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Question:

In an isobaric process, when temperature changes from T1 to T2, ΔS is equal to:

2.303Cplog(T2/T1)

2.303Cpln(T2/T1)

Cpln(T1/T2)

Cvln(T2/T1)

Solution:

The entropy change for a process, when T and P are the variable is given by
ΔS = Cp ln(T2/T1) - R ln(P2/P1)
For an isobaric process P1 = P2. Hence the above equation reduces to
Cp ln(T2/T1) = ΔS
or
ΔS = 2.303Cplog(T2/T1)