Question

# 1 mole of an ideal gas at 25∘C is subjected to expand reversibly 10 times of its initial volume. Calculate the change in entropy of expansions.

A

25.15 J K-1mol-1

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B

30.15 J K-1mol-1

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C

19.15 J K-1mol-1

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D

29.15 J K-1mol-1

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Solution

## The correct option is C 19.15 J K-1mol-1 Here, we know that ΔT = 0 Therefore, it is an isothermal process. We know for an isothermal process ΔU = 0 Now, Entropy change for an isothermal process ΔU=q+w ΔU=0 (Isothermal process) ⇒q=−w But w=−2.30 nRT log V2V1 or −nRT lnV2V1 [From first law of thermodynamics] ⇒q =2.303 nRT logV2V1=nRT lnV2V1 or for a reversible process qrev =2.303 nRT logV2V1 = nRT lnV2V1 ∴ΔsysS=qrevT=nRT lnV2V1T=2.303nRT logV2V1T Also V∝1P (Boyle's law) Substituting this relation in above equation, we get ⇒ΔS=2.303 nR logP1P2=nR lnP1P2 We have, ΔS = Given, n = 1, R = 8.314 J, T = 298 K, V1 = V, V2 = 10 V ∴ΔS=2.303×1×8.314 log1010VV = 19.15 J K−1 mol−1

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