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

Correct Answer: 400

To find the temperature of vaporization at one atmosphere using the given enthalpy of vaporization (\(\Delta H_{\text{vap}}\)) and entropy of vaporization (\(\Delta S_{\text{vap}}\)), we apply the formula derived from the Gibbs free energy relation at equilibrium: \[ \Delta G = \Delta H - T\Delta S = 0 \] Solving for temperature (\(T\)), we get: \[ T = \frac{\Delta H_{\text{vap}}}{\Delta S_{\text{vap}}} \] Convert \(\Delta H_{\text{vap}} = 30 \text{ kJ/mol}\) to joules: \[ 30 \text{ kJ/mol} = 30,000 \text{ J/mol} \] Then, substitute the values into the formula: \[ T = \frac{30,000 \text{ J/mol}}{75 \text{ J mol}^{-1} \text{K}^{-1}} = 400 \text{ K} \] The temperature is calculated to be 400 K, which fits within the provided range of 400,400. Therefore, the temperature of vaporization at one atmosphere is confirmed to be 400 K.

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)