JEE Main 2026 — Physics Thermodynamics

2026mcqmedium

An ideal gas undergoes a process maintaining relation between pressure $(P)$ and volume $(V)$ as $P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}$ , where $P_o$ and $V_o$ are constants. If two samples $A$ and $B$ (two moles each) with initial volumes $V_o$ and $3V_o$ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, $T_B - T_A$ is _____. ( $R =$ gas constant)

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{9P_o V_o}{8R}$
B
$\dfrac{11P_o V_o}{10R}$
C
$\dfrac{7P_o V_o}{6R}$
D
$\dfrac{13P_o V_o}{11R}$

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Solution

The given relation between pressure and volume is:

$P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}$

For sample A, the initial volume is $V_A = V_o$ . Substituting this into the given relation, the initial pressure $P_A$ is:

$P_A = P_o\left(1 + \left(\dfrac{V_o}{V_o}\right)^2\right)^{-1} = \dfrac{P_o}{2}$

Using the ideal gas equation $PV = nRT$ for $n = 2$ moles, the initial temperature $T_A$ is:

$T_A = \dfrac{P_A V_A}{nR} = \dfrac{\left(\dfrac{P_o}{2}\right) V_o}{2R} = \dfrac{P_o V_o}{4R}$

For sample B, the initial volume is $V_B = 3V_o$ . Substituting this into the relation, the initial pressure $P_B$ is:

$P_B = P_o\left(1 + \left(\dfrac{V_o}{3V_o}\right)^2\right)^{-1} = P_o\left(1 + \dfrac{1}{9}\right)^{-1} = \dfrac{9P_o}{10}$

The initial temperature $T_B$ for $n = 2$ moles is:

$T_B = \dfrac{P_B V_B}{nR} = \dfrac{\left(\dfrac{9P_o}{10}\right) (3V_o)}{2R} = \dfrac{27P_o V_o}{20R}$

The difference in the initial temperatures of the two samples is:

$T_B - T_A = \dfrac{27P_o V_o}{20R} - \dfrac{P_o V_o}{4R}$

$T_B - T_A = \dfrac{27P_o V_o - 5P_o V_o}{20R} = \dfrac{22P_o V_o}{20R} = \dfrac{11P_o V_o}{10R}$

Answer: $\dfrac{11P_o V_o}{10R}$

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Solution

The given relation between pressure and volume is:

$P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}$

For sample A, the initial volume is $V_A = V_o$ . Substituting this into the given relation, the initial pressure $P_A$ is:

$P_A = P_o\left(1 + \left(\dfrac{V_o}{V_o}\right)^2\right)^{-1} = \dfrac{P_o}{2}$

Using the ideal gas equation $PV = nRT$ for $n = 2$ moles, the initial temperature $T_A$ is:

$T_A = \dfrac{P_A V_A}{nR} = \dfrac{\left(\dfrac{P_o}{2}\right) V_o}{2R} = \dfrac{P_o V_o}{4R}$

For sample B, the initial volume is $V_B = 3V_o$ . Substituting this into the relation, the initial pressure $P_B$ is:

$P_B = P_o\left(1 + \left(\dfrac{V_o}{3V_o}\right)^2\right)^{-1} = P_o\left(1 + \dfrac{1}{9}\right)^{-1} = \dfrac{9P_o}{10}$

The initial temperature $T_B$ for $n = 2$ moles is:

$T_B = \dfrac{P_B V_B}{nR} = \dfrac{\left(\dfrac{9P_o}{10}\right) (3V_o)}{2R} = \dfrac{27P_o V_o}{20R}$

The difference in the initial temperatures of the two samples is:

$T_B - T_A = \dfrac{27P_o V_o}{20R} - \dfrac{P_o V_o}{4R}$

$T_B - T_A = \dfrac{27P_o V_o - 5P_o V_o}{20R} = \dfrac{22P_o V_o}{20R} = \dfrac{11P_o V_o}{10R}$

Answer: $\dfrac{11P_o V_o}{10R}$

JEE Main 2026 — Physics Thermodynamics

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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