A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2. The work done by the gas is: Given, R = 8.3 J K⁻¹ mol⁻¹. (a) 23.1 J (b) 48 J (c) 24.9 J (d) 72.9 J

Answer: (a). Helium is monatomic: Δ U = nCvΔ T = 1 × 3/2 × 8.3 × 2 = 24.9 J First law: W = Q - Δ U = 48 - 24.9 = 23.1 J So the answer is (a) 23.1 J.

A total of 48 J heat is given to one mole of helium kept in a… | JEE Main 2024 (6 Apr, Shift 2)

Q129JEE Main 2024 · 6 APR · SHIFT 2Medium

Thermodynamic Processes

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

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Answer: (a)

Helium is monatomic: $\Delta U = nC_v\Delta T = 1 \times \frac{3}{2} \times 8.3 \times 2 = 24.9\text{ J}$
First law: $W = Q - \Delta U = 48 - 24.9 = 23.1\text{ J}$
So the answer is (a) 23.1 J.

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