| Date | May 2017 | Marks available | 2 | Reference code | 17M.3.SL.TZ1.6 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Determine | 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.
Identify the two isothermal processes.
Determine the temperature of the gas at A.
The volume at B is 2.30 × 10–3\(\,\)m3. Determine the pressure at B.
Show that \({P_B}V_B^{\frac{5}{3}} = nR{T_C}V_C^{\frac{2}{3}}\)
The volume at C is 2.90 × 10–3\(\,\)m3. Calculate the temperature at C.
State a reason why a Carnot cycle is of little use for a practical heat engine.
Markscheme
«a process in which there is» no thermal energy transferred between the system and the surroundings
[1 mark]
A to B AND C to D
[1 mark]
\(T = \frac{{PV}}{{nR}}\)
\(T\left( { = \frac{{512 \times {{10}^3} \times 1.20 \times {{10}^{ - 3}}}}{{0.150 \times 8.31}}} \right) \approx 493\) «K»
The first mark is for rearranging.
[2 marks]
\({P_B} = \frac{{{P_a}{V_A}}}{{{V_B}}}\)
\({P_B} = 267{\text{ KPa}}\)
The first mark is for rearranging.
[2 marks]
«B to C adiabatic so» \({P_{\text{B}}}V_{\text{B}}^{\frac{5}{3}} = {P_{\text{C}}}V_{\text{C}}^{\frac{5}{3}}\) AND PCVC = nRTC «combining to get result»
It is essential to see these 2 relations to award the mark.
[1 mark]
\({T_{\text{C}}} = \left( {\frac{{{P_{\text{B}}}V_{\text{B}}^{\frac{5}{3}}}}{{nR}}} \right)V_{\text{C}}^{\frac{{ - 2}}{3}}\)
\({T_{\text{C}}} = \) «\(\left( {\frac{{267 \times {{10}^3} \times {{\left( {2.30 \times {{10}^{ - 3}}} \right)}^{\frac{5}{3}}}}}{{0.150 \times 8.31}}} \right){\left( {2.90 \times {{10}^{ - 3}}} \right)^{\frac{{ - 2}}{3}}}\)» = 422 «K»
[2 marks]
the isothermal processes would have to be conducted very slowly / OWTTE
[1 mark]
