Latihan Soal Energi Kinetik Gas

Problems 661

Section 21.1 Molecular Model of an Ideal Gas1. In a 30.0-s interval, 500 hailstones strike a glass window of

area 0.600 m2 at an angle of 45.0° to the window surface.Each hailstone has a mass of 5.00 g and moves with aspeed of 8.00 m/s. Assuming the collisions are elastic, findthe average force and pressure on the window.

2. In a period of 1.00 s, 5.00 % 1023 nitrogen molecules strikea wall with an area of 8.00 cm2. If the molecules move witha speed of 300 m/s and strike the wall head-on in elasticcollisions, what is the pressure exerted on the wall? (Themass of one N2 molecule is 4.68 % 10!26 kg.)

3. A sealed cubical container 20.0 cm on a side containsthree times Avogadro’s number of molecules at a tempera-ture of 20.0°C. Find the force exerted by the gas on one ofthe walls of the container.

4. A 2.00-mol sample of oxygen gas is confined to a 5.00-Lvessel at a pressure of 8.00 atm. Find the average transla-tional kinetic energy of an oxygen molecule under theseconditions.A spherical balloon of volume 4 000 cm3 contains heliumat an (inside) pressure of 1.20 % 105 Pa. How many molesof helium are in the balloon if the average kinetic energyof the helium atoms is 3.60 % 10!22 J?

6. Use the definition of Avogadro’s number to find the massof a helium atom.(a) How many atoms of helium gas fill a balloon having adiameter of 30.0 cm at 20.0°C and 1.00 atm? (b) What isthe average kinetic energy of the helium atoms? (c) Whatis the root-mean-square speed of the helium atoms?

8. Given that the rms speed of a helium atom at a certaintemperature is 1 350 m/s, find by proportion the rmsspeed of an oxygen (O2) molecule at this temperature.The molar mass of O2 is 32.0 g/mol, and the molar massof He is 4.00 g/mol.

A cylinder contains a mixture of helium and argongas in equilibrium at 150°C. (a) What is the averagekinetic energy for each type of gas molecule? (b) What isthe root-mean-square speed of each type of molecule?

10. A 5.00-L vessel contains nitrogen gas at 27.0°C and a pres-sure of 3.00 atm. Find (a) the total translational kinetic en-ergy of the gas molecules and (b) the average kinetic en-ergy per molecule.

11. (a) Show that 1 Pa # 1 J/m3. (b) Show that the density inspace of the translational kinetic energy of an ideal gas is3P/2.

Section 21.2 Molar Specific Heat of an Ideal Gas

Note: You may use data in Table 21.2 about particulargases. Here we define a “monatomic ideal gas” to havemolar specific heats CV # 3R/2 and CP # 5R/2, and a“diatomic ideal gas” to have CV # 5R/2 and CP # 7R/2.

9.

7.

5.

12. Calculate the change in internal energy of 3.00 mol ofhelium gas when its temperature is increased by 2.00 K.

A 1.00-mol sample of hydrogen gas is heated at con-stant pressure from 300 K to 420 K. Calculate (a) theenergy transferred to the gas by heat, (b) the increase inits internal energy, and (c) the work done on the gas.

14. A 1.00-mol sample of air (a diatomic ideal gas) at 300 K,confined in a cylinder under a heavy piston, occupies avolume of 5.00 L. Determine the final volume of the gasafter 4.40 kJ of energy is transferred to the air by heat.

15. In a constant-volume process, 209 J of energy is transferredby heat to 1.00 mol of an ideal monatomic gas initially at300 K. Find (a) the increase in internal energy of the gas,(b) the work done on it, and (c) its final temperature.

16. A house has well-insulated walls. It contains a volume of100 m3 of air at 300 K. (a) Calculate the energy requiredto increase the temperature of this diatomic ideal gas by1.00°C. (b) What If? If this energy could be used to lift anobject of mass m through a height of 2.00 m, what is thevalue of m?

17. An incandescent lightbulb contains a volume V of argon atpressure Pi . The bulb is switched on and constant power "is transferred to the argon for a time interval "t. (a) Show that the pressure Pf in the bulb at the end of thisprocess is Pf # Pi[1 $ (""tR)/(PiVCV)]. (b) Find the pres-sure in a spherical light bulb 10.0 cm in diameter 4.00 s af-ter it is switched on, given that it has initial pressure 1.00 atm and that 3.60 W of power is transferred to the gas.

18. A vertical cylinder with a heavy piston contains air at atemperature of 300 K. The initial pressure is 200 kPa, andthe initial volume is 0.350 m3. Take the molar mass of airas 28.9 g/mol and assume that CV # 5R/2. (a) Find thespecific heat of air at constant volume in units of J/kg & )C.(b) Calculate the mass of the air in the cylinder. (c) Sup-pose the piston is held fixed. Find the energy input re-quired to raise the temperature of the air to 700 K. (d) What If? Assume again the conditions of the initialstate and that the heavy piston is free to move. Find theenergy input required to raise the temperature to 700 K.

19. A 1-L Thermos bottle is full of tea at 90°C. You pour outone cup and immediately screw the stopper back on.Make an order-of-magnitude estimate of the change intemperature of the tea remaining in the flask that resultsfrom the admission of air at room temperature. State thequantities you take as data and the values you measure orestimate for them.

20. A 1.00-mol sample of a diatomic ideal gas has pressure Pand volume V. When the gas is heated, its pressure triplesand its volume doubles. This heating process includes twosteps, the first at constant pressure and the second at con-stant volume. Determine the amount of energy trans-ferred to the gas by heat.

13.

1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide

= coached solution with hints available at http://www.pse6.com = computer useful in solving problem

= paired numerical and symbolic problems

P R O B L E M S

662 CHAPTE R 21 • The Kinetic Theory of Gases

21. A 1.00-mol sample of an ideal monatomic gas is at an ini-tial temperature of 300 K. The gas undergoes an isovolu-metric process acquiring 500 J of energy by heat. It thenundergoes an isobaric process losing this same amount ofenergy by heat. Determine (a) the new temperature of thegas and (b) the work done on the gas.

22. A vertical cylinder with a movable piston contains 1.00 molof a diatomic ideal gas. The volume of the gas is Vi, and itstemperature is Ti . Then the cylinder is set on a stove and ad-ditional weights are piled onto the piston as it moves up, insuch a way that the pressure is proportional to the volumeand the final volume is 2Vi . (a) What is the final tempera-ture? (b) How much energy is transferred to the gas by heat?

23. A container has a mixture of two gases: n1 mol of gas 1having molar specific heat C1 and n2 mol of gas 2 of molarspecific heat C2 . (a) Find the molar specific heat of themixture. (b) What If? What is the molar specific heat if themixture has m gases in the amounts n1, n2, n3 , . . . , nm ,with molar specific heats C1, C2 , C3 , . . . , Cm , respectively?

Section 21.3 Adiabatic Processes for an Ideal Gas24. During the compression stroke of a certain gasoline en-

gine, the pressure increases from 1.00 atm to 20.0 atm. Ifthe process is adiabatic and the fuel–air mixture behavesas a diatomic ideal gas, (a) by what factor does the volumechange and (b) by what factor does the temperaturechange? (c) Assuming that the compression starts with0.016 0 mol of gas at 27.0°C, find the values of Q , W, and"Eint that characterize the process.

A 2.00-mol sample of a diatomic ideal gas expands slowlyand adiabatically from a pressure of 5.00 atm and a volumeof 12.0 L to a final volume of 30.0 L. (a) What is the finalpressure of the gas? (b) What are the initial and final tem-peratures? (c) Find Q , W, and "Eint .

26. Air (a diatomic ideal gas) at 27.0°C and atmosphericpressure is drawn into a bicycle pump that has a cylinderwith an inner diameter of 2.50 cm and length 50.0 cm. Thedown stroke adiabatically compresses the air, which reachesa gauge pressure of 800 kPa before entering the tire (Fig.P21.26). Determine (a) the volume of the compressed airand (b) the temperature of the compressed air. (c) What If?The pump is made of steel and has an inner wall that is 2.00 mm thick. Assume that 4.00 cm of the cylinder’s lengthis allowed to come to thermal equilibrium with the air. Whatwill be the increase in wall temperature?

Air in a thundercloud expands as it rises. If its initial tem-perature is 300 K and no energy is lost by thermal conduc-tion on expansion, what is its temperature when the initialvolume has doubled?

28. The largest bottle ever made by blowing glass has a volumeof about 0.720 m3. Imagine that this bottle is filled with airthat behaves as an ideal diatomic gas. The bottle is heldwith its opening at the bottom and rapidly submerged intothe ocean. No air escapes or mixes with the water. No en-ergy is exchanged with the ocean by heat. (a) If the finalvolume of the air is 0.240 m3, by what factor does the inter-nal energy of the air increase? (b) If the bottle is sub-merged so that the air temperature doubles, how muchvolume is occupied by air?

27.

25.Figure P21.26

Geor

ge S

empl

e

29. A 4.00-L sample of a diatomic ideal gas with specific heatratio 1.40, confined to a cylinder, is carried through aclosed cycle. The gas is initially at 1.00 atm and at 300 K.First, its pressure is tripled under constant volume.Then, it expands adiabatically to its original pressure.Finally, the gas is compressed isobarically to its originalvolume. (a) Draw a PV diagram of this cycle. (b) Deter-mine the volume of the gas at the end of the adiabaticexpansion. (c) Find the temperature of the gas at thestart of the adiabatic expansion. (d) Find the tempera-ture at the end of the cycle. (e) What was the net workdone on the gas for this cycle?

30. A diatomic ideal gas (( # 1.40) confined to a cylinder isput through a closed cycle. Initially the gas is at Pi , Vi , andTi . First, its pressure is tripled under constant volume. Itthen expands adiabatically to its original pressure and fi-nally is compressed isobarically to its original volume.(a) Draw a PV diagram of this cycle. (b) Determine the vol-ume at the end of the adiabatic expansion. Find (c) thetemperature of the gas at the start of the adiabatic expan-sion and (d) the temperature at the end of the cycle.(e) What was the net work done on the gas for this cycle?

31. How much work is required to compress 5.00 mol of air at20.0°C and 1.00 atm to one tenth of the original volume(a) by an isothermal process? (b) by an adiabatic process?(c) What is the final pressure in each of these two cases?

Chapter 21 603

P21.2 F =× ×

=−5 00 10 2 4 68 10 300

1 0014 0

23 26. .

..

e j e jb g kg m s

s N

and PFA

= =×

=−14 0

1017 64

..

N8.00 m

kPa2 .

P21.3 We first find the pressure exerted by the gas on the wall of the container.

PNkT

VN k T

VRTV

= = = =⋅ ⋅

×= ×−

3 3 3 8 314 293

8 00 109 13 103

5A B3

N m mol K K

m Pa

.

..

b ga f

Thus, the force on one of the walls of the cubical container is

F PA= = × × = ×−9 13 10 4 00 10 3 65 105 2 4. . . Pa m N2e je j .

P21.4 Use Equation 21.2, PNV

mv=FHGIKJ

23 2

2

, so that

Kmv PV

NN nN N

KPVN

K

av A A

avA

3

av

where

atm Pa atm m

mol molecules mol

J molecule

= = = =

= =× ×

×

= ×

−

−

2

5 3

23

21

232

2

32 2

3 8 00 1 013 10 5 00 10

2 2 6 02 10

5 05 10

b ga fe je ja fe j

. . .

.

.

P21.5 PNV

KE=23d i Equation 21.2

NPV

KE

nN

N

= =× ×

×= ×

= =×

×=

−

−

32

32

1 20 10 4 00 10

3 60 102 00 10

2 00 106 02 10

3 32

5 3

2224

24

23

d ie je je j

. .

..

..

.

molecules

molecules molecules mol

molA

P21.6 One mole of helium contains Avogadro’s number of molecules and has a mass of 4.00 g. Let us call mthe mass of one atom, and we have

N mA g mol= 4 00.

or m =×

= × −4 006 02 10

6 64 102324.

..

g mol molecules mol

g molecule

m = × −6 64 10 27. kg

P21.7 (a) PV Nk TB= : NPVk TB

= =×

×= ×

−

1 013 10 0 150

1 38 10 2933 54 10

5 43

3

2323

. .

..

Pa m

J K K atoms

π a fe ja f

(b) K k TB= = × = ×− −32

32

1 38 10 293 6 07 1023 21. .e ja f J J

(c) For helium, the atomic mass is m =×

= × −4 006 02 10

6 64 102324.

..

g mol molecules mol

g molecule

m = × −6 64 10 27. kg molecule

12

32

2mv k TB= : ∴ = =vk Tm

Brms km s

31 35.

604 The Kinetic Theory of Gases

P21.8 vk Tm

B=3

vv

MM

v

O

He

He

O

O m s

m s

= = =

= =

4 0032 0

18 00

1 350

8 00477

.. .

.

P21.9 (a) K k TB= = × = ×− −32

32

1 38 10 423 8 76 1023 21. . J K K Je ja f

(b) K mvrms= = × −12

8 76 102 21. J

so vmrms =× −1 75 10 20. J

(1)

For helium, m =×

= × −4 006 02 10

6 64 102324.

..

g mol molecules mol

g molecule

m = × −6 64 10 27. kg molecule

Similarly for argon, m =×

= × −39 96 02 10

6 63 102323.

..

g mol molecules mol

g molecule

m = × −6 63 10 26. kg moleculeSubstituting in (1) above,

we find for helium, vrms = 1 62. km s

and for argon, vrms = 514 m s

P21.10 (a) PV nRTNmv

= =2

3

The total translational kinetic energy is Nmv

E2

2= trans :

E PVtrans kJ= = × × × =−32

32

3 00 1 013 10 5 00 10 2 285 3. . . .e je j

(b)mv k T RT

NB

2

2321

23

232

3 8 314 300

2 6 02 106 21 10= = =

×= × −

A J

.

..

a fa fe j

P21.11 (a) 1 11

11

1 Pa Pa N m

Pa J

1 N m J m

23=

FHG

IKJ ⋅FHG

IKJ =a f

(b) For a monatomic ideal gas, E nRTint =32

For any ideal gas, the energy of molecular translation is the same,

E nRT PVtrans = =32

32

.

Thus, the energy per volume is E

VPtrans =

32

.

Chapter 21 605

Section 21.2 Molar Specific Heat of an Ideal Gas

P21.12 E nRTint =32

∆ ∆E nR Tint mol J mol K K J= = ⋅ =32

32

3 00 8 314 2 00 74 8. . . .a fb ga f

P21.13 We us the tabulated values for CP and CV

(a) Q nC TP= = ⋅ − =∆ 1 00 420 300 3 46. . mol 28.8 J mol K K kJb ga f

(b) ∆ ∆E nC TVint mol 20.4 J mol K K kJ= = ⋅ =1 00 120 2 45. .b ga f

(c) W Q E= − + = − + = −∆ int kJ kJ kJ3 46 2 45 1 01. . .

P21.14 The piston moves to keep pressure constant. Since VnRT

P= , then

∆∆

VnR T

P= for a constant pressure process.

Q nC T n C R TP V= = +∆ ∆b g so ∆TQ

n C RQ

n R RQnRV

=+

=+

=b g b g5 227

and ∆VnRP

QnR

QP

QVnRT

= FHGIKJ = =

27

27

27

∆V =×

⋅=

27

4 40 10 5 00

1 00 8 314 3002 52

3. .

. ..

J L

mol J mol K K L

e ja fa fb ga f

Thus, V V Vf i= + = + =∆ 5 00 2 52 7 52. . . L L L

P21.15 n = 1 00. mol, Ti = 300 K

(b) Since V = constant, W = 0

(a) ∆E Q Wint J J= + = + =209 0 209

(c) ∆ ∆ ∆E nC T n R TVint = = FHGIKJ

32

so ∆∆

TEnR

= =⋅

=2

32 209

3 1 00 8 31416 8int J

mol J mol K K

a fa fb g. .

.

T T Ti= + = + =∆ 300 16 8 317 K K K.

606 The Kinetic Theory of Gases

P21.16 (a) Consider heating it at constant pressure. Oxygen and nitrogen are diatomic, so CR

P =72

Q nC T nR TPVT

T

Q

P= = = FHGIKJ

=×

=

∆ ∆ ∆72

72

72

1 013 10 100

3001 00 118

5..

N m m

K K kJ

2 3e je j a f

(b) U mgyg =

mU

gyg

= =×

= ×1 18 10

2 006 03 10

53.

..

J

9.80 m s m kg

2e j*P21.17 (a) We assume that the bulb does not expand. Then this is a constant-volume heating process.

The quantity of the gas is nPVRT

i

i= . The energy input is Q t nC TV= =P ∆ ∆ so

∆∆ ∆

Tt

nCtRT

PVCV

i

i V= =P P

.

The final temperature is T T T TtR

PVCf i ii V

= + = +FHG

IKJ∆

∆1

P.

The final pressure is P PT

TP

tRPVCf i

f

ii

i V= = +

FHG

IKJ1

P∆.

(b) Pf = +⋅ ⋅

⋅ ⋅ ×

FHG

IKJ=1 1

3 60

12 51 183 atm

J 4 s 8.314 J m 3 mol K

s mol K 1.013 10 N 4 0.05 m J atm

2

5

.

..

π a f

P21.18 (a) C RV = = ⋅FHG

IKJ = ⋅ = ⋅

52

52

8 3141 00

719 0 719..

. J mol K mol

0.028 9 kg J kg K kJ kg Kb g

(b) m Mn MPVRT

= = FHGIKJ

m =×

⋅

FHGG

IKJJ =0 028 9

200 10

8 314 3000 811

3

..

. kg mol Pa 0.350 m

J mol K K kg

3

b g e jb ga f

(c) We consider a constant volume process where no work is done.

Q mC TV= = ⋅ − =∆ 0 811 700 300 233. kg 0.719 kJ kg K K K kJb ga f

(d) We now consider a constant pressure process where the internal energy of the gas isincreased and work is done.

Q mC T m C R T mR

T mC

T

Q

P VV= = + = FHG

IKJ = FHG

IKJ

= ⋅LNM

OQP =

∆ ∆ ∆ ∆b g

b g a f

72

75

0 811 0 719 400 327. . kg75

kJ kg K K kJ

Chapter 21 607

P21.19 Consider 800 cm3 of (flavored) water at 90.0 °C mixing with 200 cm3 of diatomic ideal gas at 20.0°C:Q Qcold hot= −

or m c T T m c TP f i w w wair , air air− = −,d i a f∆

∆Tm c T T

m c

V c

V cwP f i

w w

P

w w wa f d i b g a f

b g=− −

=− ° − °air , air air air , air C C, . .ρ

ρ90 0 20 0

where we have anticipated that the final temperature of the mixture will be close to 90.0°C.

The molar specific heat of air is C RP , air =72

So the specific heat per gram is cRMP, air J mol K

mol28.9 g

J g C= FHGIKJ = ⋅

FHG

IKJ = ⋅°

72

72

8 3141 00

1 01..

.b g

∆T wa f e je j b ga fe je j b g

= −× ⋅° °

⋅°

−1 20 10 200 1 01 70 0

1 00 800 4 186

3. . .

. .

g cm cm J g C C

g cm cm J kg C

3 3

3 3

or ∆T wa f ≈ − × °−5 05 10 3. C

The change of temperature for the water is between C and C10 103 2− −° ° .

P21.20 Q nC T nC TP V= +∆ ∆b g b gisobaric isovolumetric

In the isobaric process, V doubles so T must double, to 2Ti .

In the isovolumetric process, P triples so T changes from 2Ti to 6Ti .

Q n R T T n R T T nRT PVi i i i i= FHGIKJ − + FHG

IKJ − = =

72

252

6 2 13 5 13 5b g b g . .

P21.21 In the isovolumetric process A B→ , W = 0 and Q nC TV= =∆ 500 J

50032

2 5003

3002 500

3 1 00 8 314340

J or J

K J

mol J mol K K

= FHGIKJ − = +

= +⋅

=

nR

T T T TnR

T

B A B A

B

b g a f

a fa fb g. .

In the isobaric process B C→ ,

Q nC TnR

T TP C B= = − = −∆5

2500b g J .

Thus,

(a) T TnRC B= − = −

⋅=

2 5005

3401 000

8 314316

J K

J5 1.00 mol J mol K

Ka f

a fb g.

(b) The work done on the gas during the isobaric process is

W P V nR T TBC B C B= − = − − = − ⋅ −∆ b g a fb ga f1 00 8 314 316 340. . mol J mol K K J

or WBC = +200 J

The work done on the gas in the isovolumetric process is zero, so in total

Won gas J= +200 .

608 The Kinetic Theory of Gases

*P21.22 (a) At any point in the heating process, P kVi i= and P kVPV

VnRTV

Vi

i

i

i

= = = 2 . At the end,

PnRTV

V Pfi

ii i= =2 2 2 and T

P V

nRP VnR

Tff f i i

i= = =2 2

4 .

(b) The work input is W PdVnRTV

VdVnRTV

V nRTV

V V nRTi

fi

iV

Vi

i V

Vi

ii i i

i

i

i

i

= − = − = − = − − = −z z 2

2

2

2 2

22 2

2 24

32e j .

The change in internal energy, is ∆ ∆E nC T n R T T nRTV i i iint = = − = +52

4152

b g . The heat input

is Q E W nRT RTi i= − = =∆ int mol182

9 1a f .

P21.23 (a) The heat required to produce a temperature change isQ n C T n C T= +1 1 2 2∆ ∆

The number of molecules is N N1 2+ , so the number of “moles of the mixture” is n n1 2+ andQ n n C T= +1 2b g ∆ ,

so Cn C n C

n n=

++

1 1 2 2

1 2.

(b) Q n C T n C Ti ii

m

ii

m= =

FHGIKJ= =

∑ ∑∆ ∆1 1

Cn C

n

i ii

m

ii

m= =

=

∑

∑1

1

Section 21.3 Adiabatic Processes for an Ideal Gas

P21.24 (a) PV P Vi i f fγ γ= so

V

VPP

f

i

i

f=FHGIKJ = FHG

IKJ =

1 5 71 0020 0

0 118γ

..

.

(b)T

T

P V

PV

P

P

V

Vf

i

f f

i i

f

i

f

i= =

FHGIKJFHGIKJ = 20 0 0 118. .a fa f T

Tf

i= 2 35.

(c) Since the process is adiabatic, Q = 0

Since γ = = =+

1 40.CC

R CC

P

V

V

V, C RV =

52

and ∆T T T Ti i i= − =2 35 1 35. .

∆ ∆E nC TVint mol J mol K K J= = FHGIKJ ⋅ =0 016 0

52

8 314 1 35 300 135. . .b g b g a f

and W Q E= − + = + = +∆ int J J0 135 135 .

Chapter 21 609

P21.25 (a) PV P Vi i f fγ γ=

P PVVf i

i

f=FHGIKJ = F

HGIKJ =

γ

5 0012 030 0

1 391.40

...

. atm atm

(b) TPVnRii i= =

× ×

⋅=

−5 00 1 013 10 12 0 10

2 00365

5 3. . .

.

Pa m

mol 8.314 J mol K K

3e je jb g

TP V

nRff f= =

× ×

⋅=

−1 39 1 013 10 30 0 10

2 00253

5 3. . .

.

Pa m

mol 8.314 J mol K K

3e je jb g

(c) The process is adiabatic: Q = 0

γ = = =+

1 40.CC

R CC

P

V

V

V, C RV =

52

∆ ∆

∆

E nC T

W E Q

Vint

int

mol52

J mol K K K kJ

kJ kJ

= = ⋅FHG

IKJ − = −

= − = − − = −

2 00 8 314 253 365 4 66

4 66 0 4 66

. . .

. .

b g a f

P21.26 Vi =×F

HGIKJ = ×

−−π

2 50 100 500 2 45 10

2 24.

. . m

2 m m3

The quantity of air we find from PV nRTi i i=

nPVRT

n

i i

i= =

× ×

⋅

= ×

−

−

1 013 10 2 45 10

8 314 300

9 97 10

5 4

3

. .

.

.

Pa m

J mol K K

mol

3e je jb ga f

Adiabatic compression: Pf = + =101 3 800 901 3. . kPa kPa kPa

(a) PV P Vi i f fγ γ=

V VPP

V

f ii

f

f

=FHGIKJ = × F

HGIKJ

= ×

−

−

1

45 7

5

2 45 10101 3901 3

5 15 10

γ

...

.

m

m

3

3

(b) P V nRTf f f=

T TP V

PVT

P

PPP

TPP

T

f if f

i ii

f

i

i

fi

i

f

f

= =FHGIKJ =

FHGIKJ

= FHGIKJ =

−

−

1 1 1

5 7 1

300101 3901 3

560

γ γb g

b g K K

.

.

(c) The work put into the gas in compressing it is ∆ ∆E nC TVint =

W

W

= × ⋅ −

=

−9 97 1052

8 314 560 300

53 9

3. .

.

mol J mol K K

J

e j b ga f

continued on next page

610 The Kinetic Theory of Gases

Now imagine this energy being shared with the inner wall as the gas is held at constantvolume. The pump wall has outer diameter 25 0 2 00 2 00 29 0. . . . mm mm mm mm+ + = , andvolume

π π14 5 10 12 5 10 4 00 10 6 79 103 2 3 2 2 6. . . .× − ×LNM

OQP × = ×− − − − m m m m3e j e j

and mass ρV = × × =−7 86 10 6 79 10 53 33 6. . . kg m m g3 3e je jThe overall warming process is described by

53 9

53 9 9 97 1052

8 314 300

53 3 10 448 300

53 9 0 207 23 9 300

300 2 24

3

3

.

. . .

.

. . .

.

J

J mol J mol K K

kg J kg K K

J J K J K K

K K

= +

= × ⋅ −

+ × ⋅ −

= + −

− =

−

−

nC T mc T

T

T

T

T

V

ff

ff

ff

ff

∆ ∆

e j b gd ie jb gd ib gd i

P21.27T

TVV

f

i

i

f=FHGIKJ = FHG

IKJ

−γ 1 0 40012

.

If Ti = 300 K , then Tf = 227 K .

*P21.28 (a) In PV P Vi i f fγ γ= we have P P

VVf i

i

f=FHGIKJγ

P P Pf i i=FHG

IKJ =

0 7200 240

4 661.40

.

..

m m

3

3

Then PVT

P V

Ti i

i

f f

f= T T

P V

PVTf i

f f

i ii= = =4 66

13

1 55. .a fThe factor of increase in temperature is the same as the factor of increase in internal energy,

according to E nC TVint = . Then E

Ef

i

int,

int, = 1 55. .

(b) In T

T

P V

PVVV

V

VVV

f

i

f f

i i

i

f

f

i

i

f= =

FHGIKJ =

FHGIKJ

−γ γ 1

we have

20 720

0 7202 2 5 66

0 7205 66

0 127

0 40

1 0 4 2 5

=FHG

IKJ

= = =

= =

.

..

..

.

.

. .

m

m

m m

3

3

33

V

V

V

f

f

f

Chapter 21 611

P21.29 (a) See the diagram at the right.

(b) P V P VB B C Cγ γ=

3

3 3 2 19

2 19 4 00 8 77

1 5 7

PV PV

V V V V

V

i i i C

C i i i

C

γ γ

γ

=

= = =

= =

e j e ja f

.

. . . L L

(c) P V nRT PV nRTB B B i i i= = =3 3

T TB i= = =3 3 300 900 K Ka f

(d) After one whole cycle, T TA i= = 300 K .

B

A C

V C V i = 4 L

P i

P i 3

P

V (L)

Adiabatic

FIG. P21.29

(e) In AB, Q nC V n R T T nRTAB V i i i= = FHGIKJ − =∆

52

3 5 00b g a f.

QBC = 0 as this process is adiabatic

P V nRT P V nRTC C C i i i= = =2 19 2 19. .b g a fso T TC i= 2 19.

Q nC T n R T T nRTCA P i i i= = FHGIKJ − = −∆

72

2 19 4 17. .b g a fFor the whole cycle,

Q Q Q Q nRT nRT

E Q W

W Q nRT PV

W

ABCA AB BC CA i i

ABCA ABCA ABCA

ABCA ABCA i i i

ABCA

= + + = − =

= = +

= − = − = −

= − × × = −−

5 00 4 17 0 829

0

0 829 0 829

0 829 1 013 10 4 00 10 3365 3

. . .

. .

. . .

a f a fb g

a f a fa fe je j

∆ int

3 Pa m J

P21.30 (a) See the diagram at the right.

(b) P V P VB B C Cγ γ=

3

3 3 2 191 5 7

PV PV

V V V Vi i i C

C i i i

γ γ

γ

=

= = = .

(c) P V nRT PV nRTB B B i i i= = =3 3

T TB i= 3

(d) After one whole cycle, T TA i=

P

B

Pi

Adiabatic

A C

Vi VC

3Pi

V La f

FIG. P21.30

continued on next page

612 The Kinetic Theory of Gases

(e) In AB, Q nC T n R T T nRTAB V i i i= = FHGIKJ − =∆

52

3 5 00b g a f.

Q

P V nRT P V nRT T T

Q nC T n R T T nRT

BC

C C C i i i C i

CA P i i i

=

= = = =

= = FHGIKJ − = −

0

2 19 2 19 2 19

72

2 19 4 17

as this process is abiabatic

so . . .

. .

b gb g∆

For the whole cycle,

Q Q Q Q nRT nRT

E Q W

W Q nRT PV

ABCA AB BC CA i i

ABCA ABCA ABCA

ABCA ABCA i i i

= + + = − =

= = +

= − = − = −

5 00 4 17 0 830

0

0 830 0 830

. . .

. .

a fb g∆ int

P21.31 (a) The work done on the gas is

W PdVabV

V

a

b

= − z .

For the isothermal process,

W nRTV

dV

W nRTVV

nRTVV

ab aV

V

ab ab

a

a

b

a

b

′

′′

′

= − FHGIKJ

= −FHGIKJ =

FHGIKJ

′z 1

ln ln .

Thus, Wab ′ = ⋅5 00 293 10 0. ln . mol 8.314 J mol K Kb ga f a fWab ′ = 28 0. kJ .

FIG. P21.31

(b) For the adiabatic process, we must first find the final temperature, Tb . Since air consistsprimarily of diatomic molecules, we shall use

γ air = 1 40. and CR

V , air J mol K= = = ⋅52

5 8 3142

20 8.

.a f

.

Then, for the adiabatic preocess

T TVVb a

a

b=FHGIKJ = =

−γ 10 400293 10 0 736 K K. .a f .

Thus, the work done on the gas during the adiabatic process is

W Q E nC T nC T Tab ab V ab V b a− + = − + = −∆ ∆intb g b g b g0

or Wab = ⋅ − =5 00 736 293 46 0. . mol 20.8 J mol K K kJb ga f .

continued on next page

Chapter 21 613

(c) For the isothermal process, we have

P V P Vb b a a′ ′ = .

Thus, P PVVb a

a

b′

′=FHGIKJ = =1 00 10 0. . atm 10.0 atma f .

For the adiabatic process, we have P V P Vb b a aγ γ= .

Thus, P PVVb a

a

b=FHGIKJ = =γ

1 00 25 11.40. . atm 10.0 atma f .

P21.32 We suppose the air plus burnt gasoline behaves like a diatomicideal gas. We find its final absolute pressure:

21 0 400

21 0 1 14

7 5 7 5

7 5

.

. .

atm 50.0 cm cm

atm18

atm

3 3e j e j=

= FHGIKJ =

P

P

f

f

Now Q = 0

and W E nC T TV f i= = −∆ int d i

∴ = − = −W nRT nRT P V PVf i f f i i52

52

52d i

FIG. P21.32

W

W

= −×F

HGIKJ

= −

−52

1 14 21 01

10

150

6. . atm 400 cm atm 50.0 cm1.013 10 N m

atm m cm

J

3 35 2

3 3e j e j e j

The output work is − = +W 150 J

The time for this stroke is 14

1 606 00 10 3 min

2 500 s

1 min s

FHGIKJFHGIKJ = × −.

P =−

=×

=−Wt∆

15025 03

J6.00 10 s

kW.