JEE Main 2026 — Physics Thermodynamics

2026mcqmedium

The internal energy of a monoatomic gas is 3 nRT. One mole of helium is kept in a cylinder having internal cross section area of $17 \mathrm{~cm}^{2}$ and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by $4^{\circ} \mathrm{C}$ , then the piston will move $\_\_\_\_$ cm.

(atmospheric pressure $=10^{5} \mathrm{~Pa}$ )

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

Options

A
1.45
B
15.5
C
1.55
D
14.5

Did you get this right?

Solution

The question states the internal energy of the gas is $U = 3nRT$ .

The change in internal energy is $\Delta U = 3nR\Delta T$ .

Given: $n = 1$ mole, $\Delta T = 4^{\circ}C = 4$ K, and $R \approx 8.314$ J/mol·K.

$\Delta U = 3 \times 1 \times 8.314 \times 4 = 99.768$ J.

According to the first law of thermodynamics, $\Delta Q = \Delta U + W$ .

Given heat supplied $\Delta Q = 126$ J.

Work done by the gas $W = \Delta Q - \Delta U = 126 - 99.768 = 26.232$ J.

Since the piston is light and movable, the process occurs at constant atmospheric pressure $P = 10^5$ Pa.

Work done $W = P \Delta V = P(A \times d)$ , where $A$ is the cross-section area and $d$ is the displacement of the piston.

$A = 17 \text{ cm}^2 = 17 \times 10^{-4} \text{ m}^2$ .

$26.232 = 10^5 \times (17 \times 10^{-4}) \times d$ .

$26.232 = 170 \times d$ .

$d = \frac{26.232}{170} \approx 0.1543$ m.

Converting to cm: $d \approx 15.43$ cm.

Rounding to the nearest option, we get $15.5$ cm.

Your note

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Solution

The question states the internal energy of the gas is $U = 3nRT$ .

The change in internal energy is $\Delta U = 3nR\Delta T$ .

Given: $n = 1$ mole, $\Delta T = 4^{\circ}C = 4$ K, and $R \approx 8.314$ J/mol·K.

$\Delta U = 3 \times 1 \times 8.314 \times 4 = 99.768$ J.

According to the first law of thermodynamics, $\Delta Q = \Delta U + W$ .

Given heat supplied $\Delta Q = 126$ J.

Work done by the gas $W = \Delta Q - \Delta U = 126 - 99.768 = 26.232$ J.

Since the piston is light and movable, the process occurs at constant atmospheric pressure $P = 10^5$ Pa.

Work done $W = P \Delta V = P(A \times d)$ , where $A$ is the cross-section area and $d$ is the displacement of the piston.

$A = 17 \text{ cm}^2 = 17 \times 10^{-4} \text{ m}^2$ .

$26.232 = 10^5 \times (17 \times 10^{-4}) \times d$ .

$26.232 = 170 \times d$ .

$d = \frac{26.232}{170} \approx 0.1543$ m.

Converting to cm: $d \approx 15.43$ cm.

Rounding to the nearest option, we get $15.5$ cm.

JEE Main 2026 — Physics Thermodynamics

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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