One mole of a monatomic ideal gas undergoes the process A arrow B...

According to the question,
Now, the temperature at point A and B$By,$p V=n R T$(from ideal gas equation) Temperature at point A$, T_A=$\frac{3 p_0 V_0}{R}$Temperature at point $B$, T_B=$\frac{30 p_0 V_0}{R}So, the$temperature difference,\Delta T=T_B-T_A$$\begin{array}{ll} \therefore& \Delta T=\frac{30 p_0 V_0}{R}-\frac{3 p_0 V_0}{R}\\ \text{ or, }& \Delta T=\frac{27 p_0 V_0}{R} \end{array}$$or, Now, due to $the adiabatic process, change in internal energy, \therefore \quad$\Delta U=n C_v $\Delta T$$\\ Work done by the gas undergoes the process$$\begin{aligned}& A \rightarrow \mathrm{B} .\\ & \therefore W=\text{ Area under }p \text{ - }V \text{ graph }\\ & \text{\text{ or } } \quadW=18 p_0 V_0 \end{aligned}$$<img src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions$ \\ Now, by first law of thermodynamics, Q=$\Delta U+W By putting the values from Eqs. (i) and (ii) to above Eq. we get$$\begin{array}{lcl} \therefore& Q=\frac{3}{2}R \Delta T+\frac{2}{3}R \Delta T \\ \text{ or }& C \Delta T=\frac{13}{6}R \Delta T \\ & (\because Q=n C \Delta T, \text{\text{ where } }n=1 \text{ mole })\\ \text{\text{ or } }& C=\frac{13}{6}R \end{array}$$So, the$specific heat capacity in the processA \rightarrow B$is$C=$\frac{13}{6}$ R$.$