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=−2.30nRTlogV2V1or−nRTlnV2V1 [From first law of thermodynamics] ⇒q=2.303nRTlogV2V1=nRTlnV2V1 or for a reversible process qrev=2.303nRTlogV2V1 = nRTlnV2V1 ∴ΔsysS=qrevT=nRTlnV2V1T=2.303nRTlogV2V1T Also V∝1P (Boyle's law) Substituting this relation in above equation, we get ⇒ΔS=2.303nRlogP1P2=nRlnP1P2 We have, ΔS = Given, n = 1, R = 8.314 J, T = 298 K, V1 = V, V2 = 10 V ∴ΔS=2.303×1×8.314log1010VV = 19.15 JK−1mol−1