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Question

# Calculate the entropy change when 2 mol of an ideal gas expands isothermally and reversibly from an initial volume of 10 dm3 to 100 dm3 at 300 K.

A
45 J K1
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B
50 J K1
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C
28.294 J K1
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D
38.294 J K1
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Solution

## The correct option is D 38.294 J K−1Do you remember, how we can find the entropy change for isothermal process? In an Isothermal process ΔU = 0 ΔU = q + w ⇒q = -w But w = -2.30 nRT log V2V1 or -nRT ln V2V1 (since this is a reversible process) ⇒q = 2.303 nRT log V2V1 = nRT ln V2V1 or for a reversible process qrev = 2.303 nRT log V2V1 = nRT ln V2V1 ∴ΔsysS=qrevT=nRT lnV2V1T=2.303 nRT logV2V1T ∴ΔS=nR lnV2V1=2.303 nR logV2V1 Also V ∝1P (Boyle's law) Substituting this relation in above equation, we get ⇒△S = 2.303 nR log P1P2 = nR ln P1P2 Now, we have V1=10 dm3, V2=100 dm3, n=2 mol For an isothermal process, ΔS = nR lnV2V1 V = 10 dm3, V2 = 100 dm3, n=2 mol, R = 8.314 J K−1 mol−1 ∴ΔS=(2mol)×(8.314 J K−1 mol−1)×2.303 log100 dm310 dm3 =38.294 J K−1

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