Thermodynamics - Michael Burns Practice Questions.pdf · same temperature, we need only look at ......

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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 R

P V PV

R R

P 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 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.

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

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Question 4What is true for the process D -> A?

A) U=0 Q>0.

B) U=0 Q<0

C) 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=0

B) U=0 Q>0

C) 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 K

B) 200 K

C) 300 K

D) 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 III

B) I and II only

C) II and III only

D) 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.

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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) P

B) 2P

C) 4P

D) ½ P

E) ¼ P Since an ideal gas: PV nRT

2

2

1

2

4

4

TRP

V

T RP

V

TR

V

P

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 J

B) 100 J

C) 40 J

D) 2.5 J

E) 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 decreased

B) For large systems, the probability of such a decrease is negligible

C) Mechanical equilibrium couldn’t be sustained with a decrease in entropy

D) Heat can never be made to flow from a cold object to a hot object

E) 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 J

B) 800 J

C) 400 J

D) - 400 J

E) Impossible to determine without knowing the temperature

U Q W The First Law of Thermodynamics

600 200

400

U J J

J

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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 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 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

0.1

12.5 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 2731 0.21

100 273

C Ce

C C

Cold Day:

0 2731 0.27

100 273

C Ce

C C

Ratio:

0.271.3

0.21

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Question 16In three separate experiments, a gas is transformed from P i, 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 2

C) The same for paths 1 and 3

D) The greatest for path 2

E) 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 atm

C) 1/10 atm

D) ¼ atm

E) ½ atm

Using the Ideal Gas Law PV nRT

3

2

10 8.31 300

1

24930

nRTP

V

Jmol K

mol K

m

N

m

Ratio:

2

2

24930

0.25

100000

N

mN

m

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 J

B) - 5 J

C) 0 J

D) 5 J

E) 10 J


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 absorbed

B) 240 J dissipated

C) 880 J absorbed

D) 880 J dissipated

E) None of the above


560 320

560 320

240

J Q J

Q J J

J

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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


between two reservoirs held at 300. K and 200. K, respectively. Starting at point A


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.






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 3

23.0 10 2.0 10

8.31 300.

0.24

PVn

RT

Nm

m

JK

mol K

moles

Isotherm at TH

Isotherm at TC





isothermally to point B, where the volume is 4.0 x 10-3 m3. 500 J of heat energy is







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.

500HQ J

Recall: if the temperature is constant,

then so is the internal energy. Isotherm at TH

Isotherm at TC


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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

2

30.24 8.31 200 300

2

299

U nR T

Jmoles K K

mol K

J

W= - 299J

Isotherm at TH

Isotherm at TC












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

J

Q J

Also:

2001 1 0.33

300

C

H

T Ke

T K

1 C

H

Qe

Q

Therefore:

Isotherm at TH

Isotherm at TC












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

2

30.24 8.31 300 200

2

299

U nR T

Jmoles K K

mol K

J

Isotherm at TH

Isotherm at TC


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

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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.

We require the value of Wba



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



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



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.

U=Q+W therefore Q must be negative

The system is a

heat engine,

emitting heat on

each cycle

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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->C

ii. The entire cycle

V0 2V0

P0

2P0

C

A B

D

3V0


below. All processes are performed slowly. Respond to the following in terms of P0, V0, and

R.






i. Process B->C


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 0

D

P VT

R



R.






i. Process B->C


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

2

4 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



R.






i. Process B->C


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

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R.






i. Process B->C


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

2

23

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



R.




D) Find the heat added to the gas fro the process D->A.


i. Process B->C


V0 2V0

P0

2P0

C

A B

D

3V0

i) Using:3

2U nR T

0 0 0 0

0 0

3

2

3 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 a

ii. State b

iii. 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




equation:

A) What is the temperature of:

i. State a

ii. State b

iii. State cb

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

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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

bc bc bc

p bc bc

U Q W

nC T P V

Jmol K K Pa m m

mol K

J

0 0 5400

5400

aa ab bc ca

ca

ca

U U U U

J U

U J

29.1P

JC

mol K

20.8V

JC

mol K

or

5

2

50.4 8.31 870 220

2

5400

ca

ca

U nR T

Jmol K K

mol K

U J

0.4 20.8 870 220

5400

ca v

ca

U nC T

Jmol K K

mol K

U J




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




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




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.

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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

2201

870

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

Thermodynamics - Michael Burns Practice Questions.pdf · same temperature, we need only look at ......

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