A polyatomic gas ($C_V = 3R$, $C_P = 4R$, where $R$ is the gas constant) goes from phase space point A ($P_A = 10^5\ \text{Pa}$, $V_A = 4\times10^{-6}\ \text{m}^3$) to point B ($P_B = 5\times10^4\ \text{Pa}$, $V_B = 6\times10^{-6}\ \text{m}^3$) to point C ($P_C = 10^4\ \text{Pa}$, $V_C = 8\times10^{-6}\ \text{m}^3$), as shown in the figure. A to B is an adiabatic path and B to C is an isothermal path at $450\ \text{K}$. The net heat absorbed per unit mole by the system is
Answer: (B) $450R(\ln4 - \ln3)$
A → B is adiabatic: $Q_{AB} = 0$.
B → C is isothermal at $T = 450\ \text{K}$: $\Delta U = 0$, so the heat absorbed equals the work done. Per mole:
$$Q_{BC} = RT\ln\frac{V_C}{V_B} = 450R\ln\frac{8}{6} = 450R(\ln4 - \ln3)$$
Net heat per mole $= 450R(\ln4 - \ln3)$.
Solution by Sreeraj P, M.Sc Physics