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Date May 2017 Marks available 1 Reference code 17M.3.SL.TZ1.6
Level Standard level Paper Paper 3 Time zone Time zone 1
Command term Identify Question number 6 Adapted from N/A

Question

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

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.

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

State what is meant by an adiabatic process.

[1]
a.

Identify the two isothermal processes.

[1]
b.

Determine the temperature of the gas at A.

[2]
c.i.

The volume at B is 2.30 × 10–3\(\,\)m3. Determine the pressure at B.

[2]
c.ii.

Show that \({P_B}V_B^{\frac{5}{3}} = nR{T_C}V_C^{\frac{2}{3}}\)

[1]
d.i.

The volume at C is 2.90 × 10–3\(\,\)m3. Calculate the temperature at C.

[2]
d.ii.

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

[1]
e.

Markscheme

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

[1 mark]

a.

A to B AND C to D

[1 mark]

b.

\(T = \frac{{PV}}{{nR}}\)

\(T\left( { = \frac{{512 \times {{10}^3} \times 1.20 \times {{10}^{ - 3}}}}{{0.150 \times 8.31}}} \right) \approx 493\) «K»

 

The first mark is for rearranging.

[2 marks]

c.i.

\({P_B} = \frac{{{P_a}{V_A}}}{{{V_B}}}\)

\({P_B} = 267{\text{ KPa}}\)

 

The first mark is for rearranging.

[2 marks]

c.ii.

«B to C adiabatic so» \({P_{\text{B}}}V_{\text{B}}^{\frac{5}{3}} = {P_{\text{C}}}V_{\text{C}}^{\frac{5}{3}}\) AND PCVC = nRTC «combining to get result»

 

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

[1 mark]

d.i.

\({T_{\text{C}}} = \left( {\frac{{{P_{\text{B}}}V_{\text{B}}^{\frac{5}{3}}}}{{nR}}} \right)V_{\text{C}}^{\frac{{ - 2}}{3}}\)

\({T_{\text{C}}} = \) «\(\left( {\frac{{267 \times {{10}^3} \times {{\left( {2.30 \times {{10}^{ - 3}}} \right)}^{\frac{5}{3}}}}}{{0.150 \times 8.31}}} \right){\left( {2.90 \times {{10}^{ - 3}}} \right)^{\frac{{ - 2}}{3}}}\)» = 422 «K»

[2 marks]

d.ii.

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

[1 mark]

e.

Examiners report

[N/A]
a.
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b.
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c.i.
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c.ii.
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d.i.
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d.ii.
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e.

Syllabus sections

Option B: Engineering physics » Option B: Engineering physics (Core topics) » B.2 – Thermodynamics
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