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. Level 2, P2 is equal to 1.503 millimeters mercury. We need to transfer into atmosphere. We multiply it by 1 atmosphere over 760 millimeters mercury. We have 1.503 divided by 760 equals 0.0019777. 0.0019777 atmospheres. T2 is equal to 2,227 degrees Celsius, but we need Kelvin. That would be plus 273, which is 2,500. We have one more piece of information here. The heat of vaporization at the normal boiling point of titanium. Heat of vaporization is equal to 0.0019777. It is 435. That is kilojoules. We will make it into joules right away. Joules over mole. We have 4 pieces of information. We need a formula that relates these. Notice that we are looking at liquid and vapor. We are going to be using the clasp around equations. We are looking at the vapor liquid one, which is this one right here. It has exactly everything we need in it. Good to go. We have Ln of 1 atmospheres over 0.0019777 atmospheres. A cancel. Then we have that is equal to minus 435140 joules for a small. Delighted by the gas constant R, which is 8.3145 joules for a small Kelvin. Then we have our temperatures, which is 1 over T1. T1 is what we are looking for minus 1 over T2T2 is 2500 Kelvin. Everything cancels as we need it to cancel. All the units cancel. T1 will be in Kelvin as we need it to be. Let's start to calculate. We have 1 divided by this. 1 divided by second answer. We have 1 over the answer. That is 505. We need Ln of that. Ln of the answer. We get 622586. This is 622586. 1 over T1. We have a small. Minus 435140. We have 8.3145. Minus 5235. 1 over T1 minus 1 over 2500. We need 622586. Minus 5235. 1 over 2500 equals 1 over T1. Let's find that. We can take this answer here. We can say that once we will have to take this answer here. 1 divided by second answer. This is 1 over this. Times 6.22586. Enter. Then that plus 1 divided by 2500. Enter. We get 2.8038. This is 2.8038 equals 1 over T1. We need 1 over that. 1 divided by answer equals 3558. 1 over T1 equals 3558. Minus 273. Celcius equals 273 equals 385. This is the normal boiling point of liquid titanium.