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Engines and Carnot Cycle Markscheme

1a. [1 mark]

The P–V diagram of the Carnot cycle for a monatomic ideal gas is shown.

State what is meant by an adiabatic process.

Markscheme

«a process in which there is» no thermal energy transferred between the system and the surroundings

[1 mark]

1b. [1 mark]

Identify the two isothermal processes.

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Markscheme

A to B AND C to D

[1 mark]

1c. [2 marks]

The system consists of 0.150 mol of a gas initially at A. The pressure at A is 512 k Pa and the volume is

1.20 × 10–3 m3.

Determine the temperature of the gas at A.

Markscheme

«K»

The first mark is for rearranging.

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[2 marks]

1d. [2 marks]

The volume at B is 2.30 × 10–3 m3. Determine the pressure at B.

Markscheme

The first mark is for rearranging.

[2 marks]

1e. [1 mark]

At C the volume is VC and the temperature is TC.

Show that

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Markscheme

«B to C adiabatic so» AND PCVC = nRTC «combining to get result»

It is essential to see these 2 relations to award the mark.

[1 mark]

1f. [2 marks]

The volume at C is 2.90 × 10–3 m3. Calculate the temperature at C.

Markscheme

« » = 422 «K»

[2 marks]

1g. [1 mark]

State a reason why a Carnot cycle is of little use for a practical heat engine.

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Markscheme

the isothermal processes would have to be conducted very slowly / OWTTE

[1 mark]

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.

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.

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Markscheme

« »

p2 «= » = 5.066 × 105 «Pa»

Volume may be in liters 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

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«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.

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

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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.

Explain, without any calculation, why the pressure after this change would be lower 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.

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Markscheme

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]

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3a. [1 mark]

Part 2 Gas in an engine

A fixed mass of an ideal gas is used as the working substance in an engine. The graph shows the

variation with volume V of the pressure P of the fluid.

For the cycle identify, with the letter I, an isochoric (isovolumetric) change.

Markscheme

vertical line identified;

Do not penalize candidate who identifies both vertical lines.

Award [0] if vertical and another line identified.

3b. [2 marks]

The temperature at point X is 310 K. Calculate the temperature at point Y.

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Markscheme

at point X: ;

at point Y:

Award [2] for a bald correct answer.

Must be clear what units are here, ºC unacceptable unless converted correctly.

3c. [5 marks]

The shaded area WXYZ is 610 J. The total thermal energy transferred out of the gas in one cycle is 1.3 kJ.

(i) State what is represented by the shaded area WXYZ.

(ii) Determine the efficiency of cycle WXYZ.

(iii) Explain why the total thermal energy transferred out of the gas is degraded.

Markscheme

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(i) (net) work done by engine in one cycle; } (must imply “by engine” and “one cycle”)

(ii) efficiency= ;

0.47 or 47%;

Award [2] for a bald correct answer.

Award [1] for =0.32.

Award [0] for any fraction that would exceed 1 even if later fudged.

Award [0] for efficiency =38%, question does not imply that this is a Carnot cycle.

(iii) it has gone to the surroundings / it is dissipated;

it cannot be used for further useful work;

3d. [2 marks]

The work done on the gas during the adiabatic compression XY is 210 J. Determine the change in

internal energy during the change from X to Y.

Markscheme

ΔQ = 0;

210 (J); (do not accept negative sign)

4a. [3 marks]

Part 2 Thermodynamic cycles

A gas undergoes a thermodynamic cycle. The P–V diagram for the cycle is shown below.

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In the changes of state B to C and D to A, the gas behaves as an ideal gas and the changes in state are

adiabatic.

(i) State the circumstances in which the behaviour of a gas approximates to ideal gas behaviour.

(ii) State what is meant by an adiabatic change of state.

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Markscheme

(i) low pressure;

high temperature;

(ii) no thermal/heat energy is transferred (in change of state);

Allow “heat energy” but not “heat”.

4b. [4 marks]

With reference to the first law of thermodynamics, explain for the change of state A to B, why energy is

transferred from the surroundings to the gas.

Markscheme

work is done (by the gas) because there is an increase in volume/gas expands;

so W is positive;

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ΔU is greater than zero (because P is constant and V increases);

from first law Q=ΔU+W means that Q is positive which means energy transferred into gas;

4c. [3 marks]

Estimate the total work done in the cycle.

Markscheme

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total work done = enclosed area / number of large squares ∼40(±5);

1 square=5 J;

work done=200J (±25) J;

5a. [4 marks]

Part 2 A heat engine

The piston of an engine contains a fixed mass of an ideal gas. During one cycle of the engine, the gas

undergoes the thermodynamic processes shown below.

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(i) State what is meant by an isothermal process.

(ii) Show that process AB is isothermal.

Markscheme

(i) a process in which temperature remains constant;

(ii) calculation to show that pV=16(×102) at any point from A to B;

calculation to show that pV=16(×102) at any other point from A to B;

pV is constant;

5b. [1 mark]

State the nature of process BC.

Markscheme

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isochoric / isovolumetric / occurs at constant volume;

5c. [2 marks]

During the cycle ABCD, the net work done by the gas is 550J. Calculate the net thermal energy absorbed

by the gas.

Markscheme

;

Q=W=550(J);

5d. [3 marks]

Explain why it is not possible for this engine, operating in this cycle, to be 100% efficient.

Markscheme

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the gas/system must return to original conditions (P, V, and T);

(to do that) some of the thermal energy absorbed by gas must be given off to the surroundings;

hence not all thermal energy absorbed by gas can be converted to work;

it is the second law of thermodynamics;

6a. [2 marks]

A monatomic ideal gas is confined to a cylinder with volume 2.0 x 10–3 m3. The initial pressure of the

gas is 100 kPa. The gas undergoes a three-step cycle. First, the gas pressure increases by a factor of five

under constant volume. Then, the gas expands adiabatically to its initial pressure. Finally it is

compressed at constant pressure to its initial volume.

Show that the volume of the gas at the end of the adiabatic expansion is approximately 5.3 x 10–3 m3.

Markscheme

V = 5.3 x 10–3 «m3»

Look carefully for correct use of pVγ = constant

6b. [2 marks]

Using the axes, sketch the three-step cycle.

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Markscheme

correct vertical and horizontal lines

curve between B and C

Allow tolerance ±1 square for A, B and C

Allow ECF for MP2

Points do not need to be labelled for marking points to be awarded

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6c. [2 marks]

The initial temperature of the gas is 290 K. Calculate the temperature of the gas at the start of the

adiabatic expansion.

Markscheme

use of PV = nRT OR use of = constant

T = «5 x 290 =» 1450 «K»

6d. [2 marks]

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Using your sketched graph in (b), identify the feature that shows that net work is done by the gas in this

three-step cycle.

Markscheme

area enclosed

work is done by the gas during expansion

OR

work is done on the gas during compression

the area under the expansion is greater than the area under the compression

7a. [1 mark]

A heat engine operates on the cycle shown in the pressure–volume diagram. The cycle consists of an

isothermal expansion AB, an isovolumetric change BC and an adiabatic compression CA. The volume at

B is double the volume at A. The gas is an ideal monatomic gas.

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At A the pressure of the gas is 4.00 x 106 Pa, the temperature is 612 K and the volume is 1.50 x 10–4 m3.

The work done by the gas during the isothermal expansion is 416 J.

Justify why the thermal energy supplied during the expansion AB is 416 J.

Markscheme

ΔU = 0 so Q = ΔU + W = 0 + 416 = 416 «J»

Answer given, mark is for the proof.

[1 mark]

7b. [2 marks]

Show that the temperature of the gas at C is 386 K.

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Markscheme

ALTERNATIVE 1

use to get

hence

«TC ≈ 386K»

ALTERNATIVE 2

giving PC = 1.26 x 106 «Pa»

giving «K»

«TC ≈ 386K»

Answer of 386K is given. Look carefully for correct working if answers are to 3 SF.

There are other methods:

Allow use of PB = 2 x 106 «Pa» and is constant for BC.

Allow use of n = 0.118 and TC =

[2 marks]

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7c. [2 marks]

Show that the thermal energy removed from the gas for the change BC is approximately 330 J.

Markscheme

«–332 J»

Answer of 330 J given in the question.

Look for correct working or more than 2 SF.

[2 marks]

7d. [2 marks]

Determine the efficiency of the heat engine.

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Markscheme

e = 0.20

Allow .

Allow e = 0.21.

[2 marks]

7e. [3 marks]

State and explain at which point in the cycle ABCA the entropy of the gas is the largest.

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Markscheme

entropy is largest at B

entropy increases from A to B because T = constant but volume increases so more disorder or ΔS =

and Q > 0 so ΔS > 0

entropy is constant along CA because it is adiabatic, Q = 0 and so ΔS = 0

OR

entropy decreases along BC since energy has been removed, ΔQ < 0 so ΔS < 0

[3 marks]

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