One mole of an ideal gas|KTG & THERMODYNAMICS|JEEMAIN 2021

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JEE MAIN 2021

20-07-21 S2

Question

One mole of an ideal gas at $27^{\circ} \mathrm{C}$ is taken from A to B as shown in the given PV indicator diagram. The work done by the system will be $\_\_\_\_$ $\times 10^{-1} \mathrm{~J}$.
[Given: R = 8.3 J / mole K, $1 \mathrm{n} 2=0.6931$ ]
(Round off to the nearest integer)

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Solution

Process of isothermal W = nRT ln( V₂ / V₁ ) = 1 × 8.3 × 300 × ln2 = 17258 × 10⁻¹ J

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