In an isobaric process, when temperature changes from `T_(1)" to "T_(2), DeltaS` is equal to

Options

A. `2.303 C_(p)log(T_(2)//T_(1))`
B. `2.303C_(p)ln(T_(2)//T_(1))`
C. `C_(p)ln(T_(1)//T_(2))`
D. `C_(V)ln(T_(2)//T_(1))`

Answer

A

Explanation

The entropy change for a process, when T and P are the variables is given by
`DeltaS=C_(p)ln.(T_(2))/(T_(1))-R ln.(P_(2))/(P_(1))`
For an isobaric process `P_(1)=P_(2)`. Hence the above equation reduces to
`C_(p)ln.(T_(2))/(T_(1))=DeltaS.`
`or DeltaS=2.303C_(p)log.(T_(2))/(T_(1))`