When $$\Delta H_{vap} = 30 \text{ kJ/mol}$$ and $$\Delta S_{vap} = 75 \text{ J mol}^{-1} \text{K}^{-1}$$, then the temperature of vapour, at one atmosphere is ______ K.

The enthalpy of vaporization is given as $$\Delta H_{vap} = 30 \text{ kJ/mol} = 30000 \text{ J/mol}$$ and the entropy of vaporization as $$\Delta S_{vap} = 75 \text{ J mol}^{-1} \text{K}^{-1}$$. At the boiling point under 1 atm pressure, the liquid and vapor phases are in equilibrium, so the change in Gibbs free energy for vaporization is zero: $$\Delta G = 0$$.

The Gibbs-Helmholtz equation relates these quantities by $$\Delta G = \Delta H - T\,\Delta S$$. Substituting zero for $$\Delta G$$ gives $$0 = \Delta H_{vap} - T_{bp}\,\Delta S_{vap}$$, which rearranges to $$T_{bp} = \frac{\Delta H_{vap}}{\Delta S_{vap}}$$.

Inserting the numerical values yields $$T_{bp} = \frac{30000 \text{ J/mol}}{75 \text{ J mol}^{-1} \text{K}^{-1}} = 400 \text{ K}$$. Therefore, the boiling temperature of the substance at one atmosphere is 400 K.