P21.7
3
5
4
PV 1.013  10 Pa  3   0.150 m  
N 

 3.54  1023 atom s
23
kBT
1.38  10
J K  293 K 
(a)
PV  N kBT :
(b)
K
(c)
For helium, the atomic mass is
m 




3
3
kBT  1.38  1023  293 J 6.07  1021 J
2
2
4.00 g m ol
6.02  1023 m olecules m ol
 6.64  1024 g m olecule
m  6.64  1027 kg m olecule
1 2 3
m v  kBT :
2
2
P21.10
(a)
PV  nRT 
 vrm s 
N m v2
3
The total translational kinetic energy is
Etrans 
(b)
P21.14
3kBT
 1.35 km s
m


N m v2
 Etrans :
2

3
3
PV  3.00  1.013  105 5.00  103  2.28 kJ
2
2
m v2 3kB T 3RT 3 8.314 300



 6.21 1021 J
2
2
2N A
2 6.02  1023


The piston moves to keep pressure constant. Since V 
V 
nR T
for a constant pressure process.
P
Q  nC P T  nCV  R  T so T 
and
nRT
, then
P
V 
nR  2Q  2Q 2 Q V



P  7nR  7P 7 nRT
Q
n  CV  R 

Q
2Q

n  5R 2  R  7nR
V 


4.40  103 J  5.00 L 
2
 2.52 L
7  1.00 m ol  8.314 J m ol K   300 K 
Thus, V f  Vi  V  5.00 L  2.52 L  7.52 L
P21.20
Q   nC P T isobaric   nCV T isovolum etric
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.
7 
5 
Q  n  R   2Ti  Ti  n  R   6Ti  2Ti  13.5nRTi  13.5PV
2 
2 
P21.21
In the isovolumetric process A  B , W  0 and Q  nCV T  500 J
2 500 J
 3R 
500 J n 
T  TA  orTB  TA  3nR
 2  B
2 500 J
TB  300 K 
 340 K
3 1.00 m ol  8.314 J m ol K 
In the isobaric process B  C ,
Q  nC P T 
5nR
TC  TB   500 J.
2
Thus,
2 500 J
1000 J
 340 K 
 316 K
5nR
5 1.00 m ol  8.314 J m ol K 
(a)
TC  TB 
(b)
The work done on the gas during the isobaric process is
W BC  PB V  nR TC  TB     1.00 m ol  8.314 J m ol K   316 K  340 J
or W BC  200 J
The work done on the gas in the isovolumetric process is zero, so in total
W on gas  200 J .
P21.29
(a)
(b)
See the diagram at the right.

P
B
3 Pi

PBVB  PC VC
Ad iabatic


3PV
i i  PV
i C
 
 
VC  31  Vi  35 7 Vi  2.19Vi
Pi
VC  2.19 4.00 L   8.77 L
C
A
Vi = 4 L
(c)
PBVB  nRTB  3PV
i i  3nRTi
FIG. P21.29
TB  3Ti  3 300 K   900 K
(d)
After one whole cycle, TA  Ti  300 K .
(e)
5 
In AB, Q AB  nCV V  n R   3Ti  Ti   5.00 nRTi
2 
Q BC  0 as this process is adiabatic
PC VC  nRTC  Pi 2.19Vi   2.19 nRTi
so
TC  2.19Ti
7 
Q CA  nC P T  n  R  Ti  2.19Ti   4.17 nRTi
2 
For the whole cycle,
Q A BCA  Q A B  Q BC  Q CA   5.00  4.17 nRTi   0.829 nRTi
 Eint A BCA
 0  Q A BCA  W A BCA
W A BCA  Q A BCA    0.829 nRTi    0.829 PV
i i


W A BCA    0.829 1.013  105 Pa 4.00  103 m
3

336 J
VC
V (L)
P21.31
(a)
The work done on the gas is
Vb
W ab    PdV .
Va
For the isothermal process,
V
b
 1
W ab   nRTa    dV
V 
V
a
V 
V 
W ab   nRTa ln  b   nRT ln  a  .
 Va 
 Vb 
FIG. P21.31
Thus,
W ab  5.00 m ol 8.314 J m ol K   293 K  ln 10.0
W ab  28.0 kJ .
(b)
For the adiabatic process, we must first find the final temperature, Tb . Since air
consists primarily of diatomic molecules, we shall use
 air  1.40 and CV ,air 
5R 5 8.314

 20.8 J m ol K .
2
2
Then, for the adiabatic preocess
V 
Tb  Ta  a 
V 
 1
 293 K  10.0
0.400
 736 K .
b
Thus, the work done on the gas during the adiabatic process is
W ab  Q  Eint ab   0  nCV T  ab  nCV Tb  Ta 
or W ab  5.00 m ol 20.8 J m ol K   736  293 K  46.0 kJ .
(c)
For the isothermal process, we have
PbVb  PaV a .
V 
Thus, Pb  Pa  a   1.00 atm  10.0  10.0 atm
 Vb 
.
For the adiabatic process, we have PbVb  PaV a .

V 
1.40
Thus, Pb  Pa  a   1.00 atm 10.0
 25.1 atm .
V 
b
P21.33
The heat capacity at constant volume is nCV . An ideal gas of diatomic molecules has
three degrees of freedom for translation in the x, y, and z directions. If we take the y axis along
the axis of a molecule, then outside forces cannot excite rotation about this axis, since they have
no lever arms. Collisions will set the molecule spinning only about the x and z axes.
(a)
If the molecules do not vibrate, they have five degrees of freedom. Random collisions
put equal amounts of energy
of one molecule is
1
kBT into all five kinds of motion. The average energy
2
5
kBT . The internal energy of the two-mole sample is
2
5

5

5 
N  kBT   nN A  kBT   n  R  T  nCV T .
2

2

2 
The molar heat capacity is CV 
5
R and the sample’s heat capacity is
2
5 
5

nC V  n  R   2 m ol  8.314 J m ol K 
2 
2

nC V  41.6 J K
For the heat capacity at constant pressure we have
5
 7
7

nC P  n  C V  R   n  R  R   nR  2 m ol  8.314 J m ol K 
2
 2
2

nC P  58.2 J K
(b)
In vibration with the center of mass fixed, both atoms are always moving in opposite
directions with equal speeds. Vibration adds two more degrees of freedom for two
more terms in the molecular energy, for kinetic and for elastic potential energy. We
have
7 
nCV  n  R   58.2 J K
2 
and
P21.36
(a)
9 
nC P  n  R   74.8 J K
2 
The ratio of the number at higher energy to the number at lower energy is eE kBT
where E is the energy difference. Here,
 1.60  1019
E   10.2 eV  
1 eV

J
18
  1.63  10 J

and at 0°C,


kBT  1.38  1023 J K  273 K   3.77  1021 J.
Since this is much less than the excitation energy, nearly all the atoms will be in the
ground state and the number excited is
18

 2.70  10  exp  3.1.77631010
25
21
J
25 433
  2.70  10 e .
J


This number is much less than one, so almost all of the time
(b)
no atom is excited .
At 10 000°C,


kBT  1.38  1023 J K 10 273 K  1.42  1019 J.
The number excited is

18
 2.70  10  exp  1.1.42631010
25
19
J
25 11.5
 2.70  1020 .
  2.70  10 e
J


P21.37
(a)
vav 
 nivi 
(b)
v 

N
2
av
 nivi2  54.9 m 2
N
so vrm s 
P21.45
1
1 2  2 3  3 5  4 7  3 9  212   6.80 m s
15 
v 
2
av
s2
 54.9  7.41 m s
(c)
vm p  7.00 m s
(a)
 N 
PV N A
so that
PV  
RT and N 

RT
NA 
1.00  10  1331.00 6.02 10  
N 
10
23
 8.314 300
(b)

1
nV  d 2
2 12

V
Nd 2
2 12

 3.21 10
12
3.21 1012 m olecules
1.00 m
3
 
m olecules  3.00  1010 m

2
 21 2
 779 km
P21.51
v
 6.42  104 s1
(c)
f
(a)
Pf  100 kPa
Vf 
nRT f
Pf

Tf  400 K
2.00 m ol 8.314 J m ol K   400 K 
100  103 Pa
 0.066 5 m
3
 66.5 L
Eint   3.50 nR T  3.50 2.00 m ol 8.314 J m ol K  100 K   5.82 kJ
W  PV  nRT    2.00 m ol  8.314 J m ol K  100 K   1.66 kJ
Q  Eint  W  5.82 kJ 1.66 kJ 7.48 kJ
(b)
Tf  400 K
V f  Vi 
(c)
nRTi 2.00 m ol 8.314 J m ol K   300 K 

 0.049 9 m
Pi
100  103 Pa
3
 49.9 L
 Tf 
 400 K 
Pf  Pi   100 kPa
 133 kPa
 300 K 
 Ti 
W   PdV  0
Eint  5.82 kJ as in part (a)
Q  Eint  W  5.82 kJ 0  5.82 kJ
Pf  120 kPa
Tf  300 K
P 
 100 kPa
V f  V i i   49.9 L 
 41.6 L
 120 kPa
P
 f
Eint   3.50 nRT  0
since V  constant
T  constant
Vf
W    PdV   nRTi 
Vi
P 
 V f
dV
  nRTiln     nRTiln  i 
V
 Vi 
 Pf 
 100 kPa
W    2.00 m ol  8.314 J m ol K   300 K  ln 
 909 J
 120 kPa
Q  Eint  W  0  910 J 909 J
(d)
Pf  120 kPa


PfV f  PV
i i : so

C P CV  R 3.50R  R 4.50 9




CV
CV
3.50R
3.50 7
P 
V f  V i i 
 Pf 
1
 100 kPa
 49.9 L 
 120 kPa
79
 43.3 L
since
 PfV f 
 120 kPa  43.3 L 
T f  Ti
 300 K 
 312 K

 100 kPa  49.9 L 
 PV
i i
Eint   3.50 nR T  3.50 2.00 m ol  8.314 J m ol K   12.4 K   722 J
Q  0
 adiabatic process
W  Q  Eint  0  722 J 722 J
*P21.54
(a)


W  nCV T f  Ti

3
2 500 J 1 m ol 8.314 J m ol K Tf  500 K
2

Tf  300 K
(b)


PV
i i  PfV f

 nRT f 
 nRTi
Pi
 Pf 


 Pi 
 Pf 
   1
  1
Ti   T f

Pi
Pf
 Tf 
Pf  Pi 
 Ti 
P21.67
 5 3 3 2


Ti Pi1  T f Pf1
 Tf 
Pf  Pi 
T 
   1
i
 300
 3.60 atm 
 500
52

(a)
1.013  105 Pa 5.00  103 m
PV
n

RT
 8.314 J m ol K   300 K 
(b)
P 
 3.00
TB  TA  B   300 K 
 900 K
 1.00
 PA 
 1.00 atm
3

0.203 m ol
TC  TB  900 K
T 
 900
VC  V A  C   5.00 L 
 15.0 L
 300
 TA 
FIG. P21.67
(c)
Eint,A 
3
3
nRTA   0.203 m ol  8.314 J m ol K   300 K   760 J
2
2
Eint,B  Eint,C 
P (atm)
V(L)
T(K)
Eint (kJ)
A
1.00
5.00
300
0.760
B
3.00
5.00
900
2.28
C
1.00
15.00
900
2.28
(d)
(e)
3
3
nRTB   0.203 m ol  8.314 J m ol K   900 K   2.28 kJ
2
2
For the process AB, lock the piston in place and put the cylinder into an oven at 900 K.
For BC, keep the sample in the oven while gradually letting the gas expand to lift a
load on the piston as far as it can. For CA, carry the cylinder back into the room at 300
K and let the gas cool without touching the piston.
(f)
For AB:
W  0
Eint  Eint,B  Eint,A   2.28  0.760 kJ 1.52 kJ
Q  Eint  W  1.52 kJ
For BC:
V 
Eint  0 , W   nRTB ln  C 
 VB 
W    0.203 m ol  8.314 J m ol K   900 K  ln  3.00  1.67 kJ
Q  Eint  W  1.67 kJ
For CA:
Eint  Eint,A  Eint,C   0.760  2.28 kJ 1.52 kJ
W   P V   nR T    0.203 m ol  8.314 J m ol K   600 K   1.01 kJ
Q  Eint  W  1.52 kJ 1.01 kJ 2.53 kJ
(g)
We add the amounts of energy for each process to find them for the whole cycle.
Q A BCA  1.52 kJ 1.67 kJ 2.53 kJ 0.656 kJ
W A BCA  0  1.67 kJ 1.01 kJ 0.656 kJ
 Eint A BCA
 1.52 kJ 0  1.52 kJ 0