2 mole of ideal gas at $27^{0}$ C temp. is expanded reversibly from 2 lit. to 20 lit. Find entropy change (R = 2 cal/mol K):

92.1

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9.2

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Solution

The correct option is C 9.2
Number of moles= $2$

Temperature= $27^{o} C = 27 + 273 K = 300 K$

$V_{1} = 2 L$

$V_{2} = 20 L$

$R = 2$ cal/mol K

$△ S = \frac{q}{T} = \frac{- W}{T}$ (Since it is expanded reversibly)

$= \frac{- (- 2.303 n R T log V_{2} / V_{1})}{T}$ $(∵ W = 2.303 n R T log \frac{V_{2}}{V_{1}})$

$= 9.212$ cal/K/mol.

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