| 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.
For the process BC, calculate, in J, the work done by the gas.
For the process BC, calculate, in J, the change in the internal energy of the gas.
For the process BC, calculate, in J, the thermal energy transferred to the gas.
Explain, without any calculation, why the pressure after this change would belower if the process was isothermal.
Determine, without any calculation, whether the net work done by the engine during one full cycle would increase or decrease.
Outline why an efficiency calculation is important for an engineer designing a heat engine.
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]
«W = pΔV»
«= 5.07 × 105 × (5 × 10–3 – 2 × 10–3)»
= 1.52 × 103 «J»
Award [0] if POT mistake.
[1 mark]
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]
Q «= (1.5 + 2.28) × 103 =» 3.80 × 103 «J»
Watch for ECF from (b)(i) and (b)(ii).
[1 mark]
for isothermal process, PV = constant / ideal gas laws mentioned
since VC > VB, PC must be smaller than PB
[2 marks]
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]
to reduce energy loss; increase engine performance; improve mpg etc
Allow any sensible answer.
[1 mark]
