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 .

Physics · Kinetic Theory of Gases

JEE Main 2026 — 6 April, Morning Shift — Question 8

Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure $90\mathrm{kPa}$ and temperature $400\mathrm{K}$ . Keeping the temperature of one vessel constant at $400\mathrm{K}$ the second vessel temperature is raised to $500\mathrm{K}$ . The final pressure in the vessels is $\mathrm{kPa}$ .

Option A:
100
Correct
Option B:
120
Option C:
90
Option D:
105

Answer: A

Step-by-step solution

Mole conservation: initial total moles = $2PV/(RT)$ . Final: $P'V/(R\cdot400) + P'V/(R\cdot500) = P'V/R(1/400+1/500) = P'V/R \cdot (9/2000)$ . Equate: $2P/(400) = P' (9/2000) \Rightarrow P' = (2/400) \times (2000/9) P = (10/9)\times 90 = 100$ kPa.

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Physics
Chapter: Kinetic Theory of Gases
Topic: Energy of Gas and Gas Laws and Miscellaneous Problems