Question

The pressure and volume of a monoatomic ideal gas are related as $P\propto \frac{1}{V^2}$ for a process A $\to$ C as shown in the figure. The pressure and volume at A is $3P_{0}and{V}_0$ respectively and the pressure at C is $P_0$. If the molar heat capacity of the ideal gas for the process $A\to Cin\frac{R}{x}$ Find x. (R gas constant) ___

Answer 1

The process is described by
$P=\frac{k}{v^2}$ Where k is a constant
Let volume at C be V $\Rightarrow 3P_0{V}_0^2=P_0{V}^2\Rightarrow V=\sqrt{3}{V}_0$
$W={\int }_{V_0}^{\sqrt{3}{V}_0}PdV={\int }_{V_0}^{\sqrt{3}{V}_0}\frac{k}{v^2}dV=k{[-\frac{1}{v}]}_{v_0}^{\sqrt{3}{v}_0}[\frac{1}{v_0}-\frac{1}{\sqrt{3}{v}_0}]=(3-\sqrt{3})P_0{V}_0$
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, $\Delta Q=\Delta U+W$
$\Rightarrow \Delta Q=-\frac{2}{2}(3-\sqrt{3})(P_0{V}_0+(3-\sqrt{3})P_0{V}_0=-\frac{1}{2}(3-\sqrt{3})P_0{V}_0$
$And\Delta Q=nC\Delta T\Rightarrow nC\Delta T=-\frac{1}{2}(3-\sqrt{3})P_0{V}_0$
$\Rightarrow n.c\frac{(3-\sqrt{3})P_0{V}_0}{nR}=\frac{(3-\sqrt{3})P_0{V}_0}{2}\Rightarrow c=\frac{R}{2}$