Name: ___________________________________ Student ID # ________________________
Physics 224A
Summer-2015
July 10, 2015
Midterm Exam 1
Mark your answers on the exam. Make sure that ALL pages have your name and student ID on them! Seven points
for each correct answer.
1. A Celsius thermometer and a Fahrenheit thermometer both give the same reading for a certain sample.
The corresponding Kelvin temperature is:
A)
B)
C)
D)
E)
574 K
233 K
301 K
614 K
276 K
2. The coefficient of linear expansion of steel is 11 × 10–6 per C. A steel ball has a volume of exactly 100
cm3 at 0C. When heated to 100C, its volume becomes:
A)
B)
C)
D)
E)
100.33 cm3
100.0011 cm3
100.0033 cm3
100.000011 cm3
none of these
3. Use R = 8.2 × 10–5 m3  atm/mol  K and NA = 6.02 × 1023 mol–1. The approximate number of air
molecules in a 1 m3 volume at room temperature and atmospheric pressure is:
A)
B)
C)
D)
E)
41
450
2.5 × 1025
2.7 × 1026
5.4 × 1026
4. The rms speed of an oxygen molecule at 0C is 460 m/s. If the molar mass of oxygen is 32 g and of
helium is 4 g, then the rms speed of a helium molecule at 0C is:
A)
B)
C)
D)
E)
230 m/s
326 m/s
650 m/s
920 m/s
1300 m/s
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Name: ___________________________________ Student ID # ________________________
5. An automobile tire is pumped up to a gauge pressure of 2.0 × 105 Pa when the temperature is 27C. After
the car has been running on a hot day, its gauge pressure increased to 2.2 × 105 Pa. What is the tire
temperature after running on a hot day? Assume that the volume remains fixed and take atmospheric
pressure to be 1.013 × 105 Pa.
A)
B)
C)
D)
E)
320 C
63 C
77 C
47 C
39 C
6. An ideal gas of N diatomic molecules has temperature T. If the number of molecules is doubled without
changing the temperature, the internal energy increases by:
A)
B)
C)
D)
E)
0
1/2NkT
3/2NkT
5/2NkT
3NkT
7. An ideal gas of N monatomic molecules is in thermal equilibrium with an ideal gas of the same number of
diatomic molecules and equilibrium is maintained as temperature is increased. The ratio of the internal
energies Udia/Umon is:
A)
B)
C)
D)
E)
1/2
3/5
1
5/3
2
8. When work W is done on an ideal gas of N diatomic molecules in thermal isolation the temperature
increases by:
A)
B)
C)
D)
E)
W/2Nk
W/3Nk
2W/3Nk
2W/5Nk
W/Nk
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Name: ___________________________________ Student ID # ________________________
9. The energy absorbed as heat by an ideal gas for an isothermal process equals:
A)
B)
C)
D)
E)
the work done by the gas
the work done on the gas
the change in the internal energy of the gas
the negative of the change in internal energy of the gas
zero since the process is isothermal
10. The temperature of n moles of an ideal monatomic gas is increased by T at constant pressure. The energy
Q absorbed as heat, change Uint in internal energy, and work W done by the environment are given by:
A)
B)
C)
D)
E)
Q = (5/2)nRT, Uint = 0, W = –nRT
Q = (3/2)nRT, Uint = (5/2)nRT, W = –(3/2)nRT
Q = (5/2)nRT, Uint = (5/2)nRT, W = 0
Q = (3/2)nRT, Uint = 0, W = –nRT
Q = (5/2)nRT, Uint = (3/2)nRT, W = –nRT
11. Monatomic, diatomic, and polyatomic ideal gases each undergo slow adiabatic expansions from the same
initial volume and the same initial pressure to the same final volume. The magnitude of the work done by
the environment on the gas:
A)
B)
C)
D)
E)
is greatest for the polyatomic gas
is greatest for the diatomic gas
is greatest for the monatomic gas
is the same only for the diatomic and polyatomic gases
is the same for all three gases
12. During slow adiabatic expansion of a gas:
A)
B)
C)
D)
E)
the pressure remains constant
energy is added as heat
work is done on the gas
no energy enters or leaves as heat
the temperature is constant
13. A real gas is changed slowly from state #1 to state #2. During this process no work is done on or by the
gas. This process must be:
A)
B)
C)
D)
E)
isothermal
adiabatic
isovolumic
isobaric
a closed cycle with point #1 coinciding with point #2
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Name: ___________________________________ Student ID # ________________________
14. During a reversible adiabatic expansion of an ideal gas, which of the following is NOT true?
A)
B)
C)
D)
E)
pV 1 = constant
pV = nRT
TV 1 - 1 = constant
W = – pdV
pV = constant
15. Bonus problem. Both pressure and volume of an ideal gas of diatomic molecules are doubled. The ratio
of the new internal energy to the old both measured relative to the internal energy at 0 K is:
A)
B)
C)
D)
E)
1/4
1/2
1
2
4
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Name: ___________________________________ Student ID # ________________________
Answer Key
1. B
233 K
Feedback:
Expressed in Celsius degrees, the Fahrenheit reading is F = (9/5)C + 32. If Celsius thermometer and a
Fahrenheit thermometer both give the same reading, then C = F. Solving the equation C = (9/5)C + 32 for
C, we find C = -40C. In Kelvins, the respective temperature is -40 + 273 = 233 K.
2. A
100.33 cm3
Feedback:
The coefficient of volume expansion is three times the coefficient of linear expansion,  = 3. For steel, 
= 3(11 × 10–6 per C) = 33 × 10–6 per C. Then the temperature change of the volume of the ball is V =
VT. Plugging  = 33 × 10–6 per C, V = 100 cm3, and T = (100C) - (0C) = 100C, we find V = (33
× 10–6 per C)(100 cm3)(100C) = 0.33 cm3. Then the new volume is V + V = (100 cm3) + (0.33 cm3) =
100.33 cm3.
3. C
2.5 × 1025
Feedback:
Using the ideal gas law, PV = nRT, we can find the number of moles, n = PV/(RT). Plugging P = 1 atm, V
= 1 m3, and T = 293 K, we find: n = 41.62 mol. As one mole has NA = 6.02 × 1023 molecules of gas, in
41.62 moles there are (41.62 mol)(6.02 × 1023 mol–1) = 2.5 × 1025.
4. E
1300 m/s
Feedback:
Molecule of oxygen, O2, has 7 degrees of freedom, f = 7. Three of them correspond to translational
2
motion. Therefore, 12 mvrms
 23 kT . Then vrms  3mkT . So, vrms varies inversely with the square root of
mass. Reducing molar mass from 32 g to 4 g increases the rms speed 324  8  2.83 times, from 460
m/s to 2.83(460 m/s) = 1300 m/s.
5. D
47 C
Feedback:
As the volume remains fixed, the ideal gas law simplifies: P1/T1 = P2/T2. This gives T2 = T1(P2/P1). By its
definition, gauge pressure is by Pat = 1.013 × 105 Pa less than the real pressure. Therefore, with Pg = 2.0 ×
105 Pa, the real pressure P1 = (2.0 × 105 Pa) + (1.013 × 105 Pa) = 3.0 × 105 Pa and P2 = (2.2 × 105 Pa) +
(1.013 × 105 Pa) = 3.2 × 105 Pa Plugging T1 = 300 K, P1 = 3.0 × 105 Pa, and P2 = P1 = 3.2 × 105 Pa, we
come to T2 = T1(P2/P1) = (300 K)(3.2 × 105 Pa)/(3.0 × 105 Pa) = 320 K = 47C.
6. D
5/2NkT
Feedback:
Diatomic molecules have seven degrees of freedom, f = 7. Unless temperature is extremely high, two
vibrational degrees of freedom are frozen. So, we can assume f = 5. Therefore, the internal energy of the
gas is U = (1/2)fNkT = (5/2)NkT. If the number of molecules is doubled without changing the temperature,
the internal energy becomes U = (5/2)N2kT= (5/2)(2N)kT = 5NkT. Respectively, the change, U = 5NkT (5/2)NkT = (5/2)NkT.
7. D
5/3
Feedback:
As thermal equilibrium is maintained, temperature of monatomic molecules, Tmon, remains equal to
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Name: ___________________________________ Student ID # ________________________
8.
9.
10.
11.
temperature of diatomic molecules, Tdia, namely, Tmon = Tdia = T. For monatomic molecules, the internal
energy of the gas is Umon = (1/2)fmonNkTmon = (1/2)fmonNkT. Plugging the respective number of degrees of
freedom, fmon = 3, we find Umon = (3/2)NkT. Similarly, for diatomic molecules, the internal energy of the
gas is Udia = (1/2)fdiaNkTdia = (1/2)fdiaNkT. At normal temperatures, fdia = 5.Plugging fdia = 5, we find Udia =
(5/2)NkT. Then Udia/Umon = [(5/2)NkT]/[(3/2)NkT] = 5/3.
D
2W/5Nk
Feedback:
According to 1st Law, U = Q + W. In thermal isolation, Q = 0. Then U = W. From another hand,
internal energy of the ideal gas of diatomic molecules is Udia = (1/2)fdiaNkT. At normal conditions, the
number of degrees of freedom, fdia = 5. Then Udia = (5/2)NkT. Then U = (5/2)NkT. As, from the 1st
Law, U = W, we find (5/2)NkT = W. Solving for T, we finally come to T = 2W/(5Nk).
A
the work done by the gas
Feedback:
Internal energy of an ideal gas is U = (1/2)fNkT. For an isothermal process, T = constant. Then U =
(1/2)fNkT = constant. Then U = 0. From the 1st Law, U = Q + W. This gives Q + W = 0. Then Q = -W.
In words, the energy absorbed by the ideal gas as heat equals negative work done on the gas. It is opposite
to the work done by the gas, Won = -Wby. Then Q = -(-Wby) = Wby. In words, the energy absorbed by the
ideal gas as heat equals work done by the gas.
E
Q = (5/2)nRT, Uint = (3/2)nRT, W = –nRT
Feedback:
Internal energy of an ideal gas is Uint = (1/2)fNkT = (1/2)fnRT. For a monatomic gas, f = 3. Then Uint =
(3/2)nRT. Then Uint = (3/2)nRT. With constant pressure, P = constant, the work done by the gas on the
environment, Wby = PV. From another hand, from ideal gas law, PV = nRT. Then, with constant
pressure, P = constant, we have PV = nRT. Therefore, Wby = PV = nRT. It is opposite to the work
done by the environment on the gas, Wby = -Won. Then Won = - nRT. From the 1st Law, Uint = Q +
Won. Then Q = Uint - Won. Plugging Uint = (3/2)nRT and Won = -nRT, we find Q = (3/2)nRT - (nRT) = (5/2)nRT.
A
is greatest for the polyatomic gas
Feedback:
Adiabatic expansion follows the equation PV = C. On the P-V diagram (see below), this is represented by
the graph of P = C/V. Steepness of the graph is determined by the value of  = (f + 2)/f. The greater the
value of , the steeper the graph. For monatomic gas, f = 3, and, therefore,  = 5/3 = 1.67. For diatomic
gas, f = 5, and, therefore,  = 7/5 = 1.4. For a polyatomic gas, f  6 , and, therefore,   8 / 6 = 1.33. As
it follows, the steepest graph is for monatomic molecules with the greatest value of . Among the three
graphs shown on the diagram for the three types of molecules, the one at the top corresponds to
polyatomic molecules with the least value of . 
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Name: ___________________________________ Student ID # ________________________
12.
A)
B)
C)
D)
E)
13.
14.
A)
B)
C)
D)
E)
15.
Area under the graph equals work of the gas during its expansion. From the diagram, magnitude of work
(its absolute value) is greatest for the polyatomic gas.
D
no energy enters or leaves as heat
Feedback:
An adiabatic expansion follows the equation PV = C or P = C/V. The only limitation is Q = 0.
Therefore,
During the expansion, increasing volume results in the respective decrease of pressure, P = C/V. So, P
does not remain constant.
The limitation is Q = 0. Therefore, energy is NOT added as heat.
According to 1st Law, U = Q + W. With Q = 0, we have U = W. In an adiabatic expansion,
temperature of the gas drops (see the part E below). Respectively, its internal energy, U = (1/2)fNkT,
drops as well. This means U < 0. As U = W, the work done on the gas is negative, W < 0. In other
words, the gas does some positive work on the environment, not the opposite.
As Q = 0, no energy enters or leaves as heat (true statement!).
During adiabatic expansion, we have PV = C. Then PV = C/V As  > 1, C/V decreases with V.
From the ideal gas law, PV = NkT. Decreasing PV = C/V means decreasing NkT. In an adiabatic
expansion, gas temperature drops, T is not constant.
C
isovolumic
Feedback:
On the P-V diagram, work done by the gas is expressed by area under the graph (see problem 11 above).
This area is zero under condition that the two points, (V1, P1) and (V2, P2) of the initial and final state are
on the same vertical line above one another. In other words, volume does not change, V1 = V2. The process
is isovolumic.
E
pV = constant
Feedback:
The false statement is pV = constant.
As the expansion is adiabatic, pV 1 = constant is true.
As the gas is ideal, pV = nRT is true
As the expansion is adiabatic, TV 1 - 1 = constant is true. Actually, it is equivalent to pV 1 = constant.
W = – pdV , definition of work done on the gas, is true.
During adiabatic expansion, we have PV = C. Then PV = C/V As  > 1, C/V decreases with V.
Then PV = C/V decreases as well. So, pV = constant is a false statement.
E
4
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Name: ___________________________________ Student ID # ________________________
Feedback:
Let P2 = 2P2 and V2 = 2V2. Then P2V2 = (2P2)(2V2) = 4P1V1. From ideal gas law, P1V1 = NkT1 and P2V2 =
NkT2. This gives NkT2 = 4NkT1 or, equivalently, T2 = 4T1. From another hand, the internal energy U1 =
0.5fNkT1 and U2 = 0.5fNkT2. Then the ratio of the new internal energy to the old one, U2/U1 =
(0.5fNkT2)/(0.5fNkT1) = T2/T1. Plugging T2 = 4T1, we come to U2/U1 = 4.
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