The temperature of gas at state ‘C’ is — Thermodynamics and Thermochemistry Chemistry Question
Question
The temperature of gas at state ‘C’ is
💡 Solution & Explanation
Since DA is isothermal expansion, $T_A = T_D = 800\text{ K}$ (max temp). The minimum temp occurs at B (after isobaric compression from A), so $T_B = 800 / 4 = 200\text{ K}$. $V_A$ is max volume, $V_C$ is min volume (after adiabatic compression). $V_A/V_C = 8\sqrt{2}$. Since $P_A = P_B$, $V_A/V_B = T_A/T_B = 4$. Thus $V_B/V_C = (V_A/4) / (V_A/8\sqrt{2}) = 2\sqrt{2} = 2^{1.5}$. For adiabatic BC, $T_B V_B^{\gamma-1} = T_C V_C^{\gamma-1}$. Monoatomic $\gamma-1 = 2/3$. $T_C = T_B (V_B/V_C)^{2/3} = 200 \times (2^{1.5})^{2/3} = 200 \times 2 = 400\text{ K}$.