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Question

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


A

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B

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C

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D

None of these

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Solution

The correct option is C


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=CPT2T1dTT
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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