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Thermodynamics
Practice Questions
Question 1
1 mole of ideal gas is brought to a final state F by one of three processes that
have different initial states as shown in the figure. What is true for the
temperature change between initial and final states?
A)
B)
C)
D)
E)
It’s the same for all processes.
It’s the smallest for process 1.
It’s the smallest for process 2.
It’s the smallest for process 3.
It’s the same for processes 1 and 2.
The temperature
for gas at any
point is given by:
PV  TR
PV
T
R
Since all of the three
processes end at the
same temperature,
we need only look at
the process that
starts closest to:
P0  2V0 

PV
T
2 0 0
R
R
3p0
1
2p0
p0
2
F
3
V0
1.
 3P0 V0   3 PV
0 0
R
R
 2 P0 V0   2 PV
0 0
2.
R
R
 P V  PV
3. 0 0  0 0
R
R
2V0
Process 2
actually begins at
this temperature,
so its change is 0
Question 2
1 mole of ideal gas is brought to a final state F by one of three processes that
have different initial states as shown in the figure. What is true for the work done
by the gas?
A)
B)
C)
D)
E)
It’s positive for process 1 and 2, but negative for process 3.
It’s the smallest for process 1.
3p0
It’s the smallest for process 2.
1
It’s the smallest for process 3.
2p0
Zero work is done along process 3.
p0
2
F
3
V0
Work done by the gas is equal to the area under the curve
2V0
Question 3
An ideal gas is brought from S -> F by three different paths: SRF, SF, STF. The
temperature at S is the same as the temperature at F. Which of the following is true?
A) Process SRF occurs at constant temperarure
B) The work done by the gas along SRF is the same as the work done by the gas
along STF.
C) Net heat into the gas along STF is greater than the work done by the gas along
this path.
D) The change in the internal energy is the same for all three paths.
E) Work done by the gas along STF is greater than the work done by the gas along
SF.
The internal energy of an ideal gas depends
The internal energy of an ideal
only on temperature. If the initial and final
gas depends only on temperature
temp are the same, there is no change in
P
internal energy. U=0 for all three paths.
S
R
W= - Q for each process, which eliminates C
Area considerations eliminate B and E
T
F
Temperature at R is greater than the
temperature at S (gas law) eliminates A
V
Question 4
What is true for the process D -> A?
A)
B)
C)
D)
E)
U=0 Q>0.
U=0 Q<0
W=0
U>0
W=0
U<0
W=0
Q=0
2p0
A
B
Since the area (work) under D -> A is
zero, this eliminates A and B
E is eliminated because if Q = 0 and
W=0, then U would have to be zero,
thus no change in P could occur.
The internal energy must have
increased to double the pressure,
therefore U>0.
p0
D
V0
C
2V0
Question 5
What is true for the two step process A -> B -> C?
A)
B)
C)
D)
E)
U=0
U=0
W=0
W=0
W>0
Q=0
Q>0
Q>0
Q<0
Q<0
2p0
A
B
Since the area (work) under A -> B -> C
is not zero, this eliminates C and D
A is eliminated because if U=0 and
Q=0 then W must be 0, but it is not.
p0
C
D
V0
2V0
The work done on the gas is negative (the gas does positive work when
expanding), therefore Q>0, this eliminates D and E
The ideal gas law tells us that the temperature does not change
between A and C, so U=0. and since expanding Q>0.
Question 6
200 J enters a Carnot engine from the hot reservoir, held at 400 K. During the
entire engine cycle, 50 J of useful work is performed by the engine. What is the
temperature of the cold reservoir?
A)
B)
C)
D)
E)
100 K
200 K
300 K
250 K
150 K
e
From the definition of efficiency:
Wby
TC
e  1
TH
QH
e
Since this is a Carnot engine, we have
Wby
QH

50 J
 0.25
200 J
e  1
TC
TH
TC
400
TC  300 K
0.25  1 
Question 7
For an ideal gas in a container with fixed volume and constant temperature,
which of the following is true?
I. Pressure results from molecular collisions with the walls.
II. The molecules all have the same speed.
III. The average kinetic energy is directly proportional to the temperature
A)
B)
C)
D)
E)
I, II, and III
I and II only
II and III only
II only
I and III only
The distribution of speeds among the molecules,
leading to an average value, but the molecules
certainly do not all have the same speed.
Question 8
1 mole of ideal gas is in a container of volume V, temperature T, and pressure P.
If the volume is halved and the temperature doubled, the pressure will be:
A)
B)
C)
D)
E)
P
2P
4P
½P
¼P
Since an ideal gas:
PV  nRT
TR
V
2T  R

P2 
1 
 V
2 
TR
4
V
 4P
P
Question 9
Two identical containers contain 1 mole each of two different monatomic ideal
gases, gas A and gas B, with the mass of gas B 4 times the mass of gas A. Both
gases are at the same temperature, and 10 J of heat is added to gas A,
resulting in a temperature change T. How much heat must be added to gas B
to cause the same T?
A)
B)
C)
D)
E)
10 J
100 J
40 J
2.5 J
1600 J
For the internal
Energy of Ideal Gas
U 
3
nRT
2
For the same number of moles, T
depends only on the energy added, not
the mass of the gas molecules
Question 10
The entropy of a closed macroscopic system will never decrease because
A)
B)
C)
D)
E)
Energy wouldn’t be conserved if entropy decreased
For large systems, the probability of such a decrease is negligible
Mechanical equilibrium couldn’t be sustained with a decrease in entropy
Heat can never be made to flow from a cold object to a hot object
Molecular motions reach their minimum only at absolute 0
The probability of such a macro
state occurring is essentially 0
Question 11
During a certain process, 600 J of heat is added to a system. While this occurs,
the system performs 200 J of work on the surroundings. The change in internal
energy of the system is:
A)
B)
C)
D)
E)
200 J
800 J
400 J
- 400 J
Impossible to determine without knowing the temperature
The First Law of Thermodynamics
U  Q  W
U  600 J   200 J 
 400 J
Question 12
Which of the following is not true about thermal radiation?
A) It’s a mechanism of energy exchange between systems at different
temperatures
B) It involves atomic excitations
C) Ice at 00C will emit thermal radiation
D) It requires some material substance to travel through
E) It’s a by product of the food we eat
Electromagnetic radiation doesn’t need a material medium to travel through.
In E, we eat to maintain body temperature, and we emit a lot of energy as
thermal radiation
Question 13
To increase the diameter of an aluminium ring from 50.0 mm to 50.1 mm, the
temperature of the ring must increase by 800C. What temperature change would
be necessary to increase the diameter of an aluminium ring from 100.0 mm to
100.1 mm?
A)
B)
C)
D)
E)
200C
400C
800C
1100C
1600C
We apply the formula for length expansion:
0.1mm
80C 
  50.0mm 

0.1mm
80C  50.0mm 
 2.5 105
1
C
T 
Therefore
0.1mm

5 1 
 2.5 10
 100mm 

C


 40C
T 
L
 Lo
Question 14
A gas is enclosed in a metal container with a moveable piston on top. Heat is
added to the gas by placing a candle flame in contact with the container's
bottom. What of the following is true about the temperature of the gas?
A)
B)
C)
D)
E)
The temperature must go up if the piston remains stationary.
The temperature must go up if the piston is pulled out dramatically.
The temperature must go up no matter what happens to the piston.
The temperature must go down no matter what happens to the piston.
The temperature must go down if the piston is compressed dramatically.
Use the First Law of Thermodynamics, U=Q+W. The temperature adds
heat to the gas, so Q is positive. Internal energy is directly related to
temperature, so if U is positive, then the temperature goes up (and vice
versa). Here we can be sure that U is positive if the work done on the
gas is either positive or zero. The only possible answer is A. For B, the
work done on the gas is negative because the gas expands (Note:
because we add heat to a gas does NOT mean the temperature
automatically goes up)
Question 15
A small heat engine operates using a pan of 1000C boiling water as the high
temperature reservoir and the atmosphere as a low temperature reservoir.
Assuming ideal behaviour, how much more efficient is the engine on a cold, 00C
day than on a warm, 200C day?
A)
B)
C)
D)
E)
About 1.3 times as efficient.
2 times as efficient..
20 times as efficient.
Infinitely more efficient.
Just as efficient.
Using the Engine efficiency equation:
Hot Day:
Cold Day:
e  1
e  1
e  1
 20C  273C   0.21
100C  273C 
 0C  273C 
100C  273C 
 0.27
TC
TH
Ratio:
0.27
 1.3
0.21
Question 16
In three separate experiments, a gas is transformed from Pi, Vi to state Pf, Vf
along paths (1, 2, and 3) illustrated below. The work done on the gas is?
A)
B)
C)
D)
E)
Greatest for path 1
Least for path 2
The same for paths 1 and 3
The greatest for path 2
The same for all three paths
1
Pi
2
Pf
D
Vf
The work done on the gas during a
thermodynamic process is equal to the
area of the region in the P-V diagram
above the V axis and below the path the
system makes from its initial to its final
state.
3
Vi
Question 17
A1 m3 container contains 10 moles of ideal gas at room temperature. At what
fraction of atmospheric pressure is the gas inside the container?
A)
B)
C)
D)
E)
1/40 atm
1/20 atm
1/10 atm
¼ atm
½ atm
2p0
B
Using the Ideal Gas Law
P

A
PV  nRT
nRT
V
p0
C
D
V0
10 mol   8.31

N
 24930 2
m
J 
  300 K 
mol  K 
1m3
Ratio: 2V0
N
m 2  0.25
N
100000 2
m
24930
Question 18
An ideal gas is compressed isothermally from 20 L to 10 L. During this process,
5J of work done to compress the gas. What is the change of internal energy for
this gas?
A)
B)
C)
D)
E)
-10 J
-5J
0J
5J
10 J
The First Law of Thermodynamics
U  Q  W
During an isothermal process, U is ALWAYS ZERO
Question 19
Through a series of thermodynamic processes, the internal energy of a sample
of confined gas is increased by 560 J. If the net amount of work done on the
sample by its surroundings is 320 J, how much heat was transferred between
the gas and its environment?
A)
B)
C)
D)
E)
240 J absorbed
240 J dissipated
880 J absorbed
880 J dissipated
None of the above
The First Law of Thermodynamics
560 J  Q  320 J
Q  560 J  320 J
 240 J
U  Q  W
Question 20
In one of the steps of the Carnot cycle, the gas undergoes an isothermal
expansion. Which of the following statements is true concerning this step?
A) No heat is exchanged between the gas and its surrounding, because the
process is isothermal.
B) The temperature decreases because the gas expands.
C) This steps violates the Second Law of Thermodynamics because all the
heat absorbed is transferred into work.
D) The internal energy of the gas remains constant.
E) The internal energy of the gas decreases due to the expansion.
Statement A is wrong because “no heat exchanged between the gas and its
surroundings” is the definition of adiabatic, not isothermal.
Statement B cannot be correct since the step described in question is
isothermal; by definition, the temperature does not change.
Statement C is false, because although the heat absorbed is converted
completely to work, it does not include being returned back to it is initial state
and having 100% conversion to heat (P decreases and V increases).
Isothermal by definition states internal energy remains constant
Question 21
A Carnot engine using a monatomic ideal gas as a working substance operates
between two reservoirs held at 300 K and 200 K, respectively. Starting at point A
with pressure and volume as indicated on the graph, the gas expands
isothermally to point B, where the volume is 4 x 10-3 m3. 500 J of heat energy is
absorbed by the gas in this process.
A)
B)
C)
D)
E)
How many moles of gas are present?
Find the work done on the gas during A->B process.
Find the work done on the gas during the process B->C adiabatic expansion.
How much heat is expelled in the process C->D?
Determine the change in internal energy in the process D->A.
A
3x105
Isotherm at TH
B
Adiabat: Q=0
Adiabat: Q=0
D
C
Isotherm at TC
2x10-3
4x10-3
Question 21
A Carnot engine using a monatomic ideal gas as a working substance operates
between two reservoirs held at 300. K and 200. K, respectively. Starting at point A
with pressure and volume as indicated on the graph, the gas expands
isothermally to point B, where the volume is 4.0 x 10-3 m3. 500. J of heat energy
is absorbed by the gas in this process.
A)
B)
C)
D)
E)
How many moles of gas are present?
Find the work done on the gas during A->B process.
Find the work done on the gas during the process B->C adiabatic expansion.
How much heat is expelled in the process C->D?
Determine the change in internal energy in the process D->A.
A
Using the gas law and the given P,V, and T 3.0x105
R
PV
nT
Isotherm at TH
PV
RT

5 N 
3.0

10
2.0 103 m3 

2 
m 

J 

8.31

  300.K 
mol

K


 0.24 moles
n
B
Adiabat: Q=0
Adiabat: Q=0
D
C
Isotherm at TC
2.0x10-3
4.0x10-3
Question 21
A Carnot engine using a monatomic ideal gas as a working substance operates
between two reservoirs held at 300 K and 200 K, respectively. Starting at point A
with pressure and volume as indicated on the graph, the gas expands
isothermally to point B, where the volume is 4.0 x 10-3 m3. 500 J of heat energy is
absorbed by the gas in this process.
A)
B)
C)
D)
E)
How many moles of gas are present?
Find the work done on the gas during A->B process.
Find the work done on the gas during the process B->C adiabatic expansion.
How much heat is expelled in the process C->D?
Determine the change in internal energy in the process D->A.
A
3.0x105
Recall: if the temperature is constant,
at TH
QHIsotherm
 500 J
then so is the internal energy.
B
Adiabat: Q=0
Along A->B isotherm, there is no
change in internal energy, so the first
law gives us U = 0 = Q + W. But Q
is given as 500 J, so W = -500 J is
done ON the gas (the gas DID 500J of
work.
Adiabat: Q=0
D
C
Isotherm at TC
2.0x10-3
4.0x10-3
Question 21
A Carnot engine using a monatomic ideal gas as a working substance operates
between two reservoirs held at 300 K and 200 K, respectively. Starting at point A
with pressure and volume as indicated on the graph, the gas expands
isothermally to point B, where the volume is 4.0 x 10-3 m3. 500 J of heat energy is
absorbed by the gas in this process.
A)
B)
C)
D)
E)
How many moles of gas are present?
Find the work done on the gas during A->B process.
Find the work done on the gas during the process B->C adiabatic expansion.
How much heat is expelled in the process C->D?
Determine the change in internal energy in the process D->A.
Recall: if the process is adiabatic, there is no
heat transfer (Q=0). Therefore U=W.
A
3.0x105
Now we know the adiabat connects the two isotherms
where we know the temperature, and since the ideal
gas internal energy depends only upon temperature.
3
nRT
2
3
J 

  0.24moles   8.31
  200 K  300 K 
2
mol  K 

W= - 299J
 299 J
U 
Isotherm at TH
B
Adiabat: Q=0
Adiabat: Q=0
D
C
Isotherm at TC
2.0x10-3
4.0x10-3
Question 21
A Carnot engine using a monatomic ideal gas as a working substance operates
between two reservoirs held at 300 K and 200 K, respectively. Starting at point A
with pressure and volume as indicated on the graph, the gas expands
isothermally to point B, where the volume is 4.0 x 10-3 m3. 500 J of heat energy is
absorbed by the gas in this process.
A)
B)
C)
D)
E)
How many moles of gas are present?
Find the work done on the gas during A->B process.
Find the work done on the gas during the process B->C adiabatic expansion.
How much heat is expelled in the process C->D?
Determine the change in internal energy in the process D->A.
We are looking for QC
For the Carnot cycle: e  1 
A
3.0x105
TC
200 K
 1
 0.33
TH
300 K
Q
Also: e  1  C
QH
Isotherm at TH
B
Adiabat: Q=0
Adiabat: Q=0
D
C
Therefore:
QC
500 J
QC  333 J
0.33  1 
Isotherm at TC
2.0x10-3
4.0x10-3
Question 21
A Carnot engine using a monatomic ideal gas as a working substance operates
between two reservoirs held at 300 K and 200 K, respectively. Starting at point A
with pressure and volume as indicated on the graph, the gas expands
isothermally to point B, where the volume is 4.0 x 10-3 m3. 500 J of heat energy is
absorbed by the gas in this process.
A)
B)
C)
D)
E)
How many moles of gas are present?
Find the work done on the gas during A->B process.
Find the work done on the gas during the process B->C adiabatic expansion.
How much heat is expelled in the process C->D?
Determine the change in internal energy in the process D->A.
Once again we are connecting two
isotherms. Since the internal energy
depends only on temperature, we have
A
3.0x105
3
U  nRT
2
3
J 

  0.24moles   8.31
  300 K  200 K 
2
mol

K


 299 J
The opposite of B->C
Isotherm at TH
B
Adiabat: Q=0
Adiabat: Q=0
D
C
Isotherm at TC
2.0x10-3
4.0x10-3
Question 22
When a system is taken from state a to state b along the path acb shown in the figure
below, 70 J of heat flows into the system, and the system does 30 J of work.
A) When the system is returned from state b to state a along the curved path shown,
60 J of heat flows out of the system. Does the system perform work on its
surroundings or do the surroundings perform work on the system? How much
work is done?
B) If the system does 10 J of work in transforming from state a to state b along path
adb, does the system absorb or does it emit heat? How much heat is transferred?
C) If Ua=0 J and Ud=30 J, determine the heat absorbed in the process db and ad.
D) For the process adbca, identify each of the following quantities as positive,
negative, or zero.
W = _________________
Q
P
b
c
= _________________
U = _________________
d
a
V
Question 22
When a system is taken from state a to state b along the path acb shown in the figure
below, 70 J of heat flows into the system, and the system does 30 J of work.
A) When the system is returned from state b to state a along the curved path shown,
60 J of heat flows out of the system. Does the system perform work on its
surroundings or do the surroundings perform work on the system? How much
work is done?
We require the value of Wba
P
b
c
First let’s calculate Uabc
U acb  Qacb  Wacb
 70 J   30 J 
U acb  40 J
-30 J because
system does work
d
a
V
Because Ua->b does
not depend on path
taken from a to b, then
since Uab = 40J then
Uba = -Uab = – 40 J
U ba  Qba  Wba
40 J  60 J  Wba
Wba  20 J
Since the value is + 20J,
the surroundings does the
work on the gas (system).
Question 22
When a system is taken from state a to state b along the path acb shown in the figure
below, 70 J of heat flows into the system, and the system does 30 J of work.
B. If the system does 10 J of work in transforming from state a to state b along path
adb, does the system absorb or does it emit heat? How much heat is transferred?
P
b
c
Again, using the fact that Ua->b
does not depend on the path taken,
we know that Uadb = 40 J
Now we require Qadb
U adb  Qadb  Wadb
40 J  Qadb   10 J 
Qadb  50 J
d
a
Because the system
does 10 J of work
Therefore the system absorbs 50 J of heat
V
Question 22
When a system is taken from state a to state b along the path acb shown in the figure
below, 70 J of heat flows into the system, and the system does 30 J of work.
C) If Ua=0 J and Ud=30 J, determine the heat absorbed in the process db and ad.
For the process db, there is no change
in volume, therefore Wdb = 0J.
P
b
c
U db  Qdb  Wdb
U db  Qdb
d
a
Now, Uab = 40 J, and Ua = 0 J, tells us
that Ub = 40 J, therefore:
U ad  Qad  Wad
U db  U b  U d
 40 J  30 J
 10 J
Qdb  10 J
V
For ad:
Wadb=Wad+Wdb=Wad+0J=Wad
Since Wadb =-10 J, then Wad =-10J
U d  U a  Qad   10 J 
30 J  0 J  Qad   10 J 
Qad  40 J
Question 22
When a system is taken from state a to state b along the path acb shown in the figure
below, 70 J of heat flows into the system, and the system does 30 J of work.
D. For the process adbca, identify each of the following quantities as positive,
negative, or zero.
W = _________________
Q
P
b
c
= _________________
U = _________________
d
a
The process adbca is cyclic, so U = zero
Because the process is traversed counterclockwise in the PV diagram , we know that
W is positive (on the system).
U=Q+W therefore Q must be negative
V
The system is a
heat engine,
emitting heat on
each cycle
Question 23
1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted
below. All processes are performed slowly. Respond to the following in terms of P0, V0,
and R.
A)
B)
C)
D)
E)
Find the temperature at each vertex.
Find the heat added to the gas for the process A->B.
Find the work done on the gas for the process C->D.
Find the heat added to the gas fro the process D->A
Find the change in internal energy for
i. Process B->C
ii. The entire cycle
2P0
P0
A
B
C
D
V0
2V0
3V0
Question 23
1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted
below. All processes are performed slowly. Respond to the following in terms of P0, V0, and
R.
A) Find the temperature at each vertex.
B) Find the heat added to the gas for the process A->B.
C) Find the work done on the gas for the process C->D.
D) Find the heat added to the gas fro the process D->A
E) Find the change in internal energy for
i. Process B->C
ii. The entire cycle
From the Ideal Gas Law: T 
TA
2 P0 V0 


TB
 2 P0  2V0   4 PV
0 0

TC
P0  3V0  3PV


 0 0
PV
nR
2P0
A
B
R
R
TD 
R
 P0 V0 
R
P0
C
D
R
R
V0
2V0
3V0
Question 23
1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted
below. All processes are performed slowly. Respond to the following in terms of P0, V0, and
R.
A) Find the temperature at each vertex.
B) Find the heat added to the gas for the process A->B.
C) Find the work done on the gas for the process C->D.
D) Find the heat added to the gas fro the process D->A
E) Find the change in internal energy for
i. Process B->C
B
A
2P0
ii. The entire cycle
Process A->B occurs at constant pressure.
So we can use the First Law
U  Q  W
Q  U  W
3

  nRT   W
2

3
 nR TB  TA   W
2
3  4 PV 2 PV 
 R  0 0  0 0  W
2  R
R 
P0
Since Work
done on the
gas is negative
of the area
under graph
C
D
V0
2V0
3V0
2 PV
3  4 PV

0 0
Q  R
 0 0    2 PV
0 0
2  R
R 
 5PV
0 0
Question 23
1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted
below. All processes are performed slowly. Respond to the following in terms of P0, V0, and
R.
A) Find the temperature at each vertex.
B) Find the heat added to the gas for the process A->B.
C) Find the work done on the gas for the process C->D.
D) Find the heat added to the gas fro the process D->A
E) Find the change in internal energy for
i. Process B->C
B
A
2P0
ii. The entire cycle
The area under the process C->D
W   P V
  P0 V0  3V0 
 2 PV
0 0
The sign of the
net work is
negative for
cycles that run
clockwise
The decrease in volume corresponds
to positive work done on the gas
P0
C
D
V0
2V0
3V0
Question 23
1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted
below. All processes are performed slowly. Respond to the following in terms of P0, V0, and
R.
A) Find the temperature at each vertex.
B) Find the heat added to the gas for the process A->B.
C) Find the work done on the gas for the process C->D.
D) Find the heat added to the gas fro the process D->A
E) Find the change in internal energy for
i. Process B->C
B
A
2P0
ii. The entire cycle
Process D->A occurs at constant volume.
So we can use the First Law
U  Q  W
Q  U
Since at constant
volume, no work is done
3

  nRT 
2

3
 nR TA  TD 
2
3  2 PV PV 
 R 0 0  0 0 
2  R
R 
P0
V0
3
Q  PV
0 0
2
C
D
2V0
3V0
Question 23
1 mole of monatomic ideal gas is brought through a cycle A->B->C->D->A as depicted
below. All processes are performed slowly. Respond to the following in terms of P0, V0, and
R.
A) Find the temperature at each vertex.
B) Find the heat added to the gas for the process A->B.
C) Find the work done on the gas for the process C->D.
D) Find the heat added to the gas fro the process D->A.
E) Find the change in internal energy for
i. Process B->C
B
A
2P0
ii. The entire cycle
i) Using:
U 
3
nRT
2
3
R TC  TB 
2
3  3PV 4 PV 
 R 0 0  0 0 
2  R
R 
3
  PV
0 0
2
U 
P0
C
D
V0
2V0
ii) For the entire cycle, you end up back in the same state, so U = 0
3V0
Question 24
A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to
state b to state c and back to a along the cycle shown. Path ab is an isotherm and the
work done by the gas as it changes isothermally from state a to state b is given by the
equation:
5
V 
Wab  nRT ln  b 
 Va 
P (x10 Pa)
CP  29.1
J
mol  K
CV  20.8
J
mol  K
a
2.4
A) What is the temperature of:
i. State a
ii. State b
iii. State c
b
0.6
c
B) Determine the change in internal energy of the gas for:
i. Step ab.
ii. Step bc.
48
iii. Step ca.
12
C) How much work, Wab is done by the gas during step ab?
V (x10-3 m3)
D) What is the total work done over the cycle abca?
E) Is heat absorbed or discarded step ab?
F) If so how much?
G) What is the maximum possible efficiency (without violating the Second Law of
Thermodynamics) for a cyclical heat engine that operates between the
temperatures of state a and c?
Question 24
A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to
state b to state c and back to a along the cycle shown. Path ab is an isotherm and the
work done by the gas as it changes isothermally from state a to state b is given by the
equation:
5
V 
Wab  nRT ln  b 
 Va 
P (x10 Pa)
a
2.4
A) What is the temperature of:
i. State a
ii. State b
iii. State c
PV
T

From the Ideal Gas Law:
nR
2.4 10 Pa 12 10


5
TA
 0.4 mol   8.31

TB  870 K
m3 
J 

mol  K 
 870 K
State b is on the Isotherm with state a
0.6 10 Pa 12 10


5
TC
3
3
m3 
J 
 0.4 mol   8.31

mol  K 

 220 K
0.6
c
12
b
48
V (x10-3 m3)
Question 24
A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to
state b to state c and back to a along the cycle shown. Path ab is an isotherm and the
work done by the gas as it changes isothermally from state a to state b is given by the
equation:
5
V 
Wab  nRT ln  b 
 Va 
P (x10 Pa)
CP  29.1
J
mol  K
CV  20.8
J
mol  K
a
2.4
B) Determine the change in internal energy of the gas for:
i. Step ab.
ii. Step bc.
iii. Step ca.
Since step a takes place along an isotherm, Uab = 0
U bc  Qbc  Wbc
  nC p Tbc     PVbc 
0.6
c
12
b
48
V (x10-3 m3)
J 

5
3 3
3 3
  0.4mol   29.1
  220 K  870 K    0.6 10 Pa 12 10 m  48 10 m 
mol  K 

55
U
TvTT 
UbccaU canR
nRnC
 5400 J
22
U aa  U ab  U bc  U ca
55   0.4mol
   20.8 JJ J    870 K  220 K 
or
 0.4
mol
K 870
220
0.4mol8.31
8.31
870
220K
K
K
mol



or
2
mol

K
2
mol  K 
0  0  5400 J  U ca

U ca  5400 J
 5400
ca
U
540
0JJ J
UbccaU5400
Question 24
A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to
state b to state c and back to a along the cycle shown. Path ab is an isotherm and the
work done by the gas as it changes isothermally from state a to state b is given by the
equation:
5
V 
Wab  nRT ln  b 
 Va 
P (x10 Pa)
C) How much work, Wab is done by the gas during step ab?
 48 103 m3 
J 

   0.4mol   8.31
  870 K  ln 
3 3 
mol  K 
12

10
m 


 4000 J
J
mol  K
CV  20.8
J
mol  K
a
2.4
V 
Wab  nRT ln  b 
 Va 
CP  29.1
0.6
c
12
b
48
V (x10-3 m3)
Question 24
A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to
state b to state c and back to a along the cycle shown. Path ab is an isotherm and the
work done by the gas as it changes isothermally from state a to state b is given by the
equation:
5
V 
Wab  nRT ln  b 
 Va 
P (x10 Pa)
D) What is the total work done over the cycle abca?
0.6
c
12
Wcycle  Wab  Wbc  Wac
  4000 J    2200 J   0 J
 1800 J
J
mol  K
CV  20.8
J
mol  K
a
2.4
The total work done over a cycle is equal
to the sum of the values of the work done
over each step.
CP  29.1
b
48
V (x10-3 m3)
Question 24
A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to
state b to state c and back to a along the cycle shown. Path ab is an isotherm and the
work done by the gas as it changes isothermally from state a to state b is given by the
equation:
5
V 
Wab  nRT ln  b 
 Va 
P (x10 Pa)
J
mol  K
CV  20.8
J
mol  K
a
2.4
E) Is heat absorbed or discarded step ab?
F) If so how much?
By the First Law of Thermodynamics
CP  29.1
U  Q  W
0.6
U ab  Qab  Wab
0  Qab  Wab
Qab  Wab
 4000 J
Since Q is positive, this represents heat absorbed by the gas.
c
12
b
48
V (x10-3 m3)
Question 24
A 0.4 mol sample of an ideal diatomic gas undergoes slow changes from state a to state
b to state c and back to a along the cycle shown. Path ab is an isotherm and the work
done by the gas as it changes isothermally from state a to state b is given by the
equation:
V 
Wab  nRT ln  b 
 Va 
G) What is the maximum possible efficiency (without violating the Second Law of
Thermodynamics) for a cyclical heat engine that operates between the temperatures
of state a and c?
P (x105 Pa)
The maximum possible efficiency is the efficiency
of the Carnot engine.
eC  1 
TC
TH
220 K
870 K
 75%
a
2.4
0.6
c
b
 1
12
48
V (x10-3 m3)
Question 25
You have just purchased a cup of coffee (45 g), but you notice that the coffee is cold, so
you call over your host. “It’s 45oC”, you say, “But I like it 65oC. “. The coffee in the pot is
95oC, how much coffee should the host pour to raise the temperature of the 45g of
coffee in your cup?
Look at the heat lost by the pot’s coffee and the heat gained by your coffee
Q  cmT
Coffee in cup:
Coffee in pot:
Q1  cm1 T1 f  T1i 
Q2  cm2 T2 f  T2i 
Now,
Q1  Q2
cm1  65  45   cm2  65  95 
 45 g  20   m2  30 
m2  30 g