In this video, we're going to focus on the Clausius-Clapeyron equation. We're going to go over the different forms of the equation that you need to know, and also a few examples and practice problems, so you can apply this information. So you're probably familiar with this form of the equation, the natural log of the second vapor pressure divided by the first, which is equal to negative the enthalpy of vaporization.
Divided by R and then times 1 over T2 minus 1 over T1 Now the enthalpy of vaporization it has the units Joules per mole not kilojoules per mole has been joules So if you need to convert kilojoules into joules multiplied by thousand one kilojoule is about a thousand joules So going from kilojoules to joules and multiply by a thousand, but if you need to go from joules to kilojoules, divide by a thousand. Now R is the energy constant 8.3145, and it has the units joules per mole per Kelvin. So that's why delta H has to be in joules per mole, because R is in joules per mole. T is the Kelvin temperature, not Celsius, and if you need to convert to Kelvin, simply add 273 to the Celsius temperature.
So that's the first form of the equation. By the way, sometimes you'll see this negative sign here, and other times you won't. So you might see this equation in this form.
Sometimes you'll see ln P1 over P2, and this is going to be equal to positive delta H of vaporization over R times 1 over T2 minus 1 over T1. So if you flip... P2 and P1, you basically add in a negative sign, which will cancel with this one, making the equation positive.
Or, if you reverse T2 and T1, you're going to introduce a new negative sign. So sometimes, if you see that negative sign, or if you don't see it, it's probably because the vapor pressures were reversed, or the temperatures were reversed. But both of these equations can give you the same answer.
Now, there are some other forms of the equation that you want to be familiar with. If you need to find the vapor pressure at a new temperature, use this equation. P2 is equal to P1 times E raised to the negative delta H of vaporization divided by R times 1 over T2 minus 1 over T1.
So keep in mind, this entire term right here is the exponent of E. And if you're wondering what E is, E is a number. E is like 2.718. It's the inverse of the natural log function.
So if you go to your calculator and type in second natural log or shift natural log, you're going to get the E button. So this is the form of the equation that you want to use if you need to calculate the second vapor pressure at a different temperature. Now sometimes, you may need to calculate the heat of vaporization.
And if you need to do that... This is the form of the equation that you want to use. The enthalpy of vaporization is equal to negative R times the natural log of P2 divided by P1 all divided by 1 over T2 minus 1 over T1. So that's the second form of the equation.
Now sometimes you may need to find the second temperature at a different vapor pressure. So if you need to do that, use this equation. T2 is equal to 1 divided by T1 minus R ln P2 divided by P1 divided by the enthalpy of vaporization in joules per mole and not kilojoules per mole.
And this is all raised to the negative 1. So if you type this in your calculator, you should get the Kelvin temperature T2. So now you have the three most important forms of this equation. So let's work on some example problems.
Okay, so let's go ahead and try this problem. The vapor pressure of a substance is 21 Torr at 300 Kelvin. Calculate the vapor pressure at 310 Kelvin if the enthalpy of vaporization is 24 kilojoules per mole. So, we have the first pressure, that's 21 Torr, and we know that the first temperature is 300 Kelvin.
The second temperature is 310 Kelvin, and we're looking for the second pressure, and we have the enthalpy of vaporization. So the equation that we need is P2, which is equal to P1 times E raised to the negative enthalpy of vaporization over R times 1 over T2 minus 1 over T1. So let's plug in the information that we have. P1 is 21. And the enthalpy of vaporization, it's in kilojoules per mole, so we've got to convert that to joules per mole. So we need to multiply it by 1,000.
24 times 1,000 is 24,000. And we're going to divide that by 8.3145, the energy constant. And T2 is 310, and T1 is 300. So I'm going to plug this in the calculator one step at a time. So 1 divided by 310, you should get 3. 2 times 10 to negative 3. And if you subtract that by 1 over 300, you should now have negative 1.075 times 10 to negative 4. Then multiply that by negative 24,000.
You should now have positive 2.58. And then divide that by 8.3145. And you should have 0.31037.
So I have e raised to the 0.31037. By the way... Even though I'm saying rounded answers, I'm actually typing in the exact answer in my calculator.
So, I use the previous answer to get to the next answer. I don't round in between. So, if you're rounding in between, your answers might be slightly different from what I'm saying.
Now, E raised to the 0.31037, you should get about 1.3639. And if we multiply that by 21, the new vapor pressure at... 310 is going to be 28.6 torr.
So that's the answer for this particular problem. So as you can see, as the temperature increases, the vapor pressure goes up as well. The vapor pressure is proportional to the temperature. Not like in a linear proportion, but the fact that it just goes up whenever the temperature goes up.
So let's try this problem. Feel free to pause the video and see if you can get the answer yourself. So the vapor pressure of a substance is 30 Torr at 250 Kelvin.
At what temperature will the substance have a vapor pressure of 150 Torr? And so we have the enthalpy of vaporization. Let's make a list of the facts that we have to know.
So P1 is 30 Torr and T1 is 250 Kelvin. Now we have P2, that's 150 Torr. And we're looking for T2, which we don't know what it is. And we have delta H, that's 45 kilojoules per mole, and we're going to have to convert that into joules per mole. So we need to use a different form of the equation.
Let's use this form. T2 is equal to 1 over T1 minus R ln P2 divided by P1 over delta H of vaporization. all raised to the negative 1 power. So let's go ahead and plug in the data. So T1 is 250 Kelvin and we know R is 8.3145 and then natural log, the second pressure is 150, the first pressure is 30, divided by the entropy of vaporization.
So to convert kilojoules to joules we need to multiply by a thousand so 45 times a thousand. is 45,000. Okay, so let's make some space. So in the calculator, let's start with natural log 150 over 30. 150 divided by 30 is 5, and the natural log of 5 is 1.609437. Multiply that by 8.3145, you should get 13.38167.
Then divide that by 45,000, and what you should have is negative 2.9737 times 10 to negative 4. And 1 over 250, that's about... 4 times 10 to negative 3 so 4 times 10 to negative 3 minus 2.97 37 times 10 to negative 4 you should get 3.7 0 to 6 3 times 10 to negative 3 and then if you raise that to the negative 1 power you should get 270.1 Kelvin So that's going to be the new temperature. So we can see that at a higher vape pressure, the temperature is going to be higher. So whenever the temperature goes up, the vape pressure goes up. So the answer has to be above 250 Kelvin.
If it's not, you know, something is wrong. And so now you know how to calculate the new temperature if you're given another vape pressure.