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CGP EDU Academic Team
Published on: September 12, 2026
At N.T.P. one mole of diatomic gas is compressed adiabatically to half of its volume $\gamma = 1.41$ . The work done on gas will be
Text Solution
Verified by ExpertsThe correct answer is:
C
$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 273(2)^{0.41} = 273 \times 1.328 = 363K$ $W = \frac{R(T_1 - T_2)}{\gamma - 1} = \frac{8.31(273 - 363)}{1.41 - 1} = -1824$ $\Rightarrow |W| \approx 1815 \text{ J}$
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