Calculate the work done when $2 m o l e s$ of hydrogen expand isothermally and reversibly at $25^{\circ} C$ from $15 t o 45 l i t r e s$ .

- 1309 c a l o r i e s

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- 14.36 c a l o r i e s

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- 2872 c a l o r i e s

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- 28.72 c a l o r i e s

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Solution

The correct option is A $- 1309 c a l o r i e s$
Work done in an isothermal reversible process is given as,
$w = - 2.303 n R T l o g \frac{V_{2}}{V_{1}}$
Given : $V_{1} = 15 L$
$V_{2} = 45 L$
$n = 2 m o l$
$R = 2 c a l / K . m o l$
$T = 298 K$
So,
$w = - 2.303 \times 2 \times 2 \times 298 \times l o g \frac{45}{15}$
$w = - 2745.1 \times l o g 3$
$w = - 1309.7 c a l$

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