 The problem reads, the vapor pressure of liquid titanium at 2227 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. So we're going to consider normal to be our level one. So for level one information, we have P1 is equal to one atmosphere and T1 is what we are looking for, the normal boiling point. And then for level two, P2 is equal to 1.503 millimeters mercury, which we need to transfer into atmosphere. So we multiply it by one atmosphere over 760 millimeters mercury. And taking out our calculator and we have 1.503 divided by 760 equals 0.0019777. So 0.0019777. 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 five. 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 Clapeyron 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 one atmosphere over 0.0019777. Atmospheres, they cancel. And then we have that's equal to minus 435140 joules versus mole divided by the gas constant R, which is 8.3145 joules versus mole Kelvin. And then we have our temperatures, which is one over T1. T1 is what we're looking for, minus one over T2. T2 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 one divided by this. So one divided by second answer. So we have one 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 52335.1. And then we have our one over T1 minus one over 2,500. So we need 62586 divided by minus 52335.1. Plus one over 2,500 equals one over T1. So let's find that. So we have, we can take this answer here. We didn't save that one, so we'll have to take this answer here. So one divided by second answer gives us one over this. Times 6.22586, enter. And then that plus one divided by 2,500, enter. And we get 2.81038. So this is 2.81038 equals one over T1. So we need one over that. So one divided by answer equals 3558 kelvin. So T1 equals 3558 kelvin minus 273. Celsius equals minus 273 equals 3285 Celsius. This is the normal boiling point of liquid titanium.