JEE Advanced 2023 — Paper 1 — Question 41

$\mathrm{A} \rightarrow \mathrm{B}$ (reversible adiabatic), $\gamma=\frac{5}{3}$ $\mathrm{q}_{\text {total }}=\mathrm{E}_{\mathrm{AB}}+\mathrm{q}_{\mathrm{BC}}=0+\mathrm{q}_{\mathrm{BC}}=\mathrm{q}_{\mathrm{BC}}$ $\mathrm{q}_{\mathrm{BC}}=\mathrm{RT}_{2} \ell \mathrm{n} 10$ For the process A to B $\mathrm{T}_{1} \times \mathrm{V}_{1}^{\gamma-1}=\mathrm{T}_{2} \times \mathrm{V}_{2}^{\gamma-1}$ $\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\gamma-1}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{600}{60}=10$ $\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)^{\frac{5}{3}-1}=10$ $\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=(10)^{\frac{3}{2}}$ $\mathrm{V}_{2}=10 \times(10)^{\frac{3}{2}}=(10)^{\frac{5}{2}}$ $\mathrm{V}_{2}=(10)^{\frac{5}{2}}$ $-\mathrm{RT}_{2} \ln 10=-\mathrm{RT}_{2} \ln \frac{\mathrm{~V}_{3}}{\mathrm{~V}_{2}}$ $10=\frac{\mathrm{V}_{3}}{\mathrm{~V}_{2}}$ $\mathrm{V}_{3}=10 \mathrm{~V}_{2}=10 \times(10)^{\frac{5}{2}}=(10)^{\frac{7}{2}}$ $2 \ln \mathrm{~V}_{3}=2 \times \ln 10^{\frac{7}{2}}$ $2 \log \mathrm{~V}_{3}=2 \times \frac{7}{2} \times 1=7$