One mole of a monatomic ideal gas undergoes the process $\mathrm{A} \rightarrow \mathrm{B}$ in the given $p$…
- $\frac{13 R}{3}$
- $\frac{13 R}{6}$
- $\frac{7 R}{3}$
- $\frac{2 R}{3}$
Solution
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 { or } \quad W=18 p_0 V_0 \end{aligned} $
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 { where } n=1 \text { mole }) \\ \text { or } & C=\frac{13}{6} R \end{array} $ So, the specific heat capacity in the process $A \rightarrow B$ is $C=\frac{13}{6} R$
Asked in: AP EAMCET 2019 (20 Apr Shift 2)