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Question

# Which of the following is the correct expression of ΔS for an isobaric process?

A

ΔS=nCPln T2T1+nR ln V2V1

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B

ΔS=nCp lnT2T1

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C

ΔS=nCp lnT1T2

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D

None of these

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Solution

## The correct option is C ΔS=nCp lnT1T2 So, what is the first thing which comes to our mind? Yes, Isobaric ⇒ constant pressure Now, as we have already seen that Entropy change in an isobaric process (at constant pressure) dS = dqrev(P)T also dqrev(P) = dH But dH=CPdT, substituting for relationship with entropy, We get, ⇒dS=CPdTT Integrating both sides ∫S2S1ds=CP∫T2T1dTT ΔS=Cp[ln T]T2T1=Cp lnT2T1 ΔS=2.303Cp logT2T1=ΔS=Cp lnT2T1 For n moles: ΔS=nCPln T2T1=2.303nCPlog T2T1

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