A sample of a monatomic ideal gas is taken from state A, with volume 1.0 L and pressure 3P₀, to…

Step 1. Work (a). The area under the line is a trapezium: $W = \tfrac12(3P_0 + P_0)(2V_0) = 4P_0V_0 = 400\ \text{J}$. At A and B, $PV = 3P_0V_0$, so the temperature at B is also $300\ \text{K}$, and $\Delta U = 0$.

Step 2. Highest temperature (b). $PV = aV - bV^2$ is largest at $V = \dfrac{a}{2b} = 2V_0 = 2.0\ \text{L}$, where $P = 2P_0$ and $PV = 4P_0V_0$, against $3P_0V_0$ at A. So $T_{\max} = 300\ \text{K} \times \dfrac{4}{3} = 400\ \text{K}$.

Step 3. Where heat stops going in (c). Along the line,

This is positive until $V = \dfrac{5a}{8b} = 2.5V_0 = 2.5\ \text{L}$ (where $P = 1.5P_0$), and negative after it.

Step 4. Heat in and out (d). Integrating $\left(\tfrac52 a - 4bV\right)dV$: from $V_0$ to $2.5V_0$ it gives $450\ \text{J}$ taken in, and from $2.5V_0$ to $3V_0$ it gives $-50\ \text{J}$, that is $50\ \text{J}$ given out. (In units of $P_0V_0 = 100\ \text{J}$: $4.5$ in and $0.5$ out.)