One mole of an ideal diatomic gas expands from volume V to 2V… | JEE Main 2026 (23 Jan, Shift 2)

Q43JEE Main 2026 · 23 JAN · SHIFT 2Medium

Thermodynamic Processes

One mole of an ideal diatomic gas expands from volume $V$ to $2V$ isothermally at a temperature $27^\circ\text{C}$ and does $W$ joule of work. If the gas undergoes same magnitude of expansion adiabatically from $27^\circ\text{C}$ doing the same amount of work $W$ , then its final temperature will be (close to) $\underline{\hspace{3em}}\ ^\circ\text{C}.\ (\log_e 2 = 0.693)$

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Answer: (a)

Isothermal work: $W = nRT\ln 2 = 1 \times 8.314 \times 300 \times 0.693 \approx 1728$ J.
Adiabatic work: $W = nC_v(T_1 - T_2)$ with $C_v = \frac{5}{2}R = 20.8$ J mol $^{-1}$ K $^{-1}$ .
$T_1 - T_2 = \frac{1728}{20.8} \approx 83$ K
$T_2 \approx 27 - 83 = -56^\circ$ C
So the answer is (a) $-56$ .

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