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Entropy IB Practice Qs

1a. [2 marks]

The pV diagram of a heat engine using an ideal gas consists of an isothermal expansion A → B, an

isobaric compression B → C and an adiabatic compression C → A.

The following data are available:

Temperature at A = 385 K

Pressure at A = 2.80 × 106 Pa

Volume at A = 1.00 × 10–4 m3

Volume at B = 2.80 × 10–4 m3

Volume at C = 1.85 × 10–4 m3

Show that at C the pressure is 1.00 × 106 Pa.

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Markscheme

ALTERNATIVE 1:

✔

= «= 1.00 × 106 Pa» ✔

ALTERNATIVE 2:

✔

«= 1.00 × 106 Pa» ✔

1b. [2 marks]

Show that at C the temperature is 254 K.

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Markscheme

ALTERNATIVE 1:

Since TB = TA then Tc = ✔

= «= 254.4 K» ✔

ALTERNATIVE 2:

«K»✔

«= 254.4 K» ✔

1c. [3 marks]

Show that the thermal energy transferred from the gas during the change B → C is 238 J.

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Markscheme

work done = «pΔV = 1.00 × 106 × (1.85 × 10−4 − 2.80 × 10−4 =» −95 «J» ✔

change in internal energy = « pΔV = − × 95 =» −142.5 «J» ✔

Q = −95 − 142.5 ✔

«−238 J»

Allow positive values.

1d. [2 marks]

The work done by the gas from A → B is 288 J. Calculate the efficiency of the cycle.

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Markscheme

net work is 288 −238 = 50 «J» ✔

efficiency = « =» 0.17 ✔

1e. [1 mark]

State, without calculation, during which change (A → B, B → C or C → A) the entropy of the gas

decreases.

Markscheme

along B→C ✔

2a. [2 marks]

The pressure–volume (pV) diagram shows a cycle ABCA of a heat engine. The working substance of the

engine is 0.221 mol of ideal monatomic gas.

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At A the temperature of the gas is 295 K and the pressure of the gas is 1.10 × 105 Pa. The process from A

to B is adiabatic.

Show that the pressure at B is about 5 × 105 Pa.

Markscheme

« »

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p2 «= » = 5.066 × 105 «Pa»

Volume may be in litres or m3.

Value to at least 2 sig figs, OR clear working with substitution required for mark.

[2 marks]

2b. [1 mark]

For the process BC, calculate, in J, the work done by the gas.

Markscheme

«W = pΔV»

«= 5.07 × 105 × (5 × 10–3 – 2 × 10–3)»

= 1.52 × 103 «J»

Award [0] if POT mistake.

[1 mark]

2c. [1 mark]

For the process BC, calculate, in J, the change in the internal energy of the gas.

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Markscheme

ΔU = pΔV = 5.07 × 105 × 3 × 10–3 = 2.28 × 10–3 «J»

Accept alternative solution via Tc.

[1 mark]

2d. [1 mark]

For the process BC, calculate, in J, the thermal energy transferred to the gas.

Markscheme

Q «= (1.5 + 2.28) × 103 =» 3.80 × 103 «J»

Watch for ECF from (b)(i) and (b)(ii).

[1 mark]

2e. [2 marks]

The process from B to C is replaced by an isothermal process in which the initial state is the same and

the final volume is 5.00 × 10–3 m3.

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Explain, without any calculation, why the pressure after this change would belower if the process was

isothermal.

Markscheme

for isothermal process, PV = constant / ideal gas laws mentioned

since VC > VB, PC must be smaller than PB

[2 marks]

2f. [2 marks]

Determine, without any calculation, whether the net work done by the engine during one full cycle

would increase or decrease.

Markscheme

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the area enclosed in the graph would be smaller

so the net work done would decrease

Award MP2 only if MP1 is awarded.

[2 marks]

2g. [1 mark]

Outline why an efficiency calculation is important for an engineer designing a heat engine.

Markscheme

to reduce energy loss; increase engine performance; improve mpg etc

Allow any sensible answer.

[1 mark]

3a. [2 marks]

A cylinder is fitted with a piston. A fixed mass of an ideal gas fills the space above the piston.

The gas expands isobarically. The following data are available.

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Show that the final volume of the gas is about 53 m3.

Markscheme

ALTERNATIVE 1

«Using »

V2 =

V2 = 52.7 «m3»

ALTERNATIVE 2

«Using PV = nRT»

V =

V = 52.6 «m3»

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[2 marks]

3b. [2 marks]

Calculate, in J, the work done by the gas during this expansion.

Markscheme

W «= PΔV» = 11.2 × 103 × (52.7 – 47.1)

W = 62.7 × 103 «J»

Accept 66.1 × 103 J if 53 used

Accept 61.6 × 103 J if 52.6 used

[2 marks]

3c. [3 marks]

Determine the thermal energy which enters the gas during this expansion.

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Markscheme

ΔU «= nRΔT» = 1.5 × 243 × 8.31 × (19 – (–12)) = 9.39 × 104

Q «= ΔU + W» = 9.39 × 104 + 6.27 × 104

Q = 1.57 × 105 «J»

Accept 1.60 × 105 if 66.1 × 103 J used

Accept 1.55 × 105 if 61.6 × 103 J used

[3 marks]

3d. [2 marks]

The gas returns to its original state by an adiabatic compression followed by cooling at constant

volume.

Sketch, on the pV diagram, the complete cycle of changes for the gas, labelling the changes clearly. The

expansion shown in (a) and (b) is drawn for you.

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Markscheme

concave curve from RHS of present line to point above LHS of present line

vertical line from previous curve to the beginning

[2 marks]

3e. [1 mark]

Outline the change in entropy of the gas during the cooling at constant volume.

Markscheme

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energy is removed from the gas and so entropy decreases

OR

temperature decreases «at constant volume (less disorder)» so entropy decreases

OWTTE

[1 mark]

3f. [1 mark]

There are various equivalent versions of the second law of thermodynamics. Outline the benefit gained

by having alternative forms of a law.

Markscheme

different paradigms/ways of thinking/modelling/views

allows testing in different ways

laws can be applied different situations

OWTTE

[1 mark]

4a. [6 marks]

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A fixed mass of an ideal monatomic gas undergoes an isothermal change from A to B as shown.

The temperature at A is 350 K. An identical mass of the same ideal monatomic gas undergoes an

isobaric change from A to C.

(i) Calculate the temperature at C.

(ii) Calculate the change in internal energy for AC.

(iii) Determine the energy supplied to the gas during the change AC.

(iv) On the graph, draw a line to represent an adiabatic expansion from A to a state of volume

4.0×10−3m3 (point D).

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Markscheme

(i) 1400 «K»

(ii)

1800 J

(iii) 1800+PΔV=1800+4×105×3×10−3 OR use of

3000 J

(iv) curve starting at A ending on line CB AND between B and zero pressure

4b. [4 marks]

(i) State the change in entropy of a gas for the adiabatic expansion from A to D.

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(ii) Explain, with reference to the concept of disorder, why the entropy of the gas is greater at C than B.

Markscheme

(i) 0

(ii)

ALTERNATIVE 1

C has the same volume as B OR entropy is related to disorder

higher temperature/pressure means greater disorder

therefore entropy at C is greater «because entropy is related to disorder»

ALTERNATIVE 2

to change from B to C, ΔQ > 0

so ΔS > 0

ΔS related to disorder

5. [1 mark]

Which of the following can be deduced from the second law of thermodynamics?

A. Thermal energy cannot spontaneously transfer from a low temperature region to a high temperature

region.

B. Thermal energy cannot spontaneously transfer from a high temperature region to a low temperature

region.

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C. The entropy of an isolated system always decreases with time.

D. The entropy of an isolated system is the measure of the internal energy of the system.

Markscheme

A

6. [1 mark]

A system consists of a refrigerator with its door open operating in a thermally insulated room. What are

the change in the entropy of the system and the change in temperature of the room?

Markscheme

D

7. [1 mark]

A block of ice at 0°C is placed on a surface and allowed to melt completely to give water at 0°C. During

this process the entropy of the

A. molecules in the block has decreased.

B. surroundings has increased.

C. universe has increased.

D. universe has decreased.

Markscheme

C

8a. [3 marks]

This question is about an ideal gas cycle.

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The P–V diagram shows a cycle ABCDA for a fixed mass of an ideal gas.

Estimate the total work done in the cycle.

Markscheme

use of area under the curve;

each (1 cm × 1 cm) square has energy of 250 J or each small square has energy of 10 J;

estimate (14 to 16 × 250) = 3500 to 4000 J;

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8b. [3 marks]

The change AB is isothermal and occurs at a temperature of 420K. Calculate the number of moles of

gas.

Markscheme

clear use of value on AB; (must see correct values)

use of PV = nRT;

0.56 to 0.60 mol;

8c. [2 marks]

Identify and explain the change, if any, in the entropy of the gas when it has completed one cycle.

Markscheme

entropy unchanged;

gas returned to original state;

9. [1 mark]

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The pressure–volume (P–V) graph shows an adiabatic compression of a fixed mass of an ideal gas.

Which of the following correctly describes what happens to the temperature and the internal energy of

the gas during the compression?

Markscheme

D

10. [1 mark]

Which of the following correctly describes the entropy changes of the water molecules and the universe

when a sample of water freezes?

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Markscheme

B

11. [1 mark]

The entropy of a system

A. will decrease if the system’s temperature is increased.

B. is related to the degree of disorder in the system.

C. must always increase.

D. is always conserved.

Markscheme

B

12a. [3 marks]

Part 2 Thermodynamics

A fixed mass of an ideal gas undergoes the three thermodynamic processes, AB, BC and CA, represented

in the P–V graph below.

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State which of the processes is isothermal, isochoric (isovolumetric) or isobaric.

Process AB:

Process BC:

Process CA:

Markscheme

process AB: isobaric;

process BC: isochoric / isovolumetric;

process CA: isothermal;

12b. [2 marks]

The temperature of the gas at A is 300K. Calculate the temperature of the gas at B.

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Markscheme

use of ;

to give 750 K;

12c. [2 marks]

The increase in internal energy of the gas during process AB is 4100J. Determine the heat transferred to

the gas from the surroundings during the process AB.

Markscheme

W=(PΔV=9×105×[5-2]×10-3=)2700;

Q=ΔU+W=4100+2700=6800J;

12d. [3 marks]

The gas is compressed at constant temperature. Explain what changes, if any, occur to the entropy of

(i) the gas.

(ii) the surroundings.

(iii) the universe.

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Markscheme

(i) same number of molecules occupy smaller volume, so disorder and hence entropy decrease;

(ii) heat flows from gas to surroundings, so temperature of surroundings increases, so entropy

increases;

(iii) from second law, entropy of universe increases during any process;

Award [0] for simple statement of entropy increases/decreases.

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