JEE Main 2026 Apr 6 Shift 1, Physics Q10: Kinetic Theory of an Ideal Gases

JEE Main2026Apr 6, Shift 1Physics

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.

A
$100$
B
$120$
C
$90$
D
$105$

Solution

The vessels are connected by a narrow tube so the pressure equalises; total moles of gas are conserved. Let the final common pressure be $P$ .

Initial total moles: $n_{i} = \frac{P _{0} ( 2 V )}{R T _{0}} = \frac{90 \cdot 2 V}{R \cdot 400}$ .

Final total moles: $n_{f} = \frac{P V}{R \cdot 400} + \frac{P V}{R \cdot 500}$ .

Setting $n_{i} = n_{f}$ :

$\frac{180}{400} = P (\frac{1}{400} + \frac{1}{500}) = P \cdot \frac{9}{2000}$ .

$P = \frac{180 \cdot 2000}{400 \cdot 9} = 100$ kPa.

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JEE Main 2026 Apr 6 Shift 1, Physics Q10: Kinetic Theory of an Ideal Gases

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