According to the question
Now, the temperature at point $A$ and $B$
By, $p V=n R T \quad$ (from ideal gas equation) Temperature at point $A$,
$T_{A}=\frac{3 p_{0}\, V_{0}}{R}$
Temperature at point $B$,
or, $T_{B}=\frac{30\, p_{0}\, V_{0}}{R}$
So, the temperature difference, $\Delta T=T_{B}-T_{A}$
$\therefore \Delta T=\frac{30 \,p_{0} \,V_{0}}{R}-\frac{3 \,p_{0} \,V_{0}}{R} $
$\Delta T=\frac{27\, p_{0}\, V_{0}}{R}$
Now, due to the adiabatic process, change in internal energy,
$\therefore \Delta U=n C_{v} \,\Delta T$
Or,$\Delta U=\frac{3}{2} R\, \Delta T\,\,\,...(i)$
Work done by the gas undergoes the process $A \rightarrow B$
$\therefore W=$ Area under $p-V$ graph
or$W=18 \,p_{0} \,V_{0}$
or$W=\frac{2}{3} R \Delta T\,\,\,...(ii)$
Now, by first law of thermodynamics,
$Q=\Delta U+W$
By putting the values from Eqs. (i) and (ii) to above Eq. we get
$\therefore Q=\frac{3}{2} R \,\Delta T+\frac{2}{3} R\, \Delta T$
or $C \Delta T=\frac{13}{6} R \,\Delta T$
or $(\because Q=n C \Delta T, $ where $ n=1$ mole $) $
$ C=\frac{13}{6} R$
So, the specific heat capacity in the process $A \rightarrow B$ is $C=\frac{13}{6} R$
