Thermodynamics: PYQ Walkthrough (8)

For the isothermal expansion at $T$ from $V$ to $2V$:

$W_1 = nRT \ln\frac{2V}{V} = RT \ln 2$

For isochoric cooling: $W_2 = 0$ (no volume change).

For adiabatic compression returning to start: $W_3 = -\Delta U_3 = -nC_v(T - T/2) = -\frac{3}{2}R \cdot \frac{T}{2} \cdot (-1) \cdot (-1)$

Wait — let’s be careful. $W_{\text{adiabatic}} = -\Delta U = -nC_v\Delta T$. The gas goes from $T/2$ back up to $T$, so $\Delta T = T/2$ and $W_3 = -\frac{3}{2}R \cdot \frac{T}{2} = -\frac{3RT}{4}$.

$W_{\text{net}} = W_1 + W_2 + W_3 = RT\ln 2 + 0 - \frac{3RT}{4} = RT\left(\ln 2 - \frac{3}{4}\right)$

Numerically, $\ln 2 \approx 0.693$, so $W_{\text{net}} \approx RT(0.693 - 0.75) \approx -0.057\, RT$.

Heat is absorbed only during the isothermal expansion (step 1), where $Q_1 = W_1 = RT\ln 2$ (since $\Delta U = 0$ in isothermal). Heat is released in the isochoric cooling.

$\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = \frac{RT(\ln 2 - 3/4)}{RT\ln 2} = 1 - \frac{3}{4\ln 2} \approx 1 - 1.082 \approx -0.082$

The negative net work signals that the cycle as described actually requires net work input — it’s not a heat engine. This is the trap in the question.