please help

as P is the equilibrium constant of the equation H₂O(l) -><- H₂O(g) Now, ΔG⁰ = -RTln(K) where K=equilibrium constant ==> -RTln(K) = ΔH⁰ - TΔS° ==> lnK = ΔS°/R - ΔH°/RT So, taking logP = log(K) as y and 1/T as x we have, y = c - ΔH°Rx So, it is a straight line. If ΔH°>0 then slope is negative and if ΔH°<0 then slope is positive. This reaction is endothermic. So, ΔH>0 hence slope -ve. Hence answer is (C) cheers !!

C