QJEEM24Che240: JEE Main 2024 Chemistry - Class 11 JEEM24Che240 Solution: When and , then the temperature of vapour, at one atmosphere is …

To find the temperature of vaporization at one atmosphere, we can use the Clausius-Clapeyron equation, which relates the enthalpy of vaporization ( $\Delta H_{vap}$ ) to the change in entropy ( $\Delta S_{vap}$ ) during the phase transition at a particular temperature (T). The relationship can be simplified under the assumption that both the enthalpy of vaporization and the entropy change of vaporization are constant with temperature to the form:

$\Delta H_{vap} = T \cdot \Delta S_{vap}$

This equation states that the enthalpy change of vaporization is equal to the product of the temperature at which the phase change occurs and the entropy change of vaporization. We rearrange this equation to solve for the temperature (T):

$T = \frac{\Delta H_{vap}}{\Delta S_{vap}}$

However, it's crucial to ensure the units are consistent. Given that $\Delta H_{vap}$ is in kJ/mol and $\Delta S_{vap}$ is in J/(mol·K), we need to convert $\Delta H_{vap}$ from kJ/mol to J/mol to match units:

$\Delta H_{vap} = 30 \, \text{kJ/mol} = 30 \times 10^3 \, \text{J/mol}$

Substituting the given values into the equation, we obtain:

$$T=\frac{30\times {10}^3\text{J/mol}}{75\text{J/(mol·K)}}$$

$$T=\frac{30000\text{J/mol}}{75\text{J/(mol·K)}}$$

$T = 400 \, \text{K}$

Therefore, the temperature of vaporization at one atmosphere is 400 K.