 So Closhes-Clapeyron equation The normal boiling point of benzene is 80.1 degrees Celsius and the heat of vaporization is 30.8 kilojoules per mole What is the boiling point of benzene in degrees Celsius on the top of Mount Everest where the pressure is 260 millimeters of mercury okay, so let's write down what the problem gives so it says We've got T1 normal boiling point is 80 point Yeah, okay, so T1 is 80.1 degrees C and It says the normal boiling point so that means the pressure do you guys know what the pressure would be? 180 right okay, but the other pressure is given to us in millimeters of mercury So let's put this one in millimeters of mercury to so that'd be 760 millimeters Okay, so pressure 1 is 760 It gives us the belted heat of vaporization is 30.8 kilojoules per mole Also gave us pressure 2 at the top of Mount Everest 260 millimeters of mercury And it wants to know well, what would be the new boiling point so that would be T2 Everybody okay with all of that stuff? So now you have to remember well, how do I do this problem use the Clausius-Clapeyron equation okay, so ln P2 over P1 equals negative Delta H that you have vaporization Over R remember, that's the new R that we just described or we just converted 1 divided by T2 Find this 1 divided by T1 So what are we looking for? We're looking for T2 here Okay, so you can go ahead and solve for this Like isolate that variable or you can just plug and then isolate the variable It doesn't matter to me so let's Semi-isolate the variable. How about that? Okay, so what we'll do is just put 1 over T2 minus 1 over T1 and set that equal to ln of P2 over P1 So that's all times R divided by the H, right? So it's gonna be 1 over T1 to everything so we get 1 over T2 equals We'll say negative ln P2 over P1 times R divided by Delta H that Right, and we're adding so that to What's all for that? Everybody okay with that? face the top two given to us as 8.314 joules or So what does that mean? For this temperature here what we got to do to it? We're going to Kelvin. Okay, so how do we do that? Give T2 and degrees Celsius, so we got to watch out and convert that back from kelvin when we get 760 millimeters of mercury 314 Joules per 1 All kelvin 8 of vaporization well it's 1 over the either vaporization so it's going to be 1 mole 30.8 kilojoules. So you see that we've got joules here and kilojoules here So we're going to have to convert one to the other right so let's just convert kilojoules or we'll convert So now let's cancel everything out so that cancels with that that cancels with that That cancels with that that cancels with that and we're left with now 1 over kelvin here added to 1 over kelvin Okay, and we're looking for 1 over temperature so that's a good unit. Does everybody understand what I'm saying? Okay, so now let's just plug and chug so First thing I would do is do 260 divided by 760 and then take the natural log of that number multiply that by 8.314 Divide that by 30.8 Divide that by 1000 do point a bunch of sig figs for right now That is 4 and it's at 2 this number 1 1 divided by 353.1 I did that in the calculator. So what I got was 0.003 1 2 So what I'm going to do now is take that number and say 1 divided by That number there What do I get? 820.3 But it wants it in degrees Celsius. So what do we have to do? subtract 273 kelvin from that 47.3 Any questions about doing something like this a couple more a few more I think recorded So if you're having trouble there a solving for different variables, okay? Questions before I kill it Take it