A total of 48J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^{\circ} \mathrm{C}$. The work done by the gas is :\\ Given, $\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$.

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Heat & Thermodynamics

A total of 48J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by

2^{\circ} C

. The work done by the gas is :\\
Given,

R = 8.3 JK^{- 1} mol^{- 1}

(A)72.9 J
(B)24.9 J
(C)48 J
(D)23.1 J✓ Correct

Answer: (D)

Solution

By the first law,

Q = Δ U + W

. For one mole of a monatomic gas (helium),

Δ U = \frac{3}{2} R Δ T = \frac{3}{2} (8.3) (2) = 24.9 J

. With

Q = 48 J

W = 48 - 24.9 = 23.1 J

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option

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