Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 34D / b

For $N$ particles of a classical monatomic ideal gas, the equation of state and internal energy areOnly translational degrees of freedom contribute; the particle number is fixed throughout the thermodynamic cycle.

For a reversible change, the first law of thermodynamics is $TdS=dU+pdV$. Using the monatomic internal energy,Along the constant-volume leg, $Q_{AB}=\frac{3}{2}{V}_0(2p_0-p_0)=\frac{3}{2}{p}_0{V}_0>0$, so heat is absorbed. Along the constant-pressure return leg, $Q_{CA}=\frac{5}{2}{p}_0(V_0-2V_0)=-\frac{5}{2}{p}_0{V}_0$, so heat is emitted. The work signs are zero on $AB$ and negative, or work on the gas, on $CA$.

On the straight sloping leg, $p=p_0(3-V/V_0)$ and $dp=-p_{0}dV/V_0$. Put $v=V/V_0$, which increases from $1$ to $2$. The reversible heat transfer isThus heat is absorbed for $1<v<15/8$ and emitted for $15/8<v<2$, with zero instantaneous heat at $v=15/8$. The switching point has $(p,V)=(9p_0/8,15V_0/8)$. In particular the net heat on this leg, $\frac{3}{2}{p}_0{V}_0$, is not the total positive heat on it.

The total work by the gas is the enclosed triangular area, $W=\frac{1}{2}{p}_0{V}_0$. The positive heat on the sloping leg isAdd $Q_{AB}=48p_0{V}_0/32$, excluding the emitting part of $BC$ and all of $CA$. HenceAs an independent conservation of energy check, the heat emitted on late $BC$ is $p_0{V}_0/32$ and on $CA$ is $80p_0{V}_0/32$; thus $Q_{\mathrm{in}}-Q_{\mathrm{out}}=16p_0{V}_0/32=W$. Using only the net heat on $BC$ would give an incorrect thermal efficiency.