The vapour pressure of a liquid at 20 deg C is 0.57 atm but it raises to 0.65 at when the temperature raises by 3 deg C . Calculate the enthalpy of vaporisation of this liquid (in \( \mathrm{kJ} / \mathrm{mol} \) ).

To calculate the enthalpy of vaporization (\( \Delta H_{vap} \)) of the liquid, we can use the Clausius-Clapeyron equation given by: \[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] Where: - \( P_1 = 0.57 \, \text{atm} \) (pressure at \( T_1 = 20 \, \text{°C} = 293.15 \, \text{K} \)) - \( P_2 = 0.65 \, \text{atm} \) (pressure at \( T_2 = 23 \, \text{°C} = 296.15 \, \text{K} \)) - \( R = 8.314 \, \text{J/(mol·K)} \) First, we convert the pressures to a natural logarithm: \[ \ln\left(\frac{0.65}{0.57}\right) = \ln(1.14035) \approx 0.131 \] Now, substituting the values into the Clausius-Clapeyron equation: \[ 0.131 = -\frac{\Delta H_{vap}}{8.314} \left(\frac{1}{293.15} - \frac{1}{296.15}\right) \] Now calculate the difference in the inverse of temperatures: \[ \frac{1}{293.15} \approx 0.003412 \quad \text{and} \quad \frac{1}{296.15} \approx 0.003375 \] \[ \frac{1}{293.15} - \frac{1}{296.15} \approx 0.000037 \] Thus, substituting this into the equation gives: \[ 0.131 = -\frac{\Delta H_{vap}}{8.314} (0.000037) \] Now we can rearrange to find \( \Delta H_{vap} \): \[ \Delta H_{vap} = -\frac{0.131 \cdot 8.314}{0.000037} \] \[ \Delta H_{vap} \approx -\frac{1.089854}{0.000037} \approx 29466.86 \, \text{J/mol} \approx 29.47 \, \text{kJ/mol} \] So, the enthalpy of vaporization of the liquid is approximately **29.47 kJ/mol**.