the gas
Solution:(Q = 3000 J
W = P ( V = (2(105 N/m2) (10 (10-3m3)
= 2 ( 103 J
(U = (Q W = 3000 ( 2000= 1000 J.
Specific Heat Capacities of Gases
S =
where (Q = amount of heat required for (T temperature change.
m = mass of the gas.
In case of gases, the concept of a molar heat capacity is useful. Molar heat capacity is the amount of heat required to raise the temperature of one mole of the gas by one degree.
So, if (Q amount of heat goes to change the temperature of n moles of a gas in a particular process, molar heat capacity C can be mathematically given by:
C =
In terms of differentials,
C =
Two special cases are:-
(i)If volume is kept constant during the process then
CV =
This is the molar heat capacity of the gas at constant volume
Note: Since (U is independent of the process. (U = n Cv (T is true for all processes.
(ii)If pressure remains constant, then
Cp =
This is the molar heat capacity of the gas at constant pressure
Relation Between Cp and Cv
Cp ( Cv = R
This is known as Mayers relation.
The Values of Cp and CvIf f is the number of degrees of freedom of a gas molecule then the internal energy of n moles of that gas is given as
U = f/2 n RT
((U = f/2 n R(T = n Cv(T
(Cv = f/2 R
From Mayers Relation
Cp = Cv + R
Cp = (f/2+1)R
And the ratio of specific heats
( = =
( = =
Example 10: Find the molar heat capacity of an ideal gas with adiabatic
exponent ( for the polytropic process P V ( = constant.
Solution:We have, from first law of thermodynamics
C = Cv +
(n = number of moles)
We have, P V( = constant
From Ideal gas equation P V = n RT
Taking ratio, T V((1 = Constant
Differentiating we get = (
Putting it in the equation for C.
C = Cv ( = Cv (
= Cv (
C =
Second law of thermodynamics
(i) Kelvin Statement:- It is impossible to derive a continuous supply of work by cooling a body to a temperature lower than that of the coldest of its surroundings.
(ii) Clausius Statement:-It is impossible for a self acting machine, unaided by an external agency to transfer heat from a body to another at higher temperature.
Reversible Process:A process which can be made to proceed in the reverse direction by variations in its conditions so that all changes occurring in any part of the direct process are exactly reversed in the corresponding part of the reverse process is called a reversible processes.
Irreversible Process:A process which can not be made to proceed in the reverse direction is called an irreversible process.
Heat Engine:It is a device which continuously converts heat energy into the mechanical energy in a cyclic process.
Efficiency of heat engine:
( = =
Where Q1 is the heat supplied by the source and Q2 is the heat rejected to the sink.
Carnot Engine:
It is an ideal heat engine which is based on Carnot's reversible cycle. It works in four steps viz. Isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. The efficiency of a Carnot engine is given by
( = 1 ( = 1 (
where T1 and T2 are the temperatures of source and sink respectively.
Example 11:The efficiency of a Carnot cycle is 1/6. If on reducing the
temperature of the sink by 650C, the efficiency becomes 1/3,
find the initial and final temperatures between which the cycle
is working.
Solution:Given (1 = , (2 =
If the temperatures of the source and the sink between which the
cycle is working are T1 and T2, then the efficiency in the first case
will be
(1 = 1 - =
In the second case (2 = 1 (
=
Solving T1 = 390 K andT2 = 325 K.
OBJECTIVE1:Calculate the root mean square speed of smoke particles of mass
kg in Brownian motion in air at NTP. Boltzmann constant
(A)1.5 cm/s(B) 2.2 cm/s
(C) 2.3 cm/s(D)4.4 cm/s
Ans. (a)
Solution:PV =
(
=
where ( = mass of one molecule.
(
= cm/s
2:During an experiment an ideal gas is found to obey an additional law
= constant. The gas is initially at temp T and volume V. What will be the temperature of the gas when it expands to a volume 2V?
(A)
(B)
(C)
(D)
Ans. (b)
Solution:According to the given problems
VP2 = constant
From the gas law
PV = nRT
(
(
(
i.e,
(
Q.3-5 We have two vessels of equal volume, one filled with hydrogen and the other with equal mass of Helium. The common temperature is 27oC.
3:What is the relative number of molecules in the two vessels ?
(A)
(B)
(C)
(D)
Ans. (C)4: If pressure of Hydrogen is 2 atm, what is the pressure of Helium ?
(A) pHe = 2 atm.
(B) pHe = 3 atm.
(C) pHe = 4 atm.
(D) pHe = 1 atm.
Ans. (d)
5: If the temperature of Helium is kept at 27o C and that of hydrogen is changed, at what temperature will its pressure become equal to that of helium ? The molecular weights of hydrogen and helium are 2 and 4 respectively.
(A) (123oC(B) (140oC
(C) (160oC(D) (183oC
Ans. (a)
Solution 3-5: 3. The masses of hydrogen and helium gases in the vessels are equal. This means that the product of the number of molecules and the mass of a molecule must be same for H2 and He gases. Since molecular masses of H2 and He are in the ratio 1: 2, their number of molecules nH and nHe in the vessels must be in the reverse ratio, that is,
4. The equation of state for one mole of a gas is
pV = RT = NkT
Where N is Avogadros number (no. of molecules in one mole) and k is Boltzmanns constant. If a gas has n molecules, the equation of state will be
pV = nkT
For a given volume and a given temperature, we have
p ( n.
Since H2 and He have same volume and same temperature (27oC), we have
HerepH = 2 atm.
(pHe = 1 atm.
5.Again, we have
pV = nkT
H2 and He have equal volumes. For having equal pressure, we must have
nHTH = nHeTHe
or
HereTHe = 27 + 273 = 300 K
(TH = THe = 150 K
= 150 ( 273 = (123oC
6: A vessel contains a mixture of 7 gm of nitrogen and 11 gm of carbon dioxide at temperature T = 290 K. If pressure of the mixture P = 1 atm, calculate its density (R = 8.31 J/mol k)
(A) 2.5 kg/m3(B)1.5 kg/m3
(C) 4.5 kg/m3(D)7.5 kg/m3
Ans. (b)
Solution:As molecular weight of N2 and CO2 are 28 and 44, and ,
So,
Now, according to gas law PV =
=
and m = 7+11 = 18 gm = 18 ( 10-3 kg
so, ( =
Q.7-10.The pressure of a monoatomic gas increases linearly from
N/m2 to N/m2 when its volume increases from 0.2 m3 to 0.5 m3. Calculate7:work done by the gas
(A) 2.8105 J(B)1.8106 J
(C) 1.8105 J(D)1.8102 J
Ans. (C) 8: increase in internal energy
(A)
(B)
(C)
(D)
Ans. (a)
9: amount of heat supplied
(A)
(B)
(C)
(D)
Ans. (C)10:molar heat capacity of the gas [R = 8.31 J/mol k]
(A)20.1 J/molK(B)17.14 J/molK
(C)18.14 J/molK(D) 20.14 J/molK
Ans. (b)
Solution 7- 10:7. Work done by the gas,
= area under P-v curve
=
=
=
8.Change in internal energy of a gas is given by
As the gas is monoatomic, ( = 5/3
So
(
9.From 1st law of thermodynamics
=
10.Molar heat capacity is defined as
i.e, C = J/molK Q.11-13.Two moles of Helium gas (( = 5/3) are initially at temperature 27oC and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.11:What are the final volume.
(A)
(B)
(C)
(D)
Ans. (A)
12:What are the final pressure of gas?
(A)
(B)
(C)
(D)
Ans. (a)
13:What is the work done by the gas? (Gas constant R = 8.3 T/mole K)
(A)
(B)
(C)
(D)
Ans. (d)
Solution:11.From ideal gas equation
PV = nRT
initial pressure
EMBED Equation.2
When volume of gas is doubled at constant pressure, its temperature is also doubled. This process is shown on P-V curve by line AB. The gas then cools to temperature T adiabatically. This is shown by curve BC. The whole process is represented by curve ABC.
At point B, pressure Volume , Temperature
Now from adiabatic equation = constant
We have
= 23/2
Final volume
12.final pressure
EMBED Equation.2 13.The work done by gas in isobaric process AB
EMBED Equation.2
The work done by gas during adiabatic process BC
EMBED Equation.2
EMBED Equation.2 .
Net work done by gas
14-15.When 1 gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, the volume increases from 1 cm3 to 1671 c.c. The heat of vaporization at this pressure is 540 cal/gm. Find
14:The work done (in J) in change of phase
(A)
Joule(B)
Joule
(C) 190. 78 Joule(D)
Joule
Ans. (d)
15:Increase in internal energy of water.
(A)2099.33 J(B) 3099.33 J
(C) 4099.33 J(D) 5099.33 J
Ans. (a)
Solution:14.As the process is isobaric
=
= Joule [1 erg = 10-7J]
15.From 1st law of thermodynamics
cal
= 2268 J, [ 1 cal = 4.2J]
so,
=
= 2099.33 J
16:A glass flask of volume one litre at is filled level full of mercury at this temperature. The flask and mercury are now heated to 100oC. How much mercury will spill out if coefficient of volume expansion of mercury is and linear expansion of glass is
respectively?
(A) 14.2 c.c.(B)15.2 c.c.
(C) 18.2 c.c.(D)20.2 c.c.J
Ans. (b)
Solution:In case of thermal expansion of liquid, change in volume of liquid relative to container is given by
Here V = 1 litre = 1000 c.c. = 3 (glass
=
So,
= 15.2 c.c.
17:Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K. The piston A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30K, then the rise in temperature of the gas in B is.
(A) 30K(B) 18K
(C) 50K (D) 42K
Solution: For cylinder A. For cylinder B
dQ = nCPdT1 dQ = nCvdT2 (nCPdT1 = nCvdT2
From (I) and (II)
For diatomic gas
.
18:80 gm of water at is poured on a large block of ice at . The mass of ice that melts is
(A) 30 gm(B) 80 gm
(C) 150 gm(D) 1600 gm
Solution: Since the block of ice at is large, the whole of ice will not melt, hence final temperature is .
(
= heat given up by water in cooling up to
=
= 2400 cal
If m gm be the mass of ice melted, then
= ML =
Here A is correct.
19:A gas at pressure is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be
(A)4Po(B)2Po
(C)Po(D)
Solution:
where m = mass of one gas molecules
n = total no. of gas molecules
i.e, P ( m and P ( Vrms
Here m is halved and Vrms is doubled
(pressure will be doubled
Hence, (B) is correct
20:The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressures P1 and P2 are shown in the figure. Here
(A) P1 < P2(C) P1 = P2
(B) P1 > P2(D) cant be
Solution:For a perfect gas,
(
So, the slope of the graph is
Slope (
Hence P1 > P2
Hence, (C) is correct
21: At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is
(A)
(B)
(C)
(D)
Solution:
EMBED Equation.2
It is molecular weight of hydrogen
.
22: The latent heat of vaporization of water is 2240 J. If the work done in the process of vaporization of 1 gm is 168 J, then increase in internal energy is
(A)2408 J(B)2240 J
(C)2072 J(D)1904 J
Solution: L = 2240 J, m = 1 gm
dW = 168 J
dQ = mL = dU + dW
or
dU = 2072 J
Hence, (C) is correct
23:For a gas, y = 1.286. What is the number of degrees of freedom of the moleculas of this gas ?
(A)3(B)5
(C)6(D)7
Ans. (d)
Solution: (D)
24:Which of the following temperatures is the highest?
(A)100 K(B)13oF
(C) 20oC(D)30oC
Solution: (B ) 13oF is (13+32)o below ice point on F scale.
25:An ideal gas (( = 1.5) is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of molecules 2.0 times
(A)4 times (B)16 times
(C)8 times(D)2 times
Ans. (B)
Solution:
(
Vrms is to reduce two times i.e, temperature of the gas will have to reduce four times or
During adiabatic process
or,
(
Hence, (B) is correct
26:A thin copper wire of length L increases in length by 1% when heated from to . If a thin copper plate of area is heated from to , the percentage increase in its area will be
(A)1%(B)2%
(C)3%(D)4%
Ans. (b)Solution: L = Lo
( = percentage increase in length =
(
Hence
or
Hence, (B) is correct
27: Gas at pressure Po is contained in a vessel. If the masses of all the molecules are doubled and their speed is halved, the resulting pressure P will be equal to
(A)2Po(B)Po/4
(C)Po(D)Po/2
Ans. (d)Solution:Po =
EMBED Equation.3
P =
EMBED Equation.3
where m( = 2m,
putting the value
(
=
P = Po/2
28: The molar heat capacity in a process of a diatomic gas if it does a work of Q/4, when Q amount of heat is supplied to it is
(A)
(B)
(C)
(D)
Ans. (C)Solution: dU = CV dT =
From 1st law of thermodynamics dU = dQ dW or dU =
Now molar heat capacity
Hence (C) is correct
29:For an ideal gas:
(A)the change in internal energy in a constant pressure process
from temperature T1 to T2 is equal to nCv (T2 - T1), where Cv is
the molar specific heat at constant volume and
n the number of moles of the gas.
(B)the change in internal energy of the gas and the work done by
the gas are equal in magnitude in an adiabatic process.
(C)the internal energy does not change in an isothermal process.
(D)no heat is added or removed in an adiabatic process.
(A) A, B(B) A, B, C
(C) A, B, C, D(D) A, CSolution:(C) Change in internal energy depends only on change in temperature since internal energy is a function of state only i.e. dU = nCv,dT.
In adiabatic process, dQ = 0,
Hence, dU + dW = 0 ( dU = ( dW
i.e. magnitude of change in internal energy is equal to magnitude of work done.
30:Heat is supplied to a diatomic gas at constant pressure. The ratio of
is
(A) 5:3:2(B) 5:2:3
(C) 7:5:2(D) 7:2:5
Ans. (C)Solution:
,
,
and
(
Hence, C is correct
SRI BHARADWAJ EDU GENEAPage 183
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A
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V
p
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