Question

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

A

ΔS=nCp lnT2T1

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B

ΔS=nCp lnT1T2

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C

ΔS=nCv lnT2T1

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D

ΔS=nCv lnT1T2

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Solution

## The correct option is C ΔS=nCv lnT2T1 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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