Test 2 Review
Ideal Gas
—
PV=nRT
—
Watch units
W=P
V; work done by gas vs. on gas
Adiabatic, isothermal, constant V, and P
Thermal equilibrium
—
Entropy
Heat

Internal energy
—
kinetic vs. T
1
st
and 2
nd
laws of thermodynamics
U = Q + W (W
on gas
=+; W
done b
y gas
=

)
Heat transfer: Q=mc
T
Phase change: Q=mL
Radiation, convection, conduction
Multiple choice practice
1.
A container holds 1mol of an ideal gas. 500J of heat are added to the gas. The
temperature increases most
(a) if the volume is held c
onstant.
(b) if the pressure is held constant.
(c) Both (a) and (b) give the same increase in temperature.
(d) There is insufficient information given to make a determination.
2.
A pipe organ is manufactured near sea level with barometric pressure of P=
760torr. It
is then transported to a region well above sea level with a barometric pressure of
P=670torr. Assuming the organ is housed in a room at the same temperature, the
frequency of individual pipes will be
(a) less than before
(b) greater than befo
re
(c) the same as before
HINT: Recall that the speed of sound in air is
v
P
and the frequency is
f
v
2
L
(or
v
4
L
if a closed pipe).
Two identical containers hold equal amounts of the sam
e gas at the same temperature. In
each case, a piston compresses the gas to half the original volume. In one container the
process takes place adiabatically, and in the other container it takes place at constant
temperature in contact with a heat reservo
ir. The piston must do more work in the case of
(a) the adiabatic compression.
(b) the constant

temperature compression.
(c) In both processes the amount of work is the same.
4.
Gas in a piston expands its volume at constant
pressure
. Choose the correct
statement:
(a) The internal energy of the gas increases.
(b) The entropy of the gas increases.
(c) The work done by the gas exceeds the amount of heat injected into it.
(d) The heat injected into the gas exceeds the work done by it.
(e) more than one is c
orrect
5.
In the path shown below, the gas returns to its original state. The net work done
on
the
gas along this path is
(a) negative.
(b) zero.
(c) positive.
P
V
Practice
problems for
Exam II
Instructor: Cheryl Davis
No time limit
,
closed book
,
equation
sheet on the test,
Calculator
permitted
Useful information
2
7
3
K
0
o
C
3
2
o
F
,
3
7
3
K
1
0
0
o
C
2
1
2
o
F
,
V
s
p
h
e
r
e
4
r
3
3
,
s
t
e
e
l
1
.
1
1
0
5
C
o
1
,
g
a
s
o
l
i
n
e
9
.
6
1
0
4
C
o
1
,
R
8
.
3
1
4
J
m
o
l
K
,
k
1
.
3
8
1
1
0
2
3
J
K
,
1
a
t
m
1
.
0
1
3
1
0
5
P
a
,
k
C
u
3
9
7
W
m
o
C
,
L
i
c
e
3
.
3
3
1
0
5
J
k
g
,
m
H
2
O
3
.
0
1
0
2
6
k
g
1. (a) Using information at the top of this sheet, derive the conversions from
Fahrenheit to Kelvin and back:
T
F
9
5
T
4
5
9
and
T
5
9
T
F
2
5
5
.
2. (a) A steel bridge is 518m long. How much does its length change
between temperature extremes of
23
C
and
37
C
?
(b) An automobile fuel tank is filled to the brim with 50L of gasoline at
14
C
.
Immediately afterward, the vehi
cle is parked in the Sun where the
temperature reaches
35
C
. How much gasoline overflows from the tank as a
result of expansion? Neglect the expansion of the tank.
3. Calculate the number of molecules in 1L of air near the earth's surface
(1atm)
(a) on a
hot day (
35
C
).
(b) and on a cold day (
30
C
).
4. Ice water inside a Styrofoam cooler contains 1.0kg of ice at
3
2
o
F
. In a
room at
7
0
o
F
the ice completely melts in 8hrs. How long would the same
amount of ice last if
the cooler is outdoors where the temperature is
1
0
0
o
F
?
HINT: You do not need to know the latent heat of melting
L
i
c
e
. Also,
Q
c
m
T
is irrelevant since there is no change in temperature.
5
. (a) How much work is done by 3.7mol of an ideal gas as it expands to 6.8
times its initial volume at a constant temperature of
72
C
?
(b) What is the heat flow into the gas during this process?
6. A vacuum chamber is contaminated with water vapor at
150
C
.
(a)
What is the rms speed for molecules in the container? Note: A water
molecule has a mass of
3
.
0
1
0
2
6
k
g
.
7. H
2
gas is allowed to expand from an initial state of
P
i
1
.
0
a
t
m
,
V
i
1
.
0
L
to
a final state of
P
f
2
.
0
a
t
m
,
V
f
2
.
0
L
a
long a path that forms a straight line on
a PV diagram.
(a) Draw a PV

diagram of the process.
(b) What is the work done by the gas?
(c) What is the change in internal energy of the gas?
(d) What is the heat injected into the gas during this process?
8. A piston holds an ideal gas consisting of
monatomic
molecules. The
initial. volume is 1.0L. 100J of heat is added to the gas. If the pressure is
held constant at 1.0
10
5
Pa during this process, what is the final volume?
9. Two separate con
tainers (label them 1 and 2) with the same initial
volumes are filled with a
diatomic
gas such as N
2
at temperature
T
o
. The
molecules do not vibrate internally but may rotate (the case near room
temperature). Container 1 is in contact with a thermal reser
voir at constant
temperature
T
o
. Container 2 is in complete thermal isolation. The volume
of each container is allowed to expand gradually to twice the initial volumes.
(a) Depict the two processes on a single PV diagram.
(b) What is the ratio of the fina
l pressure in container 1 to the final pressure
in container 2?
10. A Carnot engine does 400J of work and expels 800J of energy into the
cold reservoir on each cycle.
(a) What is the efficiency of the engine?
(b) If
3
J
K
of entropy is transferred into the c
old reservoir on each cycle,
what is the temperature of the
hot
reservoir?
11. A Carnot cycle on a P

V diagram reaches a maximum volume of 5.0L
and a minimum volume of 1.0L. Assume that there is 0.1mol of gas and that
5
3
.
(a) Find the work done per cycle
if the temperatures of the two reservoirs are
100
C
and
0
C
.
(b) How much entropy is transferred into the gas while in contact with the
hot reservoir?
SOLUTION
Multiple choice
1.
(a) To keep the pressure constant, the piston does work as the volume increases.
Since
the gas loses internal energy to do work, the temperature does not increase as much.
2.
(c) The ideal gas law shows that density of the air to be proportional to the pressure:
P
V
N
k
T
N
V
P
k
T
. Therefore, the velocity of sound in air d
oesn’t change.
3.
(a) Adiabatic compression has a steeper curve on the PV diagram with more area under
the curve. Since no heat can escape, the gas heats up during the adiabatic compression
causing a higher pressure.
4. (e)
5. (a)
1. (a)
T
F
AT
B
(linear conn
ection)
212
A
373
B
32
A
273
B
180
A
100
A
9
5
B
32
A
273
32
9
5
273
459
T
F
9
5
T
459
T
5
9
T
F
255
2. (a)
L
L
T
1
.
1
10
5
C
1
518
m
37
C
23
C
0
.
342
m
(b)
V
V
T
9
.
6
10
4
C
1
50
L
35
C
14
C
1
.
0
L
3. (a)
N
PV
kT
1
.
01
10
5
Pa
0
.
001
m
3
1
.
38
10
23
J
K
35
2
7
3
K
2
.
4
10
22
(b)
N
PV
kT
1
.
01
10
5
Pa
0
.
001
m
3
1
.
38
10
23
J
K
30
2
7
3
K
3
.
0
10
22
4.
Q
t
1
k
A
T
1
x
Q
t
2
k
A
T
2
x
t
2
t
1
T
1
T
2
t
2
T
1
T
2
t
1
t
2
7
0
o
F
3
2
o
F
1
0
0
o
F
3
2
o
F
8
.
0
h
4
.
5
h
5. (a)
W
PdV
V
1
V
2
nRT
dV
V
V
1
V
2
nRT
ln
V
2
V
1
3
.
7
mol
8
.
314
J
mol
K
72
273
K
ln
6
.
8
2
.
0
10
4
J
(b)
U
0
Q
W
2
.
0
10
4
J
6. (a)
v
rms
3
1
.
381
10
23
J
K
150
273
K
3
.
0
10
26
kg
764
m
s
7. (a)
V
P
i
f
(b) Area below curve:
W
1
.
013
10
5
Pa
2
1
.
013
10
5
Pa
2
0
.
0020
m
3
0
.
0010
m
3
152
J
.
(c)
U
5
2
Nk
T
5
2
Nk
P
f
V
f
Nk
P
i
V
i
Nk
5
2
P
f
V
f
P
i
V
i
5
2
2
1
.
013
10
5
Pa
0
.
0020
m
3
1
.
013
10
5
Pa
0
.
0010
m
3
760
J
(d)
U
Q
W
Q
U
W
760
J
152
J
912
J
8.
Q
n
C
p
T
f
T
i
,
T
f
PV
f
nR
,
T
i
PV
i
nR
,
C
p
5
2
R
.
Q
n
5
2
R
P
V
f
n
R
P
V
i
n
R
5
P
2
V
f
V
i
2
Q
5
P
V
f
V
i
V
f
V
i
i
2
Q
5
P
1
.
0
10
3
m
3
2
100
J
5
1
.
0
10
5
P
a
1
.
40
10
3
m
3
1
.
40
Note: We didn’t need to know the initial temperature or number of moles.
9.
P
i
V
i
P
f
1
V
f
P
i
V
i
P
f
2
V
f
P
i
V
i
P
i
V
i
P
f
1
V
f
P
f
2
V
f
P
f
1
P
f
2
V
f
V
i
1
2
C
p
C
v
1
2
7
5
1
1
.
3
2
10. (a)
e
W
Q
in
W
W
Q
out
400
J
400
J
800
J
1
3
(b)
S
Q
out
T
c
Q
in
T
h
W
Q
out
T
h
T
h
W
Q
out
S
400
J
800
J
3
J
K
400
K
Alternatively:
e
C
1
T
c
T
h
T
h
T
c
1
e
C
Q
out
S
1
e
C
800
J
3
J
K
1
1
3
400
K
11. (a)
P
C
nRT
c
V
C
and
P
B
nRT
h
V
B
(We need
V
B
)
P
B
V
B
P
C
V
C
nRT
h
V
B
V
B
nRT
c
V
C
V
C
T
h
V
B
1
T
c
V
C
1
V
B
V
C
T
c
T
h
1
1
Q
AB
W
AB
PdV
V
A
V
B
nRT
dV
V
V
A
V
B
nRT
ln
V
B
V
A
nRT
ln
V
C
V
A
T
c
T
h
1
1
Q
AB
0
.
1
mol
8
.
31
J
K
mol
373
K
ln
5
L
1
L
273
K
373
K
1
5
3
1
354
J
e
C
W
Q
AB
1
T
c
T
h
W
Q
AB
1
T
c
T
h
354
J
1
273
K
373
K
95
J
(b)
S
Q
AB
T
h
354
J
373
K
0
.
95
J
K
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