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Q.

The amount of work done in joules to increase the temperature of one mole of an ideal gas by 30°C if it obeys the law $V \propto T^{2 / 3}$ is
[round off to nearest integer]

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answer is 166.

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Detailed Solution

$\begin{array}{l} dw = p \cdot dv \\ dw = \frac{nRT}{v} \cdot dv \\ V = {KT}^{2 / 3} \\ dW = \frac{nRT}{{KT}^{2 / 3}} \cdot K \cdot \frac{2}{3} \cdot T^{\frac{- 1}{3}} \\ dw = \frac{2}{3} {nRT}^{1 / 3} \cdot T^{- 1 / 3} \cdot dT \\ \int dw = \frac{2}{3} nR \int dT \\ w = \frac{2}{3} \times 1 \times R \times 30 \\ W = 20 R \\ W = 20 \times 8.314 \\ W = 166 J \end{array}$

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