Thermodynamics Part 2

Let $p_{i}, p_{f}, V_{i}$ and $V_{f}$ be the initial and final pressure and volume. Given, $A B$ is isothermal $(\Delta T=0)$, $B C$ is isochoric $(\Delta V=0)$ and $C A$ is adiabatic $(\Delta Q=0)$ Since, isothermal work $(W_{AB}) = p_{1} V_{1} \ln \frac{V_{f}}{V_{i}}$where, $V_{i}$ and $V_{f}$ are volume at $A$ and $B$, respectively. $\therefore \;\; W_{AB} = p_{1} V_{1} \ln \frac{2 V_{1}}{V_{1}} = p_{1} V_{1} \ln 2$ Since, at constant volume, work done is zero. $\therefore \;\; W_{BC} = 0$ Since, $W_{CA}$ is an adiabatic work done, i.e. $W_{CA} \;\; = \frac{1}{1-\gamma} (p_{f} V_{f} - p_{i} V_{i})$ $\Rightarrow \;\; W_{CA} \;\; = \frac{1}{1-\gamma} \left(p_{1} V_{1} - \frac{p_{1}}{4} \times 2 V_{1}\right)$ $\;\; = \frac{1}{1-\gamma} \left(p_{1} V_{1} - \frac{p_{1} V_{1}}{2}\right) = \frac{1}{1-\gamma} \frac{p_{1} V_{1}}{2}$ $\therefore$ Net work done, $W_{\text{net }} = W_{AB} + W_{BC} + W_{CA}$ $\;\; = p_{1} V_{1} \ln 2 + 0 + \frac{1}{1-\gamma} \frac{p_{1} V_{1}}{2}$ $\;\; = p_{1} V_{1} \left[\ln 2 + \frac{1}{2(1-\gamma)}\right]$ From ideal gas law, $p V = n R T$ $\therefore W_{\text{net }} = R T \left[\ln 2 - \frac{1}{2(\gamma-1)}\right]$ $(\because n=1)$