Question

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

Answer 1

Using the relation at equilibrium:
$\Delta G = \Delta H - T\Delta S = 0$
Rearranging for $T$:
$T = \frac{\Delta H}{\Delta S}$
Substitute the given values:
$\Delta H_\text{vap} = 30 \, \text{kJ/mol} = 30 \times 10^3 \, \text{J/mol}$, $\Delta S_\text{vap} = 75 \, \text{J mol}^{-1} \text{K}^{-1}$
$T = \frac{30 \times 10^3}{75} = 400 \, \text{K}$
Final Answer: (400)

Answer 2

Using the relation at equilibrium:
$\Delta G = \Delta H - T\Delta S = 0$
Rearranging for $T$:
$T = \frac{\Delta H}{\Delta S}$
Substitute the given values:
$\Delta H_\text{vap} = 30 \, \text{kJ/mol} = 30 \times 10^3 \, \text{J/mol}$, $\Delta S_\text{vap} = 75 \, \text{J mol}^{-1} \text{K}^{-1}$
$T = \frac{30 \times 10^3}{75} = 400 \, \text{K}$
Final Answer: (400)