 The problem reads, the normal melting point of gallium is 30 degrees centigrade. The densities of solid and liquid gallium are 5.885 and 6.08 grams per cubic centimeter, respectively. The heat of fusion of gallium is 77.40 joules per gram. Calculate the change in the melting point of gallium for an increase in pressure of one atmosphere. Okay, so what do we have? We're looking at normal melting point. So we're going to take that as our level one. So P1 is equal to one atmosphere. That's what normal melting point means, means at pressure one atmosphere. And we have that the normal melting point, so T1 is equal to 30 degrees Celsius, is equal to 300 and 315 Kelvin. And now our level two is an increase in pressure of one atmosphere. So that means that P2 is equal to two atmospheres, one atmosphere plus one atmosphere. And T2, we don't know what it is. We need to find T2, but in the end we're interested in what the difference is between T2 and T1. So this is what we're looking for here. Our answer should be in the form delta T equals this. What else are we given? We're given that row of solid gallium is 5.885 grams per cubic centimeter. And row of liquid gallium is 6.08 grams per cubic centimeter. And we're given the heat of fusion is equal to 77.40 joules per gram. Now this is molar enthalpy, molar heat of fusion. So we're going to need to multiply this number times the molar weight would give us grams per mole. And the grams will cancel, we'll have joules per mole, which is the correct units for delta HF. Now let's find our formula and see what we need to do. So this has got solids and liquids. So again we're looking at the Clausius-Caperon equation. Which one is the solid liquid? This one. And in the interest of space we're just going to put that formula up. So here is our formula. And we have dP dt equals dHF. We've got dHF, that's the molar heat of fusion. The temperature is an absolute, we've changed to absolute. And the volumes here are also molar volumes. So let's see how we're going to find this part right here. V liquid minus V solid. Remember these are molar volumes, so we're going to need this molar weight again. So why don't we, until we need to see if we need to look it up, we'll just call it x. So we need x over liquid. So x over 6.08 minus x over 5885. Let's get a unit on that. So that would be in grams per mole. So x is grams per mole. This is grams per cubic centimeter. So we have x out front. We're going to find this difference here. And we have cc per mole. So here's our calculator. We have 1 divided by 6.08 minus 1 divided by 5.885. And that equals minus 005450. Let's store that somewhere. Store it in A. Enter. So minus 0.00545. Minus 0.00545. Okay, so let us see what we have. So we have dT is equal to 7740x. This x here, joules per mole divided by T. And then we have x times minus 00545 cubic centimeters per mole. We can see that the x is canceled, so we don't need to go find the molar weight of gallium. We can see that the moles cancel and that we have joules versus volume, which gives us pressure. And we have Kelvin here. So all our units are correct. They just need to be in the correct thing. So let's do 77.40 divided by this and then transfer it into atmospheres. So we need 1 over x. We need this on the bottom. So I found the 1 over x key. It is right there, the second one. Okay, answer. Times 77.4 equals... So this is minus 14202 divided by T and we have joules versus C. So that's minus 14202 versus T. Joules versus cubic centimeters. We have 10 to the sixth cubic centimeters over meters cubed. That will give us Pascals. And then we have 9.87 10 to the minus 6 atmospheres over Pascals. So that will give us atmospheres here. So let's find out how much that is equal to 1 over T times... And what do we have? Minus 1420, the 10 to the sixth cancels. So we just want to multiply this times 987. Times 9.87. And we get minus 140176. Minus 140176 atmospheres. Now I have a hard time deciding when they want either to integrate or not. So I just go for integration. So I have dP equals minus 140176 times DT or T. And we will integrate both sides. So we get the integral of this side is P equals minus 140176 times ln of T plus some constant. We will substitute our complete information here into this to get C and then substitute our partial information to get T2. So we have P over minus 140176 equals ln of T plus C. Substituting P1 to 1 and T to 30315 we will get C equals 1 over this minus ln of that. So let's do that with our calculator. Let's put this in a storage storage. Let's put it in A again. So we will put it in A and we need 1 over that. So just a second. Minus ln of 303.15. And we get that C equals minus 5714 minus 5714 minus 5714. And now that means that ln of T2 equals 2 over minus 140176 plus 5714 5.714. So T2 equals e to whatever this is. So let's find that. So we want to put this in our recall base. So we're going to store it in B and now we need 2 divided by R. Recall A, that gives us this part minus, because we've stored the minus in there, recall B. We'll not see what that is. So 5714 and then we need 8 of that. So second ln, second answer, enter. 30314783 and let's write that down. 303 subtract 30315 so minus 303.15 which was our T1, enter which is minus 002. So this is minus 00216. Stephanie. So this is our answer.