JEE Main 2026 — Physics Thermodynamics

2026mcqmedium

An ideal gas at pressure $P$ and temperature $T$ is expanding such that $PT^3 =$ constant. The coefficient of volume expansion of the gas is _______.

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{2}{T}$
B
$\dfrac{1}{T}$
C
$\dfrac{4}{T}$
D
$\dfrac{3}{T}$

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Solution

Given the process equation for the ideal gas:

$PT^3 = \text{constant}$

From the ideal gas equation, $PV = nRT$ , which gives $P = \dfrac{nRT}{V}$ .

Substituting $P$ into the process equation:

$\left(\dfrac{nRT}{V}\right) T^3 = \text{constant}$

$\dfrac{T^4}{V} = \text{constant}$

$V = k T^4$

Differentiating both sides with respect to temperature $T$ :

$\dfrac{dV}{dT} = 4k T^3$

The coefficient of volume expansion $\gamma$ is defined as:

$\gamma = \dfrac{1}{V} \dfrac{dV}{dT}$

Substituting the values of $V$ and $\dfrac{dV}{dT}$ :

$\gamma = \dfrac{1}{k T^4} (4k T^3) = \dfrac{4}{T}$

Answer: $\dfrac{4}{T}$

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Solution

Given the process equation for the ideal gas:

$PT^3 = \text{constant}$

From the ideal gas equation, $PV = nRT$ , which gives $P = \dfrac{nRT}{V}$ .

Substituting $P$ into the process equation:

$\left(\dfrac{nRT}{V}\right) T^3 = \text{constant}$

$\dfrac{T^4}{V} = \text{constant}$

$V = k T^4$

Differentiating both sides with respect to temperature $T$ :

$\dfrac{dV}{dT} = 4k T^3$

The coefficient of volume expansion $\gamma$ is defined as:

$\gamma = \dfrac{1}{V} \dfrac{dV}{dT}$

Substituting the values of $V$ and $\dfrac{dV}{dT}$ :

$\gamma = \dfrac{1}{k T^4} (4k T^3) = \dfrac{4}{T}$

Answer: $\dfrac{4}{T}$

JEE Main 2026 — Physics Thermodynamics

Options

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Solution

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JEE Main Thermodynamics in other years

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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