Thermodynamic cycle

The thermodynamic cycle can be represented in the pressure-volume diagram by the following curve: The points $A = (V_1, p_1)$ and $B = (V_2, p_1)$ have been given directly. The pressure at the point $C = (V_2, p_2)$ results from the fact that we reach the initial state $A$ in the last step by isothermal expansion. Thus $p_2 V_2 = p_1 V_1$ and hence $p_2 = \frac {V_1} {V_2} p_1 = 2 \, \text {bar}$ .

The mechanical work of the gas corresponds to the area enclosed by the curve in the pressure-volume diagram and is calculated by the integral $\begin{aligned} W &= \oint p dV \\ &= \int_{A \to B} p dV + \int_{B \to C} p dV + \int_{C \to A} p dV \\ &= p_1 (V_2 - V_1) + 0 + \int_{V_2}^{V_1} \frac{p_1 V_1}{V} dV \\ &= - \frac{1}{2} p_1 V_1 + p_1 V_1 \log \frac{V_1}{V_2} \\ &= p_1 V_1 \left( \log(2) - \frac{1}{2} \right) \\ &= 1 \,\text{bar} \cdot 16 \,\text{L} \cdot \left( \log(2) - \frac{1}{2} \right) \\ &= 10^5 \,\text{Pa} \cdot 0.016 \,\text{m}^3 \cdot \left( \log(2) - \frac{1}{2} \right) \\ &\approx 309 \,\text{J} \end{aligned}$