Physics 41 Chapter 21 HW
1. A cylinder contains a mixture of helium and argon gas in equilibrium at 150°C. (a) What is the average
kinetic energy for each type of gas molecule? (b) What is the root-mean-square speed of each type of molecule?
(a)
K
(b)


3
3
kBT  1.38  1023 J K  423 K   8.76  1021 J
2
2
K
1 2
m vrm s  8.76  1021 J
2
so
vrm s 
For helium,
m 
Similarly for argon,
1.75  1020 J
m
4.00 g m ol
6.02  1023 m olecules m ol
(1)
 6.64  1024 g m olecule
m  6.64  1027 kg m olecule
39.9 g m ol
m 
 6.63  1023 g m olecule
6.02  1023 m olecules m ol
m  6.63  1026 kg m olecule
Substituting in (1) above,
we find for helium,
vrm s  1.62 km s
and for argon,
vrm s  514 m s
2. A 1.00-mol sample of hydrogen gas is heated at constant pressure from 300 K to 420 K. Calculate (a) the
energy transferred to the gas by heat, (b) the increase in its internal energy, and (c) the work done on the gas.
We us the tabulated values for C P and C V
(a)
Q  nC P T  1.00 m ol 28.8 J m ol K   420  300 K  3.46 kJ
(b)
Eint  nCV T  1.00 m ol 20.4 J m ol K  120 K   2.45 kJ
(c)
W  Q  Eint  3.46 kJ 2.45 kJ 1.01 kJ
3. A 2.00-mol sample of a diatomic ideal gas expands slowly and adiabatically from a pressure of 5.00 atm and a
volume of 12.0 L to a final volume of 30.0 L. (a) What is the final pressure of the gas? (b) What are the initial and
final temperatures? (c) Find Q, W, and Eint.


1.40
V 
 12.0
Pf  Pi i   5.00 atm 
 1.39 atm

 30.0
 Vf

PV
i i  PfV f
(a)
(b)
Ti 
Tf 


5.00 1.013  105 Pa 12.0  103 m
PV
i i

nR
2.00 m ol 8.314 J m ol K 
PfV f

nR


3
1.39 1.013  105 Pa 30.0  103 m
2.00 m ol 8.314 J m ol K 

3

365 K
253 K
The process is adiabatic: Q  0
(c)
  1.40 
C P R  CV
5

, CV  R
CV
CV
2
5

Eint  nCV T  2.00 m ol  8.314 J m ol K   253 K  365 K   4.66 kJ
2

W  Eint  Q  4.66 kJ 0  4.66 kJ
4. During the power stroke in a four-stroke automobile engine, the piston is forced
down as the mixture of combustion products and air undergoes an adiabatic
expansion as shown. Assume that (1) the engine is running at 2 500 cycles/min, (2)
the gauge pressure right before the expansion is 20.0 atm, (3) the volumes of the
mixture right before and after the expansion are 50.0 and 400 cm 3, respectively (4) the
time involved in the expansion is one-fourth that of the total cycle, and (5) the
mixture behaves like an ideal gas with specific heat ratio 1.40. Find the average
power generated during the expansion.
We suppose the air plus burnt gasoline behaves like a diatomic ideal gas. We find its final absolute pressure:

21.0 atm 50.0 cm
 1
Pf  21.0 atm  
 8

3 75

 Pf 400 cm

3 75
75
 1.14 atm


W 

3
Now Q  0 and W  Eint  nCV T f  Ti
W 
5
1.14 atm 400 cm
2


5
5
5
nRT f  nRTi  PfV f  PV
i i
2
2
2
 1.013  105 N m 2 
6
 21.0 atm 50.0 cm 3  
 10 m

1 atm




W  150 J
The output work is W  150 J The time for this stroke is
P
W
150 J

 25.0 kW
t 6.00  103 s
1  1 m in   60 s 
3

  6.00  10 s


4  2 500  1 m in

3
cm
3

5. A 4.00-L sample of a diatomic ideal gas with specific heat ratio 1.40, confined to a cylinder, is carried through
a closed 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 original
volume. (a) Draw a PV diagram of this cycle. (b) Determine the volume of the gas at the end of the adiabatic
expansion. (c) Find the temperature of the gas at the start of the adiabatic expansion. (d) Find the temperature
at the end of the cycle. (e) What was the net work done on the gas for this cycle?
(a)
See the diagram at the right.
(b)
PBVB  PC VC


3PV
i i  PV
i C
 
P
B
3 Pi
Ad iabatic
 
VC  31  V i  35 7 V i  2.19V i
VC  2.19 4.00 L   8.77 L
Pi
(c)
C
A
PBVB  nRTB  3PV
i i  3nRTi
Vi = 4 L
TB  3Ti  3 300 K   900 K
VC
FIG. P21.29
(d)
After one whole cycle, TA  Ti  300 K .
(e)
5 
In AB, Q A B  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
V(L)
6. Consider 2.00 mol of an ideal diatomic gas. (a) Find the total heat capacity of the gas at constant volume and
at constant pressure assuming the molecules rotate but do not vibrate. (b) What If? Repeat, assuming the
molecules both rotate and vibrate.
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
1
equal amounts of energy kB T into all five kinds of motion. The average energy of one
2
5
molecule is kB T . 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 
5
R and the sample’s heat capacity is
2
5 
5

nCV  n  R   2 m ol  8.314 J m ol K 
2 
2

The molar heat capacity is CV 
nCV  41.6 J K
For the heat capacity at constant pressure we have
5
 7
7

nC P  n  CV  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 
9 
nCV  n  R   58.2 J K
nC P  n  R   74.8 J K
and
2 
2 
7. Fifteen identical particles have various speeds: one has a speed of 2.00 m/s; two have speeds of 3.00 m/s; three
have speeds of 5.00 m/s; four have speeds of 7.00 m/s; three have speeds of 9.00 m/s; and two have speeds of
12.0 m/s. Find (a) the average speed, (b) the rms speed, and (c) the most probable speed of these particles.
(a)
vav 
(b)
 nivi 
N
1
1 2  2 3  3 5  4 7  3 9  212   6.80 m s
15 
 

v2
av
 nivi2  54.9 m 2
so vrm s 
(c)
N
v 
2
av
vm p  7.00 m s
s2
 54.9  7.41 m s
8. As a 1.00-mol sample of a monatomic ideal gas expands adiabatically, the work done on it is –2 500 J. The
initial temperature and pressure of the gas are 500 K and 3.60 atm. Calculate (a) the final temperature, and (b)
the final pressure.


W  nCV T f  Ti
(a)

3
2 500 J 1 m ol 8.314 J m ol K T f  500 K
2



PV
i i  PfV f
(b)

 nRT f 
 nRTi
Pi
 Pf 


 Pi 
 Pf 

   1
  1
Ti   T f

Pi
Pf
 Tf 
Pf  Pi 
T 
i
 5 3 3 2
Ti Pi1  T f Pf1
 Tf 
Pf  Pi 
 Ti 
 300
 3.60 atm 
 500
   1
52
 1.00 atm
9. A heat engine using a monatomic gas follows the cycle
shown.
(a) Find the temperature, volume and pressure at each point and
the change in internal energy, work done ON the gas, and Q for
each process and the net for the cycle. Clearly label and show
ALL your work, briefly explaining each process. Put your
results in the tables provided below.
(b) Find the total work done by the engine during one cycle and
the thermal efficiency of the engine and
c) the engine’s power output if it runs at 600 rpm.
T (K)
300
900
300
1
2
3
V (cm3)
200
600
600
P(kPa)
200
200
66.67
W (J)
Q (J)
Eint (J)
12
-80
200
120
23
0
-12
-120
31
43.9
43.9
0
-36.1
36.1
0
Net
b) 18%
c) 361 W
T f  300 K