The problem reads the vapor pressure of liquid titanium at 2,227 degrees Celsius is 1.503 millimeters mercury. The heat of vaporization at the normal boiling point of titanium is 435.14 kilojoules per mole. Calculate its normal boiling point. Normal means at one atmosphere. We are going to consider normal to be our level 1. For level 1 information, we have P1 is equal to 1 atmosphere. T1 is what we are looking for, the normal boiling point. And then for level 2, P2 is equal to 1.503 millimeters mercury, which we need to transfer into atmosphere. So we multiply it by 1 atmosphere over 760 millimeters mercury. And taking it on our calculator, and we have 1.503 divided by 760 equals 0.001777. So 0.001977 atmospheres. And then we have T2 is equal to 2,227 degrees Celsius, but we need Kelvin. So that would be plus 273, which is 2,500 Kelvin. We have one more piece of information here. The heat of vaporization at the normal boiling point of titanium. So heat of vaporization, we denote by delta HV. And it is 435. That's kilojoules, so we'll just make it into joules right away. Joules over mole. So we have four pieces of information, and we're looking for number 5. We need a formula that relates these. Notice that we're looking at liquid and vapor. So we're going to be using the Clausius Clapper on equations. And we're looking at the vapor liquid one, which is this one right here. And it has exactly everything we need in it, so good to go. So we have Ln of 1 atmosphere over 0.001777 atmospheres. A cancel. And then we have that's equal to minus 435140 joules for a small divided by the gas constant R, which is 8.35 joules for a small Kelvin. And then we have our temperatures, which is 1 over T1. T1 is what we're looking for minus 1 over T. 2T2 is 2,500 Kelvin. So everything cancels as we need it to cancel. So all the units cancel. And T1 will be in Kelvin as we need it to be. So let's start to calculate. We have 1 divided by this. So 1 divided by second answer. So we have 1 over the answer, good. And that's 505. And we need Ln of that. So Ln of the answer. And we get 622586. So this is 622586 equals minus 435140 divided by 8.3145. So that's minus 52335. 0.1. And then we have our 1 over T1 minus 1 over 2,500. So we need 622586 divided by minus 523351 plus 1 over 2,500 equals 1 over T1. So let's find that. So we have, we can take this answer here. We can say that once we'll have to take this answer here. So 1 divided by second answer gives us 1 over this. Times 6.22586. Enter. And then that plus 1 divided by 2,500. Enter. And we get 2.81038. So this is 2.81038 equals 1 over T1. So we need 1 over that. So 1 divided by answer equals 3,5,5,8, kelvin. So T1 equals 3,5,5,8 kelvin minus 273. Celcius equals minus 273 equals 3285 Celcius. This is the normal boiling point of liquid titanium.