There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large…
- $4.4$
- $6$
- $24$
- $18$
Solution

Number of masses will remain constant
$\begin{aligned}
& \mathrm{n}_1+\mathrm{n}_2=\mathrm{n}_{\mathrm{f}} \\ & \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{RT}_1}+\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{RT}_2}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}_{\mathrm{f}}}{\mathrm{RT}_{\mathrm{f}}} \\ & \frac{8 \times 2 \mathrm{~V}}{\mathrm{R} \times 1000}+\frac{7 \times \mathrm{V}}{\mathrm{R} \times 500}=\frac{\mathrm{P}_{\mathrm{f}}(3 \mathrm{~V})}{\mathrm{R} \times 600} \\ & \frac{16}{1000}+\frac{14}{1000}=\frac{\mathrm{P}_{\mathrm{f}}}{\mathrm{R} \times 600} \\ & \frac{30}{1000}=\frac{\mathrm{P}_{\mathrm{f}}}{200} \\ & \mathrm{P}_{\mathrm{f}}=6 \mathrm{kPa}
\end{aligned}$
Asked in: JEE Main 2025 (04 Apr Shift 2)