thermodynamics practice questions. question 1 1 mole of ideal gas is brought to a final state f by...
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Thermodynamics
Practice Questions
Question 11 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) It’s the same for all processes.B) It’s the smallest for process 1.C) It’s the smallest for process 2.D) It’s the smallest for process 3.E) It’s the same for processes 1 and 2.
The temperature for gas at any point is given by:
V0 2V0
p0
2p0
3p0
1
2
3F
PV TR
PVT
R
Since all of the three processes end at the same temperature, we need only look at the process that starts closest to:
0 0 0 02
2P V PV
TR R
0 0 0 0
0 0 0 0
0 0 0 0
31. 3
22. 2
3.
P V PV
R RP V PV
R RP V PV
R R
Process 2 actually begins at this temperature, so its change is 0
Question 21 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) It’s positive for process 1 and 2, but negative for process 3.B) It’s the smallest for process 1.C) It’s the smallest for process 2.D) It’s the smallest for process 3.E) Zero work is done along process 3.
Work done by the gas is equal to the area under the curve
V0 2V0
p0
2p0
3p0
1
2
3F
Question 3An 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 temperarureB) 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.
W= - Q for each process, which eliminates C
F
S
T
R
V
P
Area considerations eliminate B and E
Temperature at R is greater than the temperature at S (gas law) eliminates A
The internal energy of an ideal gas depends only on temperature. If the initial and final temp are the same, there is no change in internal energy. U=0 for all three paths.
The internal energy of an ideal gas depends only on temperature
Question 4What is true for the process D -> A?
A) U=0 Q>0.B) U=0 Q<0C) W=0 U>0 D) W=0 U<0 E) W=0 Q=0
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.
V0 2V0
p0
2p0
C
A
B
D
The internal energy must have increased to double the pressure, therefore U>0.
Question 5What is true for the two step process A -> B -> C?
A) U=0 Q=0B) U=0 Q>0C) W=0 Q>0 D) W=0 Q<0 E) W>0 Q<0
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.
V0 2V0
p0
2p0
C
A
B
D
The ideal gas law tells us that the temperature does not change between A and C, so U=0. and since expanding Q>0.
The work done on the gas is negative (the gas does positive work when expanding), therefore Q>0, this eliminates D and E
Question 6200 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) 100 KB) 200 KC) 300 KD) 250 K E) 150 K
Since this is a Carnot engine, we have
From the definition of efficiency:50
0.25200
by
H
W Je
Q J
1
0.25 1400
300
C
H
C
C
Te
T
T
T K
by
H
We
Q 1 C
H
Te
T
Question 7For 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) I, II, and IIIB) I and II onlyC) II and III onlyD) II only E) 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 81 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) PB) 2PC) 4PD) ½ PE) ¼ P Since an ideal gas: PV nRT
2
2
12
4
4
TRP
VT R
PV
TR
VP
Question 9Two 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) 10 JB) 100 JC) 40 JD) 2.5 JE) 1600 J
For the same number of moles, T depends only on the energy added, not the mass of the gas molecules
3
2U nR T For the internal
Energy of Ideal Gas
Question 10The entropy of a closed macroscopic system will never decrease because
A) Energy wouldn’t be conserved if entropy decreasedB) For large systems, the probability of such a decrease is negligibleC) Mechanical equilibrium couldn’t be sustained with a decrease in entropyD) Heat can never be made to flow from a cold object to a hot objectE) Molecular motions reach their minimum only at absolute 0
The probability of such a macro state occurring is essentially 0
Question 11During 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) 200 JB) 800 JC) 400 JD) - 400 JE) Impossible to determine without knowing the temperature
U Q W The First Law of Thermodynamics
600 200
400
U J J
J
Question 12Which 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 excitationsC) Ice at 00C will emit thermal radiationD) It requires some material substance to travel throughE) 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 13To 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) 200C B) 400C C) 800C D) 1100C E) 1600C
We apply the formula for length expansion:o
LT
L
5
0.180
50.0
0.1
80 50.0
12.5 10
mmC
mm
mm
C mm
C
5
0.11
2.5 10 100
40
mmT
mmC
C
Therefore
Question 14A 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) The temperature must go up if the piston remains stationary. B) The temperature must go up if the piston is pulled out dramatically.C) The temperature must go up no matter what happens to the piston. D) The temperature must go down no matter what happens to the piston.E) 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 15A 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) About 1.3 times as efficient. B) 2 times as efficient..C) 20 times as efficient.D) Infinitely more efficient.E) Just as efficient.
Using the Engine efficiency equation: 1 C
H
Te
T
Hot Day: 20 273
1 0.21100 273
C Ce
C C
Cold Day:
0 273
1 0.27100 273
C Ce
C C
Ratio:
0.271.3
0.21
Question 16In 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) Greatest for path 1 B) Least for path 2C) The same for paths 1 and 3D) The greatest for path 2E) The same for all three paths
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.
Vf Vi
Pf
Pi
2
1
3
D
Question 17A1 m3 container contains 10 moles of ideal gas at room temperature. At what fraction of atmospheric pressure is the gas inside the container?
A) 1/40 atm B) 1/20 atmC) 1/10 atmD) ¼ atmE) ½ atm
Using the Ideal Gas Law PV nRT
3
2
10 8.31 300
1
24930
nRTP
VJ
mol Kmol Km
N
m
Ratio:
2
2
249300.25
100000
NmNm
V0 2V0
p0
2p0
C
A
B
D
Question 18An 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) -10 JB) - 5 JC) 0 JD) 5 JE) 10 J
U Q W The First Law of Thermodynamics
During an isothermal process, U is ALWAYS ZERO
Question 19Through 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) 240 J absorbedB) 240 J dissipatedC) 880 J absorbedD) 880 J dissipatedE) None of the above
U Q W The First Law of Thermodynamics
560 320
560 320
240
J Q J
Q J J
J
Question 20In 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 21A 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) How many moles of gas are present?B) Find the work done on the gas during A->B process.C) Find the work done on the gas during the process B->C adiabatic expansion.D) How much heat is expelled in the process C->D?E) Determine the change in internal energy in the process D->A.
2x10-3 4x10-3
3x105
CD
B
A
Isotherm at TH
Isotherm at TC
Adiabat: Q=0Adiabat: Q=0
Question 21A 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) How many moles of gas are present?B) Find the work done on the gas during A->B process.C) Find the work done on the gas during the process B->C adiabatic expansion.D) How much heat is expelled in the process C->D?E) Determine the change in internal energy in the process D->A.
2.0x10-3 4.0x10-3
3.0x105
CD
B
AUsing the gas law and the given P,V, and T
PVR
nT
5 3 323.0 10 2.0 10
8.31 300.
0.24
PVn
RTN
mmJ
Kmol K
moles
Isotherm at TH
Isotherm at TC
Adiabat: Q=0Adiabat: Q=0
Question 21A 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) How many moles of gas are present?B) Find the work done on the gas during A->B process.C) Find the work done on the gas during the process B->C adiabatic expansion.D) How much heat is expelled in the process C->D?E) Determine the change in internal energy in the process D->A.
2.0x10-3 4.0x10-3
3.0x105
CD
B
A
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.
500HQ JRecall: if the temperature is constant, then so is the internal energy. Isotherm at TH
Isotherm at TC
Adiabat: Q=0Adiabat: Q=0
Question 21A 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) How many moles of gas are present?B) Find the work done on the gas during A->B process.C) Find the work done on the gas during the process B->C adiabatic expansion.D) How much heat is expelled in the process C->D?E) Determine the change in internal energy in the process D->A.
2.0x10-3 4.0x10-3
3.0x105
CD
B
A
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.
Recall: if the process is adiabatic, there is no heat transfer (Q=0). Therefore U=W.
3
230.24 8.31 200 300
2
299
U nR T
Jmoles K K
mol K
J
W= - 299J
Isotherm at TH
Isotherm at TC
Adiabat: Q=0Adiabat: Q=0
Question 21A 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) How many moles of gas are present?B) Find the work done on the gas during A->B process.C) Find the work done on the gas during the process B->C adiabatic expansion.D) How much heat is expelled in the process C->D?E) Determine the change in internal energy in the process D->A.
2.0x10-3 4.0x10-3
3.0x105
CD
B
A
For the Carnot cycle:
We are looking for QC
0.33 1500
333
C
C
Q
JQ J
Also:
2001 1 0.33
300C
H
T Ke
T K
1 C
H
Qe
Q
Therefore:
Isotherm at TH
Isotherm at TC
Adiabat: Q=0Adiabat: Q=0
Question 21A 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) How many moles of gas are present?B) Find the work done on the gas during A->B process.C) Find the work done on the gas during the process B->C adiabatic expansion.D) How much heat is expelled in the process C->D?E) Determine the change in internal energy in the process D->A.
2.0x10-3 4.0x10-3
3.0x105
CD
B
AOnce again we are connecting two isotherms. Since the internal energy
depends only on temperature, we have
The opposite of B->C
3
230.24 8.31 300 200
2
299
U nR T
Jmoles K K
mol K
J
Isotherm at TH
Isotherm at TC
Adiabat: Q=0Adiabat: Q=0
Question 22When 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 = _________________
U = _________________
V
Pbc
da
Question 22When 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?
V
Pbc
da
First let’s calculate Uabc
70 30
40
acb acb acb
acb
U Q W
J J
U J
-30 J because system does work
Because Ua->b does not depend on path taken from a to b, then since Uab = 40J then Uba = -Uab = – 40 J
40 60
20
ba ba ba
ba
ba
U Q W
J J W
W J
Since the value is + 20J, the surroundings does the work on the gas (system).
We require the value of Wba
Question 22When 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?
V
Pbc
da
Again, using the fact that Ua->b does not depend on the path taken, we know that Uadb = 40 J
Now we require Qadb
40 10
50
adb adb adb
adb
adb
U Q W
J Q J
Q J
Therefore the system absorbs 50 J of heat
Because the system does 10 J of work
Question 22When 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.
V
Pbc
da
For the process db, there is no change in volume, therefore Wdb = 0J.
db db db
db db
U Q W
U Q
Now, Uab = 40 J, and Ua = 0 J, tells us that Ub = 40 J, therefore:
40 30
10
10
db b d
db
U U U
J J
J
Q J
For ad: Wadb=Wad+Wdb=Wad+0J=Wad
Since Wadb =-10 J, then Wad =-10J
10
30 0 10
40
ad ad ad
d a ad
ad
ad
U Q W
U U Q J
J J Q J
Q J
Question 22When 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 = _________________
U = _________________
V
Pbc
da
The process adbca is cyclic, so U = zero
Because the process is traversed counter-clockwise in the PV diagram , we know that W is positive (on the system).
U=Q+W therefore Q must be negative
The system is a heat engine, emitting heat on each cycle
Question 231 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->AE) Find the change in internal energy for
i. Process B->Cii. The entire cycle
V0 2V0
P0
2P0
C
A B
D
3V0
Question 231 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->AE) Find the change in internal energy for
i. Process B->Cii. The entire cycle
V0 2V0
P0
2P0
C
A B
D
3V0
From the Ideal Gas Law:PV
TnR
0 02A
P VT
R
0 0 0 02 2 4
B
P V PVT
R R
0 0 0 03 3
C
P V PVT
R R
0 0D
P VT
R
Question 231 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->AE) Find the change in internal energy for
i. Process B->Cii. The entire cycle
V0 2V0
P0
2P0
C
A B
D
3V0
Process A->B occurs at constant pressure. So we can use the First Law
U Q W
0 0 0 0
3
2
3
24 23
2
B A
Q U W
nR T W
nR T T W
PV PVR W
R R
Since Work done on the gas is negative of the area under graph
0 0 0 00 0
0 0
4 232
2
5
PV PVQ R PV
R R
PV
Question 231 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->AE) Find the change in internal energy for
i. Process B->Cii. The entire cycle
V0 2V0
P0
2P0
C
A B
D
3V0
The area under the process C->D
The decrease in volume corresponds to positive work done on the gas
0 0 0
0 0
3
2
W P V
P V V
PV
The sign of the net work is negative for cycles that run clockwise
Question 231 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->AE) Find the change in internal energy for
i. Process B->Cii. The entire cycle
V0 2V0
P0
2P0
C
A B
D
3V0
Process D->A occurs at constant volume. So we can use the First Law
U Q W
0 0 0 0
3
2
3
223
2
A D
Q U
nR T
nR T T
PV PVR
R R
0 0
3
2Q PV
Since at constant volume, no work is done
Question 231 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->Cii. The entire cycle
V0 2V0
P0
2P0
C
A B
D
3V0
i) Using:3
2U nR T
0 0 0 0
0 0
3
23 3 4
2
3
2
C BU R T T
PV PVR
R R
PV
ii) For the entire cycle, you end up back in the same state, so U = 0
Question 24A 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:
A) What is the temperature of:i. State aii. State biii. State c
B) Determine the change in internal energy of the gas for:i. Step ab.ii. Step bc.iii. Step ca.
C) How much work, Wab is done by the gas during step ab?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?
b
a
c
V (x10-3 m3)
P (x105 Pa)
2.4
0.6
12 48
ln bab
a
VW nRT
V
29.1P
JC
mol K
20.8V
JC
mol K
Question 24A 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:
A) What is the temperature of:i. State aii. State biii. State c
b
a
c
V (x10-3 m3)
P (x105 Pa)
2.4
0.6
12 48
ln bab
a
VW nRT
V
From the Ideal Gas Law:PV
TnR
5 3 32.4 10 12 10870
0.4 8.31A
Pa mT K
Jmol
mol K
870BT K
5 3 30.6 10 12 10220
0.4 8.31C
Pa mT K
Jmol
mol K
State b is on the Isotherm with state a
Question 24A 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:
B) Determine the change in internal energy of the gas for:i. Step ab.ii. Step bc.iii. Step ca.
b
a
c
V (x10-3 m3)
P (x105 Pa)
2.4
0.6
12 48
ln bab
a
VW nRT
V
Since step a takes place along an isotherm, Uab = 0
5 3 3 3 30.4 29.1 220 870 0.6 10 12 10 48 10
5400
p bc b
bc bc bc
cnC T P V
Jmol K K Pa m m
l K
U
m
J
Q W
o
0 0 5400
5400
aa ab bc
ca
ca
caU U U U
J U
U J
29.1P
JC
mol K
20.8V
JC
mol K
or
50.4 8.31 870 220
2
5400
5
2ca
ca
Jmol K K
mol
U n T
K
U J
R
0.4 20.8 870 220
5400
ca v
ca
U nC T
Jmol K K
mol K
U J
50.4 8.31
5
220 870
0
2
2
540
bc
bc
Jmol K K
mol
U nR T
K
U J
or
Question 24A 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:
C) How much work, Wab is done by the gas during step ab?
b
a
c
V (x10-3 m3)
P (x105 Pa)
2.4
0.6
12 48
ln bab
a
VW nRT
V
29.1P
JC
mol K
20.8V
JC
mol K
3 3
3 3
ln
48 100.4 8.31 870 ln
12 10
4000
bab
a
VW nRT
V
J mmol K
mol K m
J
Question 24A 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:
D) What is the total work done over the cycle abca?
b
a
c
V (x10-3 m3)
P (x105 Pa)
2.4
0.6
12 48
ln bab
a
VW nRT
V
29.1P
JC
mol K
20.8V
JC
mol K
The total work done over a cycle is equal to the sum of the values of the work done over each step.
4000 2200 0
1800
cycle ab bc acW W W W
J J J
J
Question 24A 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:
E) Is heat absorbed or discarded step ab?F) If so how much?
b
a
c
V (x10-3 m3)
P (x105 Pa)
2.4
0.6
12 48
ln bab
a
VW nRT
V
29.1P
JC
mol K
20.8V
JC
mol K
By the First Law of Thermodynamics
0
4000
ab ab ab
ab ab
ab ab
U Q W
Q W
Q W
J
U Q W
Since Q is positive, this represents heat absorbed by the gas.
Question 24A 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:
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?
b
a
c
V (x10-3 m3)
P (x105 Pa)
2.4
0.6
12 48
ln bab
a
VW nRT
V
The maximum possible efficiency is the efficiency of the Carnot engine.
1
2201870
75%
CC
H
Te
T
K
K
Question 25You 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
1 1 1 1f iQ cm T T
2 2 2 2f iQ cm T T
Coffee in cup:
Coffee in pot:Now, 1 2Q Q
1 2
2
2
65 45 65 95
45 20 30
30
cm cm
g m
m g
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