The vapour pressure of water is 0.04atm at 27°C. The free energy change for the following process is 

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\left(0.04\mathrm{atm}{27}^0\mathrm{C}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\left(0.04\mathrm{atm}{27}^0\mathrm{C}\right)$

Given that the vapor pressure of water at  27°C is 0.04 atm and the process is also occurring at 0.04 atm. Therefore, the system is at equilibrium.

At equilibrium $\Delta G=0$