At what temperature liquid water will be in equilibrium with water vapour?

$\Delta H_{vap}=40.73KJ/mol$

$\Delta S_{vap}=0.109KJ{K}^{-1}mo{l}^{-1}$

H₂O(l) $\leftrightharpoons$H₂O (g)

Now, at equilibrium, △G = 0

Now, from Gibb's free energy change equation

         $$\begin{gathered} ∆\mathrm{G} =∆\mathrm{H}-\mathrm{T}∆\mathrm{S} \\ ∆\mathrm{H}=\mathrm{T}∆\mathrm{S} \end{gathered}$$

$\mathrm{T}=\frac{\mathrm{\Delta H}}{\mathrm{\Delta S}}=\frac{40.73}{0.109}=373.6\mathrm{K}$