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Question

Which of the following is the correct expression for ΔS in case of isochoric process?

A

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B

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C

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D

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Solution

The correct option is C


So, here what do you have?
Isochoric ΔV=0
Now, for an isochoric process we know that
dS = dqrev(P)T Also dq(rev)(V)=dU [From first law of thermodynamics]
But dU = C-V dT
dqV=CVdT
Integration both sides with proper limit, we get
ΔS=Cv[ln T]T2T1=Cv lnT2t1
ΔS=2.303Cv logT2T1=Cv lnT1t2
For n moles:
ΔS=nCv lnT2T1=2.303nCv logT2T1
So, (c) is the correct option.

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