How to Calculate Standard State Gibbs Energy of Bromine Vapor at 298K?

[Est120]

Summary: Please help me with this problem ,I can't do it

The normal boiling point of liquid bromine is 58.2°C. At 9.3°C the equilibrium vapor
pressure of liquid bromine is 100 torr. From this data,

calculate the standard state Gibbs
energy of formation of bromine vapor at room temperature, ?G? f ,298K

I know that I have to use Claudius Clapeyron équation ,but How in the world I will get an absolute Gibbs energy value?

 

[Chestermiller]

If you calculate the equilibrium vapor pressure at 298K (25 C), you can calculate the change in Gibbs free energy of the vapor in going from this pressure to the hypothetical state of 1 bar and 298K using $dG=VdP=RTd\ln P$. Neglecting the Poynting correction, the free energy of the liquid at 298 and at the equilibrium vapor pressure is 0. And this is also the free energy of the vapor at 298 and at the equilibrium vapor pressure.