Q 11-12-027JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure $90$ kPa and temperature $400$ K. Keeping the temperature of one vessel constant at $400$ K the second vessel temperature is raised to $500$ K. The final pressure in the vessels is ______ kPa.
Answer: (A) $100$
The total number of moles is fixed and the pressure becomes equal in both vessels:
$$\frac{p_0V}{R(400)} + \frac{p_0V}{R(400)} = \frac{pV}{R(400)} + \frac{pV}{R(500)}$$
$$\frac{2 \times 90}{400} = p\left(\frac{1}{400} + \frac{1}{500}\right) = p \times \frac{9}{2000} \;\Rightarrow\; p = 100\ \text{kPa}$$
Solution by Sreeraj P, M.Sc Physics