Clausius-Clapeyron Equation - Video Tutorials & Practice Problems
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The Clausius-Clapeyron Equation establishes a quantitative relationship between vapor pressure and temperature.
Examining the Clausius-Clapeyron Equation
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concept
Clausius-Clapeyron Equation Concept 1
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Before we get into calculations dealing with the Clausius Clapeyron equation, it's important to understand the equation itself establishes the relationship between vapor pressure of liquids and temperature itself. Now recall vapor pressure represents an equilibrium between condensation where 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 Clausius Clapeyron 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 Clausius Clapeyron equation and the calculations that are related to it.
Vapor Pressure looks at the equilibrium established between vaporization and condensation. By using the Clasius-Clapeyron equation, the enthalpy of vaporization can be determined.
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Clausius-Clapeyron Equation Concept 2
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Now the Clausius Clapeyron equation can be changed into different forms. The first one 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 log, 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 anytime we do a plot, remember a plot is always of y versus x. Here our y is l and p, which is why it's on the y axis, And then inverse temperature just means 1 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, 1 over t. Basically, we're gonna use the line form of the Clausius Clapeyron equation to help us determine the enthalpy of vaporization, which is delta h vape. And that's connected to the idea of our slow intercept form of the straight line, which is y equals mx plus b. So again, our y axis has l and p, our x has one over t. 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 x. Here our linear form of the Clausius Clapeyron equation becomes lnp equals negative delta h vape over r times 1 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 big over r. Now r itself, remember, is our gas constant. It's 8.314 with the units of joules over moles times Kelvin. Pressure here is p is our vapor pressure and it's in units of Torr or millimeters or mercury. When we talk about our delta h of vape, that's our enthalpy of vaporization, here it's customarily in joules per mole. Sometimes you might see it in kilojoules per mole, but just make sure units match. The units of r are in joules, so that's why we tend to see delta h of vape in joules. Now we can also say our temperature here is temperature in Kelvin, and then c is just the constant of a substance. Now again, when it comes to the linear form of the Clausius Clapeyron 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 enthalpy of vaporization. So keep that in mind when we're given the plot of temperature, and they're asking us to determine either the slope or the enthalpy of vaporization.
Linear Form of Clausius-Clapeyron Equation used when given a plot of lnP vs inverse of Temperature.
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example
Clausius-Clapeyron Equation Example 1
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The vapor pressure of a substance is measured over a range of temperatures. A plot of the natural log of the vapor pressure versus the the inverse of the temperatures in Kelvin produces a straight line with a slope of negative 2.79 times 10 to the 3 Kelvin. Find the enthalpy of vaporization of the substance. Alright. So we need to find enthalpy of vaporization, which is just our delta h of vape. Now remember, from the classic Clapeyron equation in its linear form, it's negative delta h vape over r equals your slope m. Here, what do we have? Well, we know the slope is negative 2.79 times 10 to the 3 Kelvin. We don't know what our enthalpy of vaporization is. That's what we're looking to find. R is our gas constant, so 8.314 joules over moles times k. Multiply here both signs by negative 8.314 joules over moles times k. And when I do that my kelvins cancel out and my units will be in joules per mole. Negative times a negative will give me a positive at the end, so we'll have here delta h of vaporization. Here, this is 3 sig figs, so our answer also needs 3 sig figs. This comes out to 2.32 times 10 to the 4 joules per mole. So this would represent my enthalpy of vaporization for this particular question.
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Problem
Problem
Vapor pressure measurements at various temperature values are given below. Determine the molar heat of vaporization for cyclohexane.
A
11,520 J/mol
B
72,193 J/mol
C
33,147 J/mol
D
52,968 J/mol
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concept
Clausius-Clapeyron Equation Concept 3
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In this section of the Clausius Clapeyron equation, we're gonna take a look at its 2.4. Now we use this form of the equation when 2 temperatures and or 2 pressures are mentioned, and we're going to say that we need to remember when given normal boiling point, pressure equals 760 torr or 760 millimeters of mercury. This is known as our normal pressure. Now with the two point of the Clausius Clapeyron equation, we now have l n of p 2 over p 1 equals negative delta h vape over r times 1 over t 2 minus 1 over t 1. Here we're gonna say r is our gas constant. It's 8.314 joules over moles times k, p equals our bagel pressure, again in torrs or millimeters of mercury. Delta h vape is our enthalpy of vaporization, which is in joules per mole, and the temperatures here have to be in units of Kelvin. So keep this in mind when we're talking about the Clausius Clapeyron equation and we're given 2 temperatures and or 2 pressures, we have to use a two point form of the equation.
Two-Point Form of Clausius-Clapeyron Equation used when 2 Temperatures and 2 Pressures are involved.
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example
Clausius-Clapeyron Equation Example 2
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In this example question, it says, the enthalpy of vaporization of water is 40.30 kilojoules per mole at its normal boiling point of a 100 degrees Celsius. What is the vapor pressure in millimeters of mercury of water at 6 60 degrees Celsius? Alright. So they're telling us we have 2 temperatures. Remember, our temperatures need to be in Kelvin. Since this is the first temperature that's given to us, this here would represent our t one. This is the second temperature so this would be our t 2. We add to them 273.15 to get our kelvins, 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 means that our pressure, which is p 1 in this case, would be 760 millimeters of mercury or in tours. Since they want the second pressure, p 2, to be in millimeters of mercury, We'll keep it at 760 millimeters of mercury here. Alright. So now we're gonna use our equation, ln of p 2 over p 1 equals negative delta h of vape divided by r times 1 over t 2 minus 1 over t 1. Here we have l n, we don't know what our p 2 is, that's what we're looking to find. P 1 is 760 millimeters of mercury. Now remember, r here is in joules so delta h of vaporization also needs to be in joules, so the units can match. So converting 40.30 into joules comes out to negative 40,000 and 300 joules per mole. And here we're gonna have 1 over our t 2 is 333.15 Kelvin minus 1 over 373.15 Kelvin. Alright. 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 1.559667, and that equals l n of p 2 over 760. Now to get rid of the l n on the left side, we're gonna take the inverse of the natural log of both sides. So all that means is we're gonna do e, 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 l a. So that equals p 2 over 760. All we do now is we multiply 760 times e to the negative 1.559667, and we'll isolate our p 2. When we do that, we get our p 2 as 159.76 millimeters of mercury. Here, we have all the answers in terms of 4 sig figs, so this just becomes 159.8 millimeters of mercury, which gives me option c as my correct answer. So using the 2.4 of the Clausius
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Problem
Problem
Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1°C. At what temperature does benzene boil when the external pressure is 405 torr?