Chapter 3: The Kinetic Theory of Gases - hcmiu.edu.vn 11/Chapter3_… · Chapter 3: The Kinetic...
Transcript of Chapter 3: The Kinetic Theory of Gases - hcmiu.edu.vn 11/Chapter3_… · Chapter 3: The Kinetic...
Problems: 13, 14, 18, 20, 23, 25, 27 (p. 531-532)
Chapter 3:
The Kinetic Theory of Gases
Solving problems
9. An automobile tire has a volume of 1.64 x 102 m3 and containsair at a gauge pressure (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.
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?
RT
pVnnRTpV
(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 nRT
Vp
T
VpnRTpV
b
bb
a
aa
At point b, p=7.5 kN/m2 or 7500 N/m2;V=3 m3.
(K) 01805.12
37500
nR
VpT 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.
))((2
1abcb VVppW
(J) 10520.50002
1 3W
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 from315 K to 330 K while the pressure is held constant?
V
TKJ
V
TKJp
22 )/00662.0()/9.24(
•Work done by the gas is computed by the following formula:
)( iffV
iVVVppdVW
)(00662.0)(9.24 22
ififif 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
6
2331
6
rms
A
rms
v
mN
RT
M
RTv
20. Calculate the rms speed of helium atoms at 1000 K, the molar mass of helium atoms is 4.0026 g/mol.
(m/s) 105.2100026.4
100031.833 3
3
M
RTvrms
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(3
M
RTvrms
Root-mean-square speed:
)2(n
VM
V
nM
V
M gas
(1) and (2):
p
V
nRTvrms
33
3235 kg/m1024.1g/cm1024.1
Pa1001.1atm100.1 32 p
m/s494rmsv
(a)
)2(n
VM
Equation of state:
)3(nRTpV
p
RT
n
VM
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?
kTK2
3
(a) The translational kinetic energy per molecule:
:K 3272730 T
(J) 1065.52731038.12
3 2123 K
(b) see (a): (J) 1072.73731038.12
3 2123 K
(c) The translational kinetic energy per mole:
(J) 104.31002.61065.5 32321
moleK
Amole NKK
(d)(J) 107.4 3moleK
Note: If a sample of gas has n moles (or N molecules), its total translational kinetic energy is:
KNnKnK Amoletotal
nRTkTNnKnK Amoletotal2
3
2
3
nRTKtotal2
3
Problems: 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 atmpressure and 0.00C? take the diameter of an oxygen molecule to be 3.0 x 10-8 cm.
pd
kT2MFP
2
Mean Free Path:
J/K;1038.1 23k K;273T Pa;1001.1 5p
MFP
airinsound
sound
airinsound
sound
vvf Frequency of sound in air:
:m/s343airin sound v
m1033.9 8
MFP
GHz 3.68or (Hz)1068.31033.9
343 9
8sound
f
m103cm103 108 d
32. At 200C and 750 torr pressure, the mean free paths for argon gas (Ar) and nitrogen (N2) are Ar=9.9x10
-6 cm and N2=27.5x10
-6 cm. (a) Find the ratio of the diameter of an Aratom 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?
pd
kT22
Mean Free Path:
(a) The ratio dAr to dN2:
Ar
N
N
Ar
d
d
2
2
(b):
2
2
22
1
2
11
2;
2 pd
kT
pd
kT
1
2
1
1
22
p
p
T
T
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) 5.610
65
10
11...4321
N
v
v
N
i
i
(a)
(b) (km/s) 1.71
2
2
N
v
vv
N
i
i
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
8
m
RTv
1
11
3
m
RTvrms
Average speed:
RMS speed:
Þm1
m2
=3p
8
v2
vrms1
æ
èç
ö
ø÷
2
= 4.7121 TT
Problems: 42, 44, 46, 54, 56, 78 (p. 533-535)
42. What is the internal energy of 2.0 mol of an ideal monatomic gasat 273 K?
TnCE V
1-1- K mol J 5.122
3 RCV
(J)68252735.120.2 E
(kJ)8.6E
44. One mole of an ideal diatomic gas goes from a to c along thediagonal 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 2
5int
JVPVP aacc 5000)(2
5
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
2
5
2
5 WTRnTnCQ p
(a) At constant pressure:
(J)3740.1531.80.3 TnRVpW
(c) We use the first law of thermodynamics:
)2
3(or int TnRTnCE V
(J)561374935int E
(d) For a monatomic gas: TkKkTK 2
3
2
3avgavg
(J) 101.30.151038.12
3 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
constantpV
Equation of state: nRTpV
)(m 033.01001.15.1
30031.80.2 3
5
p
nRTV
Rf
RRf
C
RC
C
C
V
V
V
p
2
2
For a diatomic gas, f=5:
5
7
))m((N/m1028.1033.01001.15.1constant 1.43235
7
5 pV
)m (N1028.1constant 2.23
56. Suppose 1.0L of a gas with =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?
;
fiVpVp fi if VV
2
1
f
iif
V
Vpp
1
f
iif
V
VTT
f
f
f
f
T
T
V
VpnRTpV
''constant,
78. (a) An ideal gas initially at pressure p0 undergoes a free expansionuntil its volume is 3.0 times its initial volume. What then is the ratio ofits pressure to p0? (b) The gas is next slowly and adiabatically compressed back to its original volume. The pressure after compressionis (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
3
13; ppVVVpVp
(b)
0111 ' VpVp
01
0
0
111 33
3
1' pp
V
Vpp
f
f
f
f
C
RC
C
C
V
V
V
p 2
2
12
3
4
3
11
polyatomic:6f
(c)
kTKavg2
3
0
1''
T
T
K
Kr
avg
avg
) since(44.13'''
0
'
1
3/1
0
1
00
'
11
0
1 VVp
p
Vp
Vp
T
Tr