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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 1
Command term State 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 5 3 = n R T C V C 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 = P V n R

T ( = 512 × 10 3 × 1.20 × 10 3 0.150 × 8.31 ) 493  «K»

 

The first mark is for rearranging.

[2 marks]

c.i.

P B = P a V A V B

P B = 267  KPa

 

The first mark is for rearranging.

[2 marks]

c.ii.

«B to C adiabatic so» P B V B 5 3 = P C V C 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 C = ( P B V B 5 3 n R ) V C 2 3

T C =  « ( 267 × 10 3 × ( 2.30 × 10 3 ) 5 3 0.150 × 8.31 ) ( 2.90 × 10 3 ) 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.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
e.

Syllabus sections

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

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