The vapour pressure of a liquid is 20.0 kPa at 300 K and 50.0 kPa at 325 K…

Step 1. The enthalpy. $\Delta H = \dfrac{8.314 \times \ln(50.0/20.0)}{1/300 - 1/325} = \dfrac{8.314 \times 0.9163}{2.564 \times 10^{-4}} = 29\,710\ \text{J/mol} = 29.7\ \text{kJ/mol}$.

Step 2. The boiling point. $\dfrac{1}{T_b} = \dfrac{1}{325} - \dfrac{8.314 \times \ln(101.325/50.0)}{29\,710} = 0.0028793$. So $T_b = 347.3\ \text{K}$, which is $74.2\ ^\circ\text{C}$.

Step 3. Trouton's ratio. $\dfrac{29\,710}{347.3} = 85.5\ \text{J K}^{-1}\ \text{mol}^{-1}$.

Step 4. The meaning. The value lies in the range of nonpolar liquids, so the liquid has no strong hydrogen bonds. A hydrogen-bonded liquid such as water gives about $109\ \text{J K}^{-1}\ \text{mol}^{-1}$.