AP EAPCET 20-Aug-2021 Shift 1 Paper

According to given graph,As we know that, $C=\frac{R}{\gamma-1}-\frac{R}{m-1}$ ...(i) where, $C_{V}$ is heat coefficient at constant volume, $R$ is universal gas constant $=8.314\,\mathrm{JK}^{-1}$ , $\gamma$ is adiabatic constant and $m$ is polytropic constant. For path $AB, pV^{2}=$ constant $\therefore m=2$ and $\gamma_{\text{monoatomic}}=\frac{5}{2}$ Put these values in Eq (i), we get $C=\frac{R}{\frac{5}{3}-1}-\frac{R}{2-1}$ $=\frac{3R}{2}-R=\frac{R}{2}$ For path $BC$, density is fixed ∴ Volume will remain unchanged. Since, $U=nC_{V}\Delta T$ where, $U$ is internal energy and $C_{V}$ (monoatomic gas) $=\frac{3}{2}R$. Isochoric process, $\Delta V=0$ But temperature is decreasing and $C_{V}=\frac{3}{2}R$. ∴ Option (c) is incorrect. For path $CA$ (isothermal) $W=nRT\ln\frac{\rho_2}{\rho_1}$ $=nRT\ln\frac{4\rho_0}{\rho_0}=nRT\ln 4$ and $U_{0}=\frac{3}{2}nRT$ ∴ $W=\frac{2}{3}U_{0}(\ln 4)$