Find the work done (in $k J$ ) when one mole of the gas is expanded reversibly and isothermally from $5 a t m$ to $1 a t m$ at $25^{o} C$ .

+ 4 k J

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- 4 k J

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- 0.4 k J

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+ 4 k J

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Solution

The correct option is B $- 4 k J$
Work done in case of reversible isothermal process is given by

$w = - 2.303 n R T l o g \frac{P_{1}}{P_{2}}$
Given : $P_{1} = 5 a t m$
$P_{2} = 1 a t m$
$n = 1$
$T = 25 + 273 = 298 K$

$w = - 2.303 \times 1 \times 8.314$ $\times 298$ $l o g \frac{5}{1}$
using $l o g 5 = 0.693$
$w = - 3.98 k J$ or $- 4 k J$

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