The internal energy of a monoatomic gas is 3nRT. One mole of helium… | JEE Main 2026 (23 Jan, Shift 2)

Q48JEE Main 2026 · 23 JAN · SHIFT 2Medium

Thermodynamic Processes

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

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

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Answer: (c)

Given $U = 3nRT$ : $\Delta U = 3 \times 1 \times 8.3 \times 4 = 99.6\text{ J}$ (taking $R = 8.3$ ).
First law: $W = Q - \Delta U = 126 - 99.6 = 26.4\text{ J}$
Light piston, so gas pressure stays $10^5$ Pa: $W = P A x$ .
$x = \frac{26.4}{10^5 \times 17\times10^{-4}} = \frac{26.4}{170} = 0.155\text{ m}$
So the piston moves 15.5 cm, answer (c).

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