JEE Main 2025 — 24 January, Morning Shift — Question 71

Given that process is isobaric $\Delta \mathrm{T}=50^{\circ} \mathrm{C}$ Q in isobaric process $=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}=\mathrm{E}_{1}$ $\Delta \mathrm{U}$ in isobaric process $=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=\mathrm{E}_{2}$ $\therefore \frac{E_{1}}{E_{2}}=\frac{C_{p}}{C_{v}}=\gamma$ Given, gas is monoatomic

$$\begin{aligned} \therefore \gamma & =1+\frac{2}{\mathrm{f}} & =1+\frac{2}{3} & =\frac{5}{3} \end{aligned}$$

Now, as per question.

$$\begin{aligned} & \frac{5}{3}=\frac{x}{9} & x=15 \end{aligned}$$