The vapour pressure of a certain  liquid is 30 kPa at 25oC and 60 kPa at 38oC. What is the enthalpy of vaporisation? (kJ mol-1)

To solve this problem, we can use the Clausius-Clapeyron equation, which is a way of characterizing the phase transition between two states of matter such as liquid and gas. The equation is:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

Where: P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, ΔHvap is the enthalpy of vaporization, R is the ideal gas constant (8.314 J mol-1 K-1), T1 and T2 are the absolute temperatures in Kelvin (K = °C + 273.15).

Given: P1 = 30 kPa, P2 = 60 kPa, T1 = 25°C = 298.15 K, T2 = 38°C = 311.15 K.

We can substitute these values into the equation:

ln(60/30) = -ΔHvap/8.314 * (1/311.15 - 1/298.15)

Solving for ΔHvap, we get:

ΔHvap = -8.314 * ln(2) / (1/311.15 - 1/298.15)

Please note that the result will be in J mol-1. To convert it to kJ mol-1, divide the result by 1000.