JEE Main 2026 — Physics Thermodynamics

2026mcqmedium

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.

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

Let the volume of each vessel be $V$ .

Initially, both vessels are at pressure $P_1 = 90$ kPa and temperature $T_1 = 400$ K.

Total number of moles initially is given by:

$n_i = \dfrac{P_1 V}{R T_1} + \dfrac{P_1 V}{R T_1} = \dfrac{2 P_1 V}{R T_1}$

Finally, the temperature of the first vessel is $T_1 = 400$ K and the second vessel is $T_2 = 500$ K. Let the final pressure be $P_f$ .

Total number of moles finally is given by:

$n_f = \dfrac{P_f V}{R T_1} + \dfrac{P_f V}{R T_2}$

Since the total number of moles remains constant, $n_i = n_f$ :

$\dfrac{2 P_1 V}{R T_1} = \dfrac{P_f V}{R T_1} + \dfrac{P_f V}{R T_2}$

Substituting the given values:

$\dfrac{2 \times 90}{400} = \dfrac{P_f}{400} + \dfrac{P_f}{500}$

$\dfrac{180}{400} = P_f \left( \dfrac{1}{400} + \dfrac{1}{500} \right)$

$\dfrac{180}{400} = P_f \left( \dfrac{5 + 4}{2000} \right)$

$\dfrac{180}{400} = P_f \left( \dfrac{9}{2000} \right)$

$P_f = \dfrac{180 \times 2000}{400 \times 9}$

$P_f = 20 \times 5 = 100$ kPa

Answer: $100$

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Solution

Let the volume of each vessel be $V$ .

Initially, both vessels are at pressure $P_1 = 90$ kPa and temperature $T_1 = 400$ K.

Total number of moles initially is given by:

$n_i = \dfrac{P_1 V}{R T_1} + \dfrac{P_1 V}{R T_1} = \dfrac{2 P_1 V}{R T_1}$

Finally, the temperature of the first vessel is $T_1 = 400$ K and the second vessel is $T_2 = 500$ K. Let the final pressure be $P_f$ .

Total number of moles finally is given by:

$n_f = \dfrac{P_f V}{R T_1} + \dfrac{P_f V}{R T_2}$

Since the total number of moles remains constant, $n_i = n_f$ :

$\dfrac{2 P_1 V}{R T_1} = \dfrac{P_f V}{R T_1} + \dfrac{P_f V}{R T_2}$

Substituting the given values:

$\dfrac{2 \times 90}{400} = \dfrac{P_f}{400} + \dfrac{P_f}{500}$

$\dfrac{180}{400} = P_f \left( \dfrac{1}{400} + \dfrac{1}{500} \right)$

$\dfrac{180}{400} = P_f \left( \dfrac{5 + 4}{2000} \right)$

$\dfrac{180}{400} = P_f \left( \dfrac{9}{2000} \right)$

$P_f = \dfrac{180 \times 2000}{400 \times 9}$

$P_f = 20 \times 5 = 100$ kPa

Answer: $100$

JEE Main 2026 — Physics Thermodynamics

Options

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Solution

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JEE Main 2026 — Physics Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics