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Date May 2018 Marks available 1 Reference code 18M.3.SL.TZ1.7
Level Standard level Paper Paper 3 Time zone Time zone 1
Command term Calculate Question number 7 Adapted from N/A

Question

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.

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.

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

[2]
a.

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

[1]
b.i.

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

[1]
b.ii.

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

[1]
b.iii.

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

[2]
c.i.

Determine, without any calculation, whether the net work done by the engine during one full cycle would increase or decrease.

[2]
c.ii.

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

[1]
d.

Markscheme

«\({p_1}V_1^{\frac{5}{3}} = {p_2}V_2^{\frac{5}{3}}\)»

\(1.1 \times {10^5} \times {5^{\frac{5}{3}}} = {p_2} \times {2^{\frac{5}{3}}}\)

p2 «= \(\frac{{1.1 \times {{10}^5} \times {5^{\frac{5}{3}}}}}{{{{2.5}^{\frac{5}{3}}}}}\)» = 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]

a.

«W = pΔV»

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

= 1.52 × 103 «J»

 

Award [0] if POT mistake.

[1 mark]

b.i.

ΔU = \(\frac{3}{2}\)pΔV = \(\frac{3}{2}\)5.07 × 105 × 3 × 10–3 = 2.28 × 10–3 «J»

 

Accept alternative solution via Tc.

[1 mark]

b.ii.

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

 

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

[1 mark]

b.iii.

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

since VC > VB, PC must be smaller than PB

[2 marks]

c.i.

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]

c.ii.

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

 

Allow any sensible answer.

[1 mark]

d.

Examiners report

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

Syllabus sections

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