| Date | May 2017 | Marks available | 2 | Reference code | 17M.3.SL.TZ2.7 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Show that | Question number | 7 | Adapted from | N/A |
Question
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.
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.
Show that the temperature of the gas at C is 386 K.
Show that the thermal energy removed from the gas for the change BC is approximately 330 J.
Determine the efficiency of the heat engine.
State and explain at which point in the cycle ABCA the entropy of the gas is the largest.
Markscheme
ΔU = 0 so Q = ΔU + W = 0 + 416 = 416 «J»
Answer given, mark is for the proof.
[1 mark]
ALTERNATIVE 1
use \(p{V^{\frac{5}{3}}} = c\) to get \(T{V^{\frac{2}{3}}} = c\)
hence \({T_{\text{C}}} = {T_{\text{A}}}{\left( {\frac{{{V_{\text{A}}}}}{{{V_{\text{C}}}}}} \right)^{\frac{2}{3}}} = 612 \times {0.5^{\frac{2}{3}}} = 385.54\)
«TC ≈ 386K»
ALTERNATIVE 2
\({P_{\text{C}}}{V_{\text{C}}}^\gamma = {P_{\text{A}}}{V_{\text{A}}}^\gamma \) giving PC = 1.26 x 106 «Pa»
\(\frac{{{P_{\text{C}}}{V_{\text{C}}}}}{{{T_{\text{C}}}}} = \frac{{{P_{\text{A}}}{V_{\text{A}}}}}{{{T_{\text{A}}}}}\) giving \({T_{\text{C}}} = 1.26 \times \frac{{612}}{2} = 385.54\) «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 \(\frac{P}{T}\) is constant for BC.
Allow use of n = 0.118 and TC = \(\frac{{{P_{\text{C}}}{V_{\text{C}}}}}{{nR}}\)
[2 marks]
\(Q = \Delta U + W = \frac{3}{2}\frac{{{P_{\text{A}}}{V_{\text{A}}}}}{{{T_{\text{A}}}}}\Delta T + 0\)
\(Q = \frac{3}{2} \times \frac{{4.00 \times {{10}^6} \times 1.50 \times {{10}^{ - 4}}}}{{612}} \times \left( {386 - 612} \right)\)
«–332 J»
Answer of 330 J given in the question.
Look for correct working or more than 2 SF.
[2 marks]
\({\text{e}} = \frac{{{Q_{{\text{in}}}} - {Q_{{\text{out}}}}}}{{{Q_{i{\text{n}}}}}} = \frac{{412 - 332}}{{416}}\)
e = 0.20
Allow \(\frac{{416 - 330}}{{416}}\).
Allow e = 0.21.
[2 marks]
entropy is largest at B
entropy increases from A to B because T = constant but volume increases so more disorder or ΔS = \(\frac{Q}{T}\) 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]
