JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Heat & Thermodynamics

WhatsApp Email

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✓ Correct
(B)120
(C)90
(D)105

Answer: (A)

Solution

The total amount of gas is fixed. Initially both equal volumes

V

are at 90 kPa and 400 K, proportional to

\frac{2 \times 90 V}{R \times 400}

. Finally one vessel stays at 400 K and the other reaches 500 K at common pressure

P

\frac{P V}{R \times 400} + \frac{P V}{R \times 500}

. Equating,

\frac{2 \times 90}{400} = P (\frac{1}{400} + \frac{1}{500}) = P \cdot \frac{9}{2000}

, so

P = \frac{0.45 \times 2000}{9}

. The final pressure is

100

kPa.

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Heat & Thermodynamics →

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Heat & Thermodynamics

Solution

More JEE Main questions on Heat & Thermodynamics