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Q.

The expression of entropy change for reversible isothermal expansion of an ideal gas from $V_{1}$ to $V_{2}$ is

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a

$∆ S = n R T I n \frac{V_{1}}{V_{2}}$

b

$∆ S = n R T \ln \frac{V_{2}}{V_{1}}$

c

$∆ S = n R ln \frac{V_{1}}{V_{2}}$

d

$∆ S = n R ln \frac{V_{2}}{V_{1}}$

answer is B.

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Detailed Solution

$d U = 0$ at constant temperature. Hence $d q = - d w = + p d V = \frac{n R T}{V} d v$

Thus $d S = \frac{d q}{T} = n R \frac{d V}{V}$

or $∆ S = n R ln \frac{V_{2}}{V_{1}}$

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