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1 mol ideal gas at $27 ° C$ is expanded reversibly from 2 ${dm}^{3}$ to 20 ${dm}^{3}$ ${dm}^{3}$ . ∆ S for this is
(R=2 cal/mol/K)
[Hint: $ΔS = n R \ln \frac{V_{2}}{V_{1}}$ ]

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-4.606 cal/mol

+4.606 cal/mol

+2.303 cal/mol

Zero

answer is B.

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Detailed Solution

$ΔS = n R \ln \frac{V_{2}}{V_{1}} ΔS = 2.303 n R \log \frac{V_{2}}{V_{1}}$

Putting the values, we get:

$ΔS = 2.303 \times 1 \times 2 \times \log \frac{20}{2} ΔS = 4.606 cal/mol$

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