PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HOMEWORK PROBLEMS Chapter 10:
THERMAL PHYSICS
HW-10
PART-A: Hand in your answers in class on scantron on Monday ….. November-2010. Write
your name, class (1401) and HW # 10 on the scantron.
1.
On a very cold day in upstate New York, the temperature is −25°C, which is equivalent
to what temperature in Fahrenheit?
(a) −46°F
(b) −77°F
(c) 18°F
(d) 298 K
(e) −13°F
TF  95 TC  32  95  25°   32°  13° F
2.
Convert 162°F to the equivalent temperature in Kelvin.
(a) 373 K
(b) 288 K
(c) 345 K
(d) 201 K
(e) 308 K
TC  95  TF  32   95 162  32   72.2°C
, then , TK  TC  273  72.2  273  345 K
3.
The Statue of Liberty is 93 m tall on a summer morning when the temperature is 20°C. If
the temperature of the statue rises from 20°C to 30°C, what is the order of magnitude of the
statue’s increase in height? Choose the best estimate, treating the statue as though it were solid
copper. (a) 0.1 mm (b) 1 mm
(c) 1 cm
(d) 10 cm
(e) 1 m
 L   Cu L0  T   17  10 6

 °C  1   93 m  10°C 
 1.6  10 2 m  1.6 cm
4.
A hole is drilled in a metal plate. When the metal is heated, what happens to the diameter
of the hole?
(a) It decreases.
(b) It increases.
(c) It remains the same.
(d) The answer depends on the initial temperature of the metal.
(e) None of these
The correct choice is (b). When a solid, containing a cavity, is heated, the cavity expands in the same way
as it would if filled with the material making up the rest of the object.
5.
A container holds 0.50 m3 of oxygen at an absolute pressure of 4.0 atm. A valve is
opened, allowing the gas to drive a piston, increasing the volume of the gas until the pressure
drops to 1.0 atm. If the temperature remains constant, what new volume does the gas occupy?
(a) 1.0 m3
(b) 1.5 m3
(c) 2.0 m3
(d) 0.12 m3 (e) 2.5 m3
From the ideal gas law, with the mass of the gas constant, P2V2 T2  PV
1 2 T1 . Thus,
 P T 
V2   1   2  V1   4   1 0.50 m3  2.0 m 3
 P2   T1 


1
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-10
6.
If the volume of an ideal gas is doubled while its temperature is quadrupled, does the
pressure (a) remain the same, (b) decrease by a factor of 2,
(c) decrease by a factor of 4,
(d) increase by a factor of 2, or
(e) increase by a factor of 4?
From the ideal gas law, with the mass of the gas constant, P2V2 T2  PV
1 2 T1 . Thus,
V  T 
 1
P2   1   2  P1     4  P1  2 P1
 2
 V2   T1 
7.
One way to cool a gas is to let it expand. When a certain gas under a pressure of 5.00 =
106 Pa at 25.0°C is allowed to expand to 3.00 times its original volume, its final pressure is 1.07
X 106 Pa. What is its final temperature?
(a) 177°C
(b) 233 K
(c) 212 K
(d) 191 K
(e) 115 K
Remember that one must use absolute temperatures and pressures in the ideal gas law. Thus, the original
temperature is TK  TC  273.15  25  273.15  298 K , and with the mass of the gas constant, the
ideal gas law gives
 P  V 
 1.07  106 Pa 
T2   2   2  T1  
 3.00   298 K   191 K
 5.00  106 Pa 
 P1   V1 
8.
What is the internal energy of 26.0 g of neon gas at a temperature of 152°C?
(a) 2 440 J
(b) 6 830 J
(c) 3 140 J
(d) 5 870 J
(e) 5 020 J
The internal energy of n moles of a monatomic ideal gas is U  23 nRT , where R is the universal gas
constant and T is the absolute temperature of the gas. For the given neon sample,
T  TC  273.15  152  273.15 K  425 K , and
n 
26.0 g
 1.29 mol
20.18 g mol
Thus, U  23  1.29 mol   8.31 J mol  K   425 K   6.83  10 3 J
9.
Find the root-mean-square speed of a methane gas molecule (CH4) at 25.0°C.
(a) 545 m/s (b) 681 m/s (c) 724 m/s (d) 428 m/s (e) 343 m/s
The root-mean-square speed of molecules in a gas with molar mass M (expressed in kilograms per mole)
and absolute temperature T is vrms  3RT M . The molar mass of methane (CH4) is
g  1 kg 
M  12.0  4 1.00  
 1.6  10 2 kg mol
mol  103 g 
and its absolute temperature is T  TC  273.15   25.0  273.15 K  298 K . Therefore,
vrms 
3  8.31 J mol  K   298 K 
1.60  10 2 kg mol
 681 m s
2
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-10
10.
Which of the assumptions below is not made in the kinetic theory of gases?
(a) The number of molecules is very small.
(b) The molecules obey Newton’s laws of motion.
(c) The collisions between molecules are elastic.
(d) The gas is a pure substance.
(e) The average separation between molecules is large compared with their dimensions.
The kinetic theory of gases does assume that the molecules in a pure substance obey Newton’s laws and
undergo elastic collisions, and the average distance between molecules is very large in comparison to
molecular sizes. However, it also assumes that the number of molecules in the sample is large so that
statistical averages are meaningful. The untrue statement included in the list of choices is (a).
11.
Suppose for a brief moment the gas molecules hitting a wall stuck to the wall instead of
bouncing off the wall. How would the pressure on the wall be affected during that brief time?
(a) The pressure would be zero.
(b) The pressure would be halved.
(c) The pressure would remain unchanged. (d) The pressure would double.
(e) The answer would depend on the area of the wall.
In a head-on, elastic collision with a wall, the change in momentum of a gas molecule is
 p  m (vf  v0 )  m (v0  v0 )  2mv0 . If the molecule should stick to the wall instead of
rebounding, the change in the molecule’s momentum would be  p  m (0  v0 )  mv0 , which is half
that in the elastic collision. Since a gas exerts a pressure on its container by molecules imparting impulses
to the walls during collisions, and the impulse imparted equals the magnitude of the change in the
molecular momentum, decreasing the change in momentum during the collisions by a factor of 2 would
halve the pressure exerted. Thus, the correct response is choice (b).
12.
If the temperature of an ideal gas is increased from 200 K to 600 K, what happens to the
rms speed of the molecules?
(a) It increases by a factor of 3.
(b) It remains the same.
(c) It is one third of the original speed.
(d) It is √3 times the original speed.
(e) It increases by a factor of 6.
The rms speed of molecules in the gas is vrms  3RT M . Thus, the ratio of the final speed to the
original speed would be
3RT f M
 vrms  f
Tf
600 K



 3
T0
200 K
3RT0 M
 vrms 0
13.
The pressure in a constant-volume gas thermometer is 0.700 atm at 100°C and 0.512 atm
at 0°C. (a) What is the temperature when the pressure is 0.040 atm?
a) -181 C
b) -201 C
c) -251 C
d) -271 C
When the volume of a low density gas is held constant, pressure and temperature are related by a linear
equation P  AT  B , where A and B are constants to be determined. For the given constant-volume gas
thermometer,
P  0.700 atm when T  100°C  0.700 atm  A 100°C   B
[1]
P  0.512 atm when T  0°C  0.512 atm  A  0   B
From Equation [2], B  0.512 atm . Substituting this result into Equation [1] yields
3
[2]
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-10
0.700 atm  0.512 atm
 1.88  10 3 atm °C
100°C
so, the linear equation for this thermometer is: P  1.88  10 3 atm °C T  0.512 atm
A 


If P  0.0400 atm , then
PB
0.0400 atm  0.512 atm
T 

  251°C
A
1.88  10-3 atm °C
(a)
.
14.
The pressure in a constant-volume gas thermometer is 0.700 atm at 100°C and 0.512 atm
at 0°C. What is the pressure at 450°C?
a) 1.06 atm
b) 1.36 atm
c) 1.66 atm
d) 1.96 atm
If T  450°C , then
P  1.88  10 3 atm °C  450°C   0.512 atm  1.36 atm


15.
Calculate the temperature −40° F into celcius. (Note the uniqueness of the result)
a) -30 C
b) -40 C
c) +40 C
d) -32 C
Start with TF  40°F and convert to Celsius.
5
5
TC   TF  32    40  32    40°C
9
9
Since Celsius and Fahrenheit degrees of temperature change are different sizes, this is the only
temperature with the same numeric value on both scales.
16.
The New River Gorge bridge in West Virginia is a 518-mlong steel arch. How much will
its length change between temperature extremes of −20°C and 35°C?
a) 21 cm
b) 31 cm
c) 41 cm
d) 51 cm
The increase in temperature is T  35°C   20°C   55°C
Thus,
1
 L   L0  T   11  10 6  °C    518 m   55°C   0.31 m  31 cm


A copper wire with length 10.0 m and cross-sectional area 2.40 X 10−5 m2 is stretched
taut between two poles under a tension of 75.0 N. What is the tension in the wire when the
temperature falls by 10.0°C?
a) 324 N
b) 424 N
c) 524 N
d) 624 N
17.
When the temperature drops by 10.0°C, the wire will attempt to contract by

 L   Cu L0  T  17  10 6  °C 
1
 10.0 m  10.0 °C   1.70  10
3
m
If the ends of the wire are held stationary, and the wire is not allowed to contract, it will develop an
additional tension force  F sufficient to keep it stretched by an additional 1.70  103 m beyond its
natural length. The required additional tension force is  F  YCu A ( L L0) , where YCu is Young’s
modulus for copper and A is the cross-sectional area of the wire. Thus,
4
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-10
 1.7  10 3 m 
 F  11  1010 Pa 2.40  10 5 m2 
 449 N
 10.0 m 
The total tension in the wire, after the decrease in temperature, is then
F  F0  F  75.0 N  449 N  524 N



18.
A gold ring has an inner diameter of 2.168 cm at a temperature of 15.0°C. Determine its
inner diameter at 100°C (agold = 1.42 X 10−5°C−1).
a) 2.17 cm
b) 2.71 cm
c) 1.71 cm
d) 1.31 cm
When the temperature of a material is raised, the linear dimensions of any cavity in that material expands
as if it were filled with the surrounding material. Thus, the final value of the inner diameter of the ring will
be given by L  L0   L0  T  as
1
Dinner  2.168 cm  1.42  10 5  °C    2.168 cm  100°C  15.0°C   2.171 cm


19.
One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If
the gas is heated at constant volume until the pressure triples, what is the final temperature?
a) 666 C
b) 696 C
c) 556 C
d) 627 C
From the ideal gas law, PV  nRT , we find P T  nR V . Thus, if both n and V are constant as the gas
is heated, the ratio P T is constant giving
Pf
 Pf 
 3P 
P
 i
T f  Ti     300 K   i   900 K  627°C
or
Tf
Ti
 Pi 
 Pi 
20. One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C. If the gas is
heated so that both the pressure and volume are doubled, what is the final temperature?
a) 927 C
b) 957 C
c) 826 C
d) 333 C
If both pressure and volume double as n is held constant, the ideal gas law gives
  2 Pi   2Vi  
 Pf V f 
T f  Ti 
 Ti 
  4Ti  4  300 K   1 200 K  927°C

PV
 PV


i i 
i i
21.
A weather balloon is designed to expand to a maximum radius of 20 m at its working
altitude, where the air pressure is 0.030 atm and the temperature is 200 K. If the balloon is filled
at atmospheric pressure and 300 K, what is its radius at liftoff?
a) 61 cm
b) 7.1 cm
c) 8.1 cm
d) 9.1 cm
With n held constant, the ideal gas law gives
 P  T 
V1
 0.030 atm   300 K 
  2  1   
 4.5  10 2
 1.0 atm   200 K 
V2
 P1   T2 
3
Since the volume of a sphere is V   4 3r 3 , V1 V2   r1 r2 
Thus,
V 
r1   1 
V 
2
13

r2  4.5  10 2
1 3  20 m  
5
7.1 m
PHYS-1401: College Physics-I
22.
CRN 55178
Khalid Bukhari
HW-10
An air bubble has a volume of 1.50 cm3 when it is released by a submarine 100 m below
the surface of a lake. What is the volume of the bubble when it reaches the surface? Assume
the temperature and the number of air molecules in the bubble remain constant during its
ascent.
a) 12 cm3
b) 16 cm3
c) 20 cm3
d) 24 cm3
The pressure 100 m below the surface is found, using P1  Patm  gh , to be

P1  1.013  105 Pa  103 kg m 3
 9.80

m s2 100 m   1.08  106 Pa
The ideal gas law, with T constant, gives the volume at the surface as
 P 
P 
 1.08  106 Pa 
V2   1  V1   1  V  
1.50 cm3  16.0 cm 3
 1.013  105 Pa 
 P2 
 Patm 

23.

What is the average kinetic energy of a molecule of oxygen at a temperature of 300 K?
a) 6.01 x 10-21 J
b) 6.21 x 10-21 J
c) 6.51 x 10-21 J
The average kinetic energy of the molecules of any ideal gas at 300 K is
1
3
3
J
KE  mv2  kBT   1.38  10 23
 300 K   6.21  10 21 J
2
2
2
K 
24.
The rms speed of an oxygen molecule (O2) in a container of oxygen gas is 625 m/s. What
is the temperature of the gas?
a) 400 K
b) 450 K
c) 480 K
d) 501 K
The rms speed of molecules in a gas of molecular weight M and absolute temperature T is
vrms  3RT M . Thus, if vrms  625 m s for molecules in oxygen (O2), for which
M  32.0 g mol  32.0  10 3 kg mol , the temperature of the gas is
T 


2
32.0  10 3 kg mol  625 m s 
M vrms

3R
3  8.31 J mol  K 
2
 501 K
25.
A 7.00-L vessel contains 3.50 moles of ideal gas at a pressure of 1.60 X 106 Pa. Find the
temperature of the gas.
a) 345 K
b) 365 K
c) 385 K
d) 405 K
The volume occupied by this gas is
V  7.00 L 103 cm 3 1 L 1 m 3 106 cm 3  7.00  10 3 m 3



Then, the ideal gas law gives
1.60  106 Pa 7.00  10 3 m 3
PV
T 

nR
 3.50 mol   8.31 J mol  K 



 385 K
26.
A 7.00-L vessel contains 3.50 moles of ideal gas at a pressure of 1.60 X 106 Pa. Find the
average kinetic energy of a gas molecule in the vessel.
a) 7.07 x 10-21 J
b) 7.47 x 10-21 J
c) 7.97 x 10-21 J
6
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
The average kinetic energy per molecule in this gas is
KE molecule 
3
3
kBT 
1.38  10 23 J K  385 K   7.97  10 21 J
2
2


7
HW-10