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.

«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 P_CV_C = nRT_C «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]