Consider the process: Br2(l)→Br2(g).
The b.p. of a liquid is the temperature at which the liquid and the pure gas coexist at equilibrium at 1 atm.
∴ △G=0.
As it is given that △H and △S do not change with temperature △H=△Ho=30.91 kJmol−1
△S=△So=93.2 Jmol−1K−1=0.0932 kJmol−1K−1
We have,
△G=△H−T△S=0
∴ T=△H△S=30.910.0932=331.6 K
This is the temperature at which the system is in equilibrium, that is, the b.p. of bromine.