Explanation: The idea here is that you can use the Clausius - Clapeyron equation to estimate the vapor pressure of a liquid at a given temperature provided that you know the vapor pressue of the liquid at another temperature and its enthalpy of vaporization, ΔHvap. Consequently, if you know the vapor pressure of a liquid at two different temperatures, you can use the Clausius - Clapeyron equation to find is enthalpy of vaporization. The Clausius - Clapeyron equation looks like thisln(P1P2)=−ΔHvapR⋅(1T1−1T2) , whereP1 - the vapor pressure of the liquid at a temperature T1P2 - the vapor pressure of the liquid at a temperature T2R - the universal gas constant, given in this contex as 8.314 J mol−1K−1Now, it's important to realize that the normal boiling point of a substance is measured at an atmoshperic pressure of 1 atm. You can express this in mmHg by using the conversion factor1 atm = 760 mmHgAlso, keep in mind that you must use absolute temperature, which is temperature expressed in Kelvin. So, rearrange the equation to solve for ΔHvapΔHvap=−ln(P1P2)⋅R(1T1−1T2)Plug in your values to getΔHvap=−ln(134mmHg760mmHg)⋅8.314J mol−1K−1(1(273.15+0)−1(273.15+40))K−1ΔHvap=30854.8 J mol−1I'll express the answer in kilojoules per mole and leave it rounded to one sig figΔHvap=30 kJ mol−1