Past exam of the physics B course of the University of Cambridge / 2014 / Paper 2 / Section C / C10 / Solution (cycle calculations)

Put $P_{\ast }=p_1{V}_1=nR{T}_1$. For the specified ideal gas, $C_V=3nR/2$, $C_p=5nR/2$ and $\gamma =5/3$. Isobaric expansion doubles temperature, so $T_2=2T_1$. Reversible adiabatic closure gives $T{V}^{\gamma -1}=\mathrm{constant}$, henceThere is no heat input on that compression. The other stages giveTherefore $W_{\mathrm{net}}^{\mathrm{by}}=P_{\ast }-W_{31}^{\mathrm{on}}=0.44494P_{\ast }$ andThe hot reservoir changes entropy by $-Q_{12}/(2.5T_1)=-P_{\ast }/T_1$. The cold reservoir changes by $-Q_{23}/(0.5T_1)=4.11012P_{\ast }/T_1$. Since the gas completes a cycle,Although the mechanical paths can be quasistatic, heat flows across finite temperature differences in both contacts. This entropy generation explains why the efficiency is well below the Carnot value $1-0.5/2.5=80\%$ between these reservoirs.

For an isothermal final compression at $T_1$, the preceding constant-volume cooling must instead end at $T_3=T_1$; keeping the old state 3 would not close the cycle. Compression then requires work $P_{\ast }\ln 2$, while the heat input remains $5P_{\ast }/2$. Thuswhich is lower: this isothermal compression occurs at a higher temperature throughout than the original adiabat and consumes more work. It also rejects heat during compression.