Homework: 13, 18, 20, 23, 25, 27 (p. 531-532)

Example problems

Chapter 3: The Kinetic Theory of Gases

9. An automobile tire has a volume of 1.64 x 102 m3 and contains air at a gauge pressure (above atmospheric pressure) of 165 kPa when the temperature is 0.00 0C. What is the gauge pressure of the air in the tires when its temperature rises to 27.0 0C and its volume increases to 1.67 x 102 m3? Assume atmospheric pressure is 1.01x 105 Pa.

nRTpV =

nRTVp iii =

nRTVp fff =

At state i:

At the final state f:

fi

fiif VT

TVpp =⇒

Gauge pressure: - p ofG pp=

13. A sample of an ideal gas is taken through the cyclic process abca shown in the figure below; at point a, T=200 K. (a) How many moles of gas are in the sample? What are (b) the temperature of the gas at point b, (c) the temperature of the gas at point c, and (d) the net energy added to the gas as heat during the cycle?

RTpVnnRTpV =⇒=

(a) Applying the equation of state:

At point a, p=2.5 kN/m2 or 2500 N/m2; V=1 m3.

(mol) 5.120031.8

12500=

××

=n

(b) 5.12===⇒= nRTVp

TVpnRTpV

b

bb

a

aa

At point b, p=7.5 kN/m2 or 7500 N/m2; V=3 m3.

(K) 01805.12

37500=

×==

nRVpT bb

b

(c) see part b; Tc=600 K;

(d) Applying the first law of thermodynamics:

WQE −=∆

For a closed cycle, ∆E=0:

WQ =

W: work done by the system.

))((21

abcb VVppW −−=

(J) 10520.500021 3×=××=W

14. In the temperature range 310 K to 330 K, the pressure p of a certain nonideal gas is related to volume V and temperature T by: How much work is done by the gas if its temperature is raised from 315 K to 330 K while the pressure is held constant?

VTKJ

VTKJp

22 )/00662.0()/9.24( −=

•Work done by the gas is computed by the following formula:

)( iffV

iV VVppdVW −=∫=

)(00662.0)(9.24 22ififif TTTTpVpVW −−−=−=

(J) 310315;330 ≈⇒== WKTKT if

18. The temperature and pressure in the Sun’s atmosphere are 2.00x106 K and 0.0300 Pa. Calculate the rms speed of free electrons (mass 9.11x10-31 kg) there, assuming they are an ideal gas.

(m/s) 105.910023.61011.9

10231.83

33

62331

6

×=×××

×××=

==

−rms

Arms

v

mNRT

MRTv

20. Calculate the rms speed of helium atoms at 1000 K, the molar mass of helium atoms is 4.0026 g/mol.

(m/s) 105.2100026.4100031.833 3

3 ×=×××

== −MRTvrms

24. At 273 K and 1.0 x 10-2 atm, the density of a gas is 1.24 x 10-5 g/cm3. (a) Find vrms for the gas molecules. (b) Find the molar mass of the gas and (c) identify the gas (hint: see Table 19-1).

)1(3MRTvrms =Root-mean-square speed:

)2(nVM

VnM

VM gas ρρ =⇒==

(1) and (2): ρρp

VnRTvrms

33==

3235 kg/m1024.1g/cm1024.1 −− ×=×=ρ

Pa1001.1atm100.1 32 ×=×= −p

m/s494≈rmsv

(a)

)2(nVM ρ

=Equation of state:

)3(nRTpV =

pRT

nVM ρρ

==⇒

g/mol28kg/mol028.0 =≈⇒ M

From Table 19.1, the gas is nitrogen (N2)

(b)

(c)

25. Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 0.000C and (b) 1000C. What is the translational kinetic energy per mole of an ideal gas at (c) 0.000C and (d) 1000C?

kTK23

=

(a) The translational kinetic energy per molecule:

:K 3272730 =+=T

(J) 1065.52731038.123 2123 −− ×=×××=K

(b) see (a): (J) 1072.73731038.123 2123 −− ×=×××=K

(c) The translational kinetic energy per mole:

(J) 104.31002.61065.5 32321 ×=×××= −moleK

Amole NKK ×=

(d) (J) 107.4 3×=moleK

Note: If a sample of gas has n moles (or N molecules), its total translational kinetic energy is:

KNnKnK Amoletotal ××=×=

nRTkTNnKnK Amoletotal 23

23

=××=×=

nRTKtotal 23

=

Homework: 28, 32, 33, 40 (page 532)

28. At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at 1.0 atm pressure and 0.00C? take the diameter of an oxygen molecule to be 3.0 x 10-8 cm.

pdkT

2MFP 2πλ =Mean Free Path:

J/K;1038.1 23−×=k K;273=T Pa;1001.1 5×=p

MFP

airinsound

sound

airinsoundsound λλ

vvf ==Frequency of sound in air:

:m/s343airin sound =v

m1033.9 8MFP

−×=λ

GHz 3.68or (Hz)1068.31033.9

343 98sound ×≈

×=f

m103cm103 108 −− ×=×=d

32. At 200C and 750 torr pressure, the mean free paths for argon gas (Ar) and nitrogen (N2) are λAr=9.9x10-6 cm and λN2=27.5x10-6 cm. (a) Find the ratio of the diameter of an Ar atom to that of an N2 molecule. What is the mean free path of Ar at (b) 200C and 150 torr, and (c) -400C and 750 torr?

pdkT

22πλ =Mean Free Path:

(a) The ratio dAr to dN2:

Ar

N

N

Ar

dd

λλ

2

2

=

(b):

22

22

12

11 2

;2 pd

kTpd

kTπ

λπ

λ ==

12

1

1

22 λλ ××=

pp

TT

33. The speeds of 10 molecules are 2.0, 3.0, 4.0,..., 11 km/s. What are their (a) average speed and (b) rms speed?

(km/s) 5.61065

1011...4321 ==

++++==

∑=

N

vv

N

ii

(a)

(b) ( ) (km/s) 1.71

2

2 ===∑

=

N

vvv

N

ii

avgrms

40. Two containers are at the same temperature. The first contains gas with pressure p1, molecular mass m1, and rms speed vrms1. The second contains gas with pressure 1.5p1, molecular mass m2, and average speed vavg2=2.0vrms1. Find the mass ratio m1/m2.

2

22

8mRTv

π=

1

11

3mRTvrms =

Average speed:

RMS speed:

21 TT =

Homework: 42, 44, 46, 54, 56, 78 (p. 533-535)

42. What is the internal energy of 2.0 mol of an ideal monatomic gas at 273 K?

TnCE V=

1-1- K mol J 5.1223

== RCV

(J)68252735.120.2 =××=E

(kJ)8.6≈E

44. One mole of an ideal diatomic gas goes from a to c along the diagonal path in Fig. 19-25. The scale of the vertical axis is set by pab = 5.0 kPa and pc = 2.0 kPa, and the scale of the horizontal axis is set by Vbc = 4.0 m3 and Va = 2.0 m3. During the transition, (a) what is the change in internal energy of the gas, (b) how much energy is added to the gas as heat? (c) How much heat is required if the gas goes from a to c along the

indirect path abc?

TnRTnCE V ∆=∆=∆25

int

JVPVP aacc 5000)(25

−=−=

a)

b) JWEQ 200070005000int =+−=+∆=

acWEQ +∆= intc)

46. Under constant pressure, the temperature of 3.0 mol of an ideal monatomic gas is raised 15.0 K. What are (a) the work W done by the gas, (b) the energy transferred as heat Q, (c) the change ∆Eint of the gas, and (d) the change ∆K in the average KE per atom?

WQE −=∆ int

(b) (J) 935

25

25

≈=∆××=∆= WTRnTnCQ p

(a) At constant pressure:

(J)3740.1531.80.3 ≈××=∆=∆= TnRVpW

(c) We use the first law of thermodynamics:

)23(or int TnRTnCE V ∆=∆=∆

(J)561374935int =−=∆E(d) For a monatomic gas: TkKkTK ∆=∆⇒=

23

23

avgavg

(J) 101.30.151038.123 2223

avg−− ×≈×××=∆K

54. We know that for an adiabatic process . Evaluate “constant” for an adiabatic process involving exactly 2.0 mol of an ideal gas passing through the state having exactly p=1.5 atm and T=300 K. Assume a diatomic gas whose molecules rotate but do not oscillate.

(Pa) 101.01 atm 1 5×=

constant=γpV

Equation of state: nRTpV =

)(m 033.01001.15.1

30031.80.2 35 ≈

××

××==

pnRTV

Rf

RRf

CRC

CC

V

V

V

p

2

2+

=+

==γ

For a diatomic gas, f=5: 57

=γ

))m((N/m1028.1033.01001.15.1constant 1.432357

5 ××=×××== γpV)m (N1028.1constant 2.23×=

56. Suppose 1.0L of a gas with γ=1.30, initially at 285 K and 1.0 atm, is suddenly compressed adiabatically to half its initial volume. Find its final (a) pressure and (b) temperature. (c) If the gas is then cooled to 273 K at constant pressure, what is its final volume?

;γγfi

VpVp fi = if VV21

=γ

=

f

iif V

Vpp

1−

=

γ

f

iif V

VTT

f

f

f

f

TT

VV

pnRTpV''

constant, =⇒==

78. (a) An ideal gas initially at pressure p0 undergoes a free expansion until its volume is 3.0 times its initial volume. What then is the ratio of its pressure to p0? (b) The gas is next slowly and adiabatically compressed back to its original volume. The pressure after compression is (3.0)1/3p0. Is the gas monatomic, diatomic, or polyatomic? (c) What is the ratio of the average kinetic energy per molecule in this final state to that in the initial state?

(a) 01011100 313; ppVVVpVp =⇒==

(b) γγ0111 ' VpVp =

01

00

111 33

31' pp

VVpp −==

= γγ

γ

ff

f

f

CRC

CC

V

V

V

p 2

2

12

34

311 +

=+

=+

===⇒=−⇒ γγ

polyatomic:6=f

(c) kTKavg 2

3=

0

1''TT

KK

ravg

avg ==

) since(44.13'''0

'1

3/1

0

1

00

'11

0

1 VVpp

VpVp

TTr ======