Let initial pressure, volume and temperature be P₀, V₀ and T₀, respectively, indicated by state A in P-V diagram. The gas is then isochorically taken to state B (2P₀, V₀, 2T₀) and then taken from state B to state C (2P₀, 2V₀, 4T₀) isobarically.


            


Total heat absorbed by 1 mol of gas

          $\Delta \text{Q}={\text{C}}_{\text{v}}\left(2{\text{T}}_0-{\text{T}}_0\right)+{\text{C}}_{\text{P}}\left(4{\text{T}}_0-2{\text{T}}_0\right)$

                  $=\frac{5}{2}{\text{RT}}_0+\frac{7}{2}\text{R}\times 2{\text{T}}_0=\frac{19}{2}{\text{RT}}_0$

Total change in temperature from series A to C is $\Delta \text{T}=3{\text{T}}_0$

Therefore,

Molar heat capacity $=\frac{\Delta \text{Q}}{\Delta \text{T}}=\frac{\frac{19}{2}{\text{RT}}_0}{3{\text{T}}_0}=\frac{19}{6}\text{R}$