In this example question, it says the enthalpy of vaporization of water is 40.30 kilojoules per mole at its normal boiling point of 100�C. What is the vapor pressure in millimeters of mercury of water at 60�C? All right, so they're telling us we have two temperatures. Remember, our temperatures need to be in Kelvin, since this is the first temperature that's given to us this year would represent our T₁. This is the second temperature, so this would be our T₂. We add to them 273.15 to get our Kelvins. So this is 373.15 Kelvin and then here this comes out to 333.15 Kelvin.

What can we say next? We can say here they're telling us normal boiling point. Remember normal boiling point means that our pressure which is P₁ in this case would be 760mm of mercury. Or in Tours since they want the second pressure P₂ to be in millimeters of mercury. We'll keep it at 760 mm of mercury here. Alright, so now we're going to use our equation. ln(P₂/P₁) = -ΔH_vap / R * (1/T₂ - 1/T₁). Here we have ln. We don't know what our P₂ is. That's what we're looking to find. P₁ is 760mm of mercury.

Now remember R here is in joules, so ΔH_vap also needs to be in joules so the units can match. So converting 40.30 into Joules comes out to -40,300 Joules per mole. And here we're going to have 1/T₂ is 1/333.15 Kelvin, -1/373.15 Kelvin. All right, so now we find what this is in our calculator and we multiply it by this in our calculator. So when we multiply those two numbers together, we're going to get as our current answer -1.559667 and that equals ln(P₂/760).

Now to get rid of the ln on the left side, we're going to take the inverse of the natural log of both sides. So all that means is we're going to do e and then this part here becomes a power, so e^-1.559667 taking the inverse of the natural log on the left side basically cancels out this ln so that equals P₂/760. All we do now is we multiply 760 * e^-1.559667 and we'll isolate P₂. When we do that we get our P₂ as 159.76mm of Mercury. Here we have all the answers in terms of four sig figs. So this just becomes 159.8mm of mercury, which gives me option C as my correct answer. So using the two point form of the Clausius-Clapeyron equation gives us this as our final answer.