Enthalpy is equal to

Options

A. `T^(@)[(del(G//T))/(delT)]_(P)`
B. `-T^(2)[(del(G//T))/(delT)]_(P)`
C. `T^(2)[(del(G//T))/(delT)]_(V)`
D. `-T^(2)[(del(G//T))/(delT)]_(V)`

Answer

B

Explanation

`G=H-TS` . . . (i)
`G=U+PV-TS` ltbr `DeltaG=DeltaU+PDeltaV+VDeltaP-TDeltaS-SDeltaT`
From the first and second laws,
`T DeltaS=DeltaU+PDeltaV`
`thereforeDeltaG=VDeltaP-SDeltaT`
At constant `DeltaP=0`
`(DeltaG)/(DeltaT)=S` . . .(ii)
From eq.s (i) and (ii)
`G=H+T(DeltaT)/(DeltaT)`
or `G=H+T((delG)/(delT))_(P)`
`-(H)/(T^(2))=-(G)/(T^(2))+(1)/(T)((delG)/(delT))_(P)`
`=[(del(G//T))/(delT)]_(P)`
`H=-T^(2)[(del(G//T))/(delT)]_(P)`.