One mole of an ideal monoatomic gas undergoes a process as shown in the figure. Find the molar specific heat of the gas in the process.

R is a gas constant 

$\mathrm{V} = \mathrm{m}\left(\frac{1}{\mathrm{T}}\right); \mathrm{m} = \mathrm{slope} \mathrm{of} \mathrm{the} \mathrm{curve}$

    $\mathrm{VT} = \mathrm{Constant}. \mathrm{TdV} + \mathrm{VdT} = 0$--------(i)

     $\mathrm{dV} = -\frac{\mathrm{V}}{\mathrm{T}}\mathrm{dT}$

      $\mathrm{Q} = ∆\mathrm{U} + \mathrm{W}$

   $\mathrm{CdT} = {\mathrm{C}}_{\mathrm{v}}\mathrm{dT} + \mathrm{PdV}$----------------(ii)

From (i) and (ii)

$\mathrm{CdT} = {\mathrm{C}}_{\mathrm{v}}\mathrm{dT} - \frac{\mathrm{P}}{\mathrm{T}}\mathrm{VdT} \Rightarrow \mathrm{C} = {\mathrm{C}}_{\mathrm{V}}-\frac{\mathrm{PV}}{\mathrm{T}}$

For monoatomic gas ${\mathrm{C}}_{\mathrm{v}} = \frac{3}{2}\mathrm{R}$

$\mathrm{C} = {\mathrm{C}}_{\mathrm{v}}-\mathrm{R} = \frac{3}{2}\mathrm{R}-\mathrm{R} \Rightarrow \mathrm{C} = \frac{\mathrm{R}}{2}$