21∂G=-ve for spontaneous,=0 at equilibrium and +ve for forced processes.
Now at normal boiling(1 atm pressure) ∂G=0 [the vapor and liquid are in equilibrium]
Now if we increase the pressure ,boiling pt increases and to make the liq. boil at that particular temp the process has to forced,Hence ∂G>0
Mathematically,
we know at boiling pt,
∂S=∂H/Tboiling pt
Now when we increase the pressure, boiling pt. increases to T'.Let earlier boiling pt. be T[ie current temp=351]
Hence ∂S=∂H/T'
Now at current temp,
∂G=∂H-T∂S [T=current temp]
=∂H-∂H(T/T') [now T<T']
Therefore ∂G>0
