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JEE Main levelNumerical answerDifficulty 2 of 3td180

Two moles of a monatomic ideal gas at $400 K$ expand adiabatically and reversibly, doing $3.0 kJ$ of work on the surroundings. Taking $R = 8.314 J mo l^{- 1} K^{- 1}$ , find the final temperature in K (nearest integer).

Numerical answer type, answer in K, tolerance 1.

Answer: 280 K

With

q = 0

,

Δ U = w = - 3000 J = n C_{V} Δ T = 2 \times 1.5 \times 8.314 Δ T

. So

Δ T = - 120.28 K

and

T_{2} = 400 - 120.28 = 279.7 K

, about 280 K.

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