Clausius-Clapeyron Equation

Before we get into calculations dealing with the classiest flaperon equation, it's important to understand that the equation itself establishes the relationship between vapor pressure of liquids and temperature itself. Now recall vapor pressure represents an equilibrium between condensation were a gas goes down into a liquid and vaporization, which is the opposite where a liquid is converted into a gas. And one important piece of information is that as our temperature increases, our vapor pressure will also increase the classes flaperon equation is just a way of us looking at this change in vapor pressure as temperature increases or vice versa. So keep this in mind as we start to look at the different forms of the classiest clapper on equation and the calculations that are related to it.

now the classiest clapper on equation can be changed into different forms. The first time we're going to talk about is the linear form of it. We're gonna say we use this form of the equation when a plot of the natural law which is Ln of pressure versus inverse temperature is given. So if we talk about this we can look at it in terms of our graph here and any time we do a plot, remember a plot is always of Y versus X. Here Ry is L. N. P. Which is why it's on the y axis. And then inverse temperature just means one over T. That represents our X. So that's why it's here on the X axis. So we have inverse of the pressure and then we have inverse temperature one over T. Okay, basically we're gonna use the line form of the classes Clapper on equation to help us determine the entropy of vaporization which is delta H Vape. And that's connected to the idea of our slope intercept form of the straight line which is why equals mx plus B. So again, our why axis has L. N. P. R. X. Has won over team here with this information we can find our slope which equals rise over run which remember is change in your y divided by change in your ex here are linear form of the classes Clapper on equation becomes L. N. P equals negative delta H. Vape over R times one over T plus C. Again this is related to the slope intercept form of a straight line. So this would be Y equals M X plus B. M. Represents our slope which represents negative delta H. Vape. Over our now our itself remembers r gas constant, it's 8.314 with the units of jewels over moles times kelvin pressure here is P is our vapor pressure and it's in units of tour or millimeters of mercury. When we talk about our delta H A Vape, that's our entropy of vaporization here. It's customarily in jewels per mole. Sometimes you might see it in killer joules per mole but just make sure units match the units of our our in jewels. So that's why we tend to see delta H. A Vape in jewels. Now we can also say our temperature here is temperature in kelvin and then see is just the constant of a substance. Now again, when it comes to the linear form of the classes Clapper on equation. The most important part is this portion here in blue. This portion here helps us to link together the concept of slope and our entropy of vaporization. So keep that in mind when we're given the plot a substance where it's LMP versus inverse temperature and they're asking us to determine either the slope or the entropy of vaporization

the vapor pressure of a substance is measured over range of temperatures. A plot of the natural log of the vapor pressure versus the inverse of the temperatures in Calvin produces a straight line with a slope of negative 2.79 times 10 to the three kelvin find the entropy of vaporization of the substance. All right. So we need to find entropy of vaporization which is just our delta H of Vape. Now remember from the classes clapper on equation in its linear form, it's negative delta H. V over R equals your slope M. Here. What do we have? Well, we know the slope is negative 2.79 times 10 to the three Kelvin. We don't know what our entropy of vaporization is. That's what we're looking to find. Our is our gas constant 8.314 jewels over moles times K. Multiply here both sides by negative jewels over moles, times K. And when I do that, my kelvin's cancel out and my units will be in jewels promote negative times A negative will give me a positive at the end. So we have here delta H of vaporization here. This is 36 fix or answer. Also need three sig figs. This comes out to 2.32 Times 10 to before Jules promote. So this would represent my entropy of vaporization for this particular question

In this section of the classiest clapper on equation, we're going to take a look at its 2.4. Now we use this form of the equation when two temperatures and or two pressures are mentioned. And we're going to say that we need to remember when given normal boiling point Pressure equals 760 tour or 760 of mercury. This is known as our normal pressure. Now with the two point of the classiest copper an equation, we now have L N A P two over P one equals negative delta H V over R times one over T two minus one over T one. Here we're gonna say ours our gas constant, it's 8.314 jewels over moles times K P equals our favorite pressure again in tours or millimeters of mercury. Delta H Vape is our entropy of vaporization, which is in jewels per mole. And the temperatures here have to be in units of kelvin. So keep this in mind when we're talking about the classiest clapper on equation and were given to temperatures and or two pressures, we have to use a two point form of the equation

In this example question, it says the entropy of vaporization of water is 40.30. Killer jewels promote at its normal boiling point of 100°C. What is the vapor pressure in mm of mercury of water at 660°C. Alright, so they're telling us we have to temperatures remember our temperatures need to be in Calvin. Since this is the first temperature that's given to us this year would represent our T one. This is the second temperature. So this would be our T. two. We add to them to 73.15 to get our Kelvin's. So this is 373-15 Kelvin. And then here This comes out to 333.15 Kelvin. What can we say next? We can say here they're telling us normal boiling point. Remember normal boiling point. Is that our pressure Which is P one in this case would be 760 of mercury. Or in tours since they want the second pressure P. Two to be in millimeters of mercury, We'll keep it at 760 of Mercury here. Alright, so now we're gonna use our equation allen of P two over P one equals negative delta H. Of Vape Divided by R. Times one over T 2 -1 over T. one. Here we have L. N. We don't know what R. P two is. That's what we're looking to find. P one is 760 of Mercury. Now remember our here is in jewels. So delta H of vaporization also needs to be in jewels. So the units can match. So converting 40.30 into jules comes out to negative 40,300 jewels per mole. And here we're gonna have one over R. T. two is 333.15 Kelvin -1/3 73.15 Kelvin. All right. So now we find what this is in our calculator and we multiply it by this in our calculator. So when we multiply those two numbers together, We're going to get as our current answer negative 1559667. And that equals a line of P two over 760. Now to get rid of the Ellen on the left side, we're gonna take the inverse of the natural log of both sides. So all that means is we're going to do uh huh. And then this part here becomes a power so e to the negative 1.559667. Taking the inverse of the natural log on the left side, basically cancel out this Ellen. So that equals P two over 760. All we do now is we multiply 760 times e to the negative 1.559667. And we'll isolate our p. two when we do that we get our P two s 59.76 millimeters of mercury. Here we have all the answers in terms of four sig figs. So this just becomes 1 59.8 millimeters of mercury, which gives me options, see as my correct answer. So using the two point form of the classes, clapper on equation gives us this as our final answer.