| Date | May 2015 | Marks available | 2 | Reference code | 15M.2.HL.TZ2.7 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Calculate, Determine, Explain, Identify, and State | Question number | 7 | Adapted from | N/A |
Question
Part 2 Gas in an engine
A fixed mass of an ideal gas is used as the working substance in an engine. The graph shows the variation with volume V of the pressure P of the fluid.
For the cycle identify, with the letter I, an isochoric (isovolumetric) change.
The temperature at point X is 310 K. Calculate the temperature at point Y.
The shaded area WXYZ is 610 J. The total thermal energy transferred out of the gas in one cycle is 1.3 kJ.
(i) State what is represented by the shaded area WXYZ.
(ii) Determine the efficiency of cycle WXYZ.
(iii) Explain why the total thermal energy transferred out of the gas is degraded.
The work done on the gas during the adiabatic compression XY is 210 J. Determine the change in internal energy during the change from X to Y.
Markscheme
vertical line identified;
Do not penalize candidate who identifies both vertical lines.
Award [0] if vertical and another line identified.
at point X: \(nR = \frac{{1.42 \times 1 \times 100}}{{310}} = 0.4581\);
at point Y: \(T = \left( {\frac{{0.44 \times 5.2 \times 100}}{{0.4581}} = } \right)500{\rm{K}}\)
Award [2] for a bald correct answer.
Must be clear what units are here, ºC unacceptable unless converted correctly.
(i) (net) work done by engine in one cycle; } (must imply “by engine” and “one cycle”)
(ii) efficiency=\(\frac{{610}}{{1300}}\);
0.47 or 47%;
Award [2] for a bald correct answer.
Award [1] for \(\frac{{610}}{{\left( {610 + 1300} \right)}}\)=0.32.
Award [0] for any fraction that would exceed 1 even if later fudged.
Award [0] for efficiency \(=\frac{{{T_C} - {T_H}}}{{{T_H}}}\)=38%, question does not imply that this is a Carnot cycle.
it cannot be used for further useful work;
ΔQ = 0;
210 (J); (do not accept negative sign)
