Physics - Error Question with Solution | TestHub

The process is described by $P=\frac{k}{V^2}$ Where k is a constant Let volume at C be $V\to 3P_0{V}_0^2=P_0{V}^2\Rightarrow V=\sqrt{3}{V}_0$ $W={\int }_{V_0}^{\sqrt{3}{V}_0}PdV={\int }_{V_0}^{\sqrt{3}{V}_0}\frac{k}{V^2}=k{\left[-\frac{1}{V}\right]}_{V_o}^{\sqrt{3}{V}_o}\left[\frac{1}{V}-\frac{1}{\sqrt{3}{V}_0}\right]-(3-\sqrt{3})P_0{V}_0$ Change in internal energy $\Delta U=\frac{f}{2}nR\Delta T=\frac{3}{2}nR\Delta t$ $\Delta T=T_C-T_A=\frac{\sqrt{3}{P}_0{V}_0}{nR}-\frac{3P_0{V}_0}{nR}=-(3-\sqrt{3})\frac{P_0{V}_0}{nR}$ $\Delta U=\frac{3}{2}(3-\sqrt{3})P_0{V}_0$ First law of thermodynamics, $\Delta Q=\Delta U+W$ $\Rightarrow \Delta Q\frac{3}{2}(3-\sqrt{3})P_0{V}_0+(3-\sqrt{3})P_0{V}_0=\frac{1}{2}(3-\sqrt{3})P_0{V}_0$ and $\Rightarrow \Delta Q=nc\Delta T\Rightarrow nc\Delta T=-\frac{1}{2}(3-\sqrt{3})P_0{V}_0\Rightarrow n.c\frac{(3-\sqrt{3})}{nR}=\frac{(3-\sqrt{3})P_0{V}_0}{2}\Rightarrow c=R/2$