Clausius-Clapeyron Equation - Online Tutor, Practice Problems & Exam Prep

The Clausius-Clapeyron equation describes the relationship between the vapor pressure of liquids and temperature, indicating that vapor pressure increases with temperature. This equation can be presented in a linear form, which is useful when plotting the natural logarithm (Ln) of vapor pressure against the inverse of temperature (1/T). The linear form,

$\mathrm{Ln}P=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T}\right)+C$

allows us to calculate the enthalpy of vaporization (ΔHvap), with R being the gas constant (8.314 J/mol·K). The slope of the plot corresponds to -ΔHvap/R, linking the concept of slope to the enthalpy of vaporization. When dealing with two temperatures or pressures, the two-point form of the equation is used:

$\mathrm{Ln}\left(\frac{P_2}{P_1}\right)=-\frac{\Delta H_{\mathrm{vap}}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

with P representing vapor pressure in torr, ΔHvap in joules per mole, and temperatures in kelvin. Remember, at normal boiling point, the pressure is 760 torr.

Clausius-Clapeyron Equation Concept 1

Before we get into calculations dealing with the Clausius-Clapeyron 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 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.

Clausius-Clapeyron Equation Concept 2

Now the classiest clapron equation can be changed into different forms. The first one we're going to talk about is the linear form of it. We're going to say we use this form of the equation when a plot of the natural log, which is lane 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&P which is why it's on the Y axis. And then inverse temperature just means 1 / 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 / T. Basically we're going to use the line form of the classes Clapper on equation to help us determine the enthalpy of vaporization which is ΔHvape. And that's connected to the idea of our slope intercept form of a straight line which is Y equals MX plus B. So again our Y axis has LNP. Our X has 1 / 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 classes Clapperon equation becomes lnP=−ΔHvapeR⋅1T+C. Again this is related to the slope intercept form of a straight line. So this would be Y equals MX plus B represents our slope, which represents −ΔHvapeR. Now R itself remembers 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 Tor or millimeters of mercury.

When we talk about our ΔHvape, that's our entropy 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 ΔHvape in Joules. Now we can also say our temperature here is temperature in Kelvin and then C is just a constant of a substance.

Now again, when it comes to the linear form of the clauses Clapperon 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 a substance where it's LMP versus inverse temperature, and they're asking us to determine either the slope or the entropy of vaporization.

Clausius-Clapeyron Equation Example 1

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 Kelvin produces a straight line with a slope of -2.79 * 10 to the three Kelvin. Find the enthalpy of vaporization of the substance.

All right, so we need to find enthalpy of vaporization, which is just our delta H of vape. Now remember from the Clausius 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 -2.79 * 10 to the three 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 sides by -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 Fix or answer also needs three sig figs. This comes out to 2.32 * 10 to the four joules per mole, so this would represent my enthalpy of vaporization for this particular question.

ln ( P ) = - ΔH _ vape R ⋅ 1 T + C ΔH _ vape = - slope ⋅ R ΔH _ vape = - ( -2.79 ⋅ 10 ^ 3 ) ⋅ 8.314 = 2.32 ⋅ 10 ^ 4 J / mol

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

Clausius-Clapeyron Equation Concept 3

In this section of the Clausius Clapeyron equation we're going to take a look at its 2.4. Now we use this form of the equation when two temperatures and Oregon 2 pressures are mentioned. And we're going to say that we need to remember when given normal boiling point pressure equals 760 Tor or 760mm of mercury. This is known as our normal pressure.

Now with the two point of the Clausius Clapper on equation we now have

ln ⁡ ( P 2 P 1 ) = − Δ H vape R ⁢ ( 1 T 2 − 1 T 1 )

Here we're going to say R is our gas constant. It's 8.314 Joules over moles times KP equals our vapor pressure. Again in tours 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 classiest Clapper on equation and we're given two temperatures and Oregon 2 pressures, we have to use a two point form of the equation.

Clausius-Clapeyron Equation Example 2

In this example question, it says the enthalpy of vaporization of water is 40.30 kilojoules per mole at its normal boiling point of 100�C. What is the vapor pressure in millimeters of mercury of water at 60�C? All right, so they're telling us we have two temperatures. Remember, our temperatures need to be in Kelvin, since this is the first temperature that's given to us this year would represent our T₁. This is the second temperature, so this would be our T₂. 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 P₁ in this case would be 760mm of mercury. Or in Tours since they want the second pressure P₂ to be in millimeters of mercury. We'll keep it at 760 mm of mercury here. Alright, so now we're going to use our equation. ln(P₂/P₁) = -ΔH_vap / R * (1/T₂ - 1/T₁). Here we have ln. We don't know what our P₂ is. That's what we're looking to find. P₁ is 760mm of mercury.

Now remember R here is in joules, so ΔH_vap also needs to be in joules so the units can match. So converting 40.30 into Joules comes out to -40,300 Joules per mole. And here we're going to have 1/T₂ is 1/333.15 Kelvin, -1/373.15 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 -1.559667 and that equals ln(P₂/760).

Now to get rid of the ln on the left side, we're going to take the inverse of the natural log of both sides. So all that means is we're going to do e and then this part here becomes a power, so e^-1.559667 taking the inverse of the natural log on the left side basically cancels out this ln so that equals P₂/760. All we do now is we multiply 760 * e^-1.559667 and we'll isolate P₂. When we do that we get our P₂ as 159.76mm of Mercury. Here we have all the answers in terms of four sig figs. So this just becomes 159.8mm of mercury, which gives me option C as my correct answer. So using the two point form of the Clausius-Clapeyron equation gives us this as our final answer.

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?

• A. 251.9 K
• B. 720.7 K
• C. 924.2 K
• D. 333.2 K

Here’s what students ask on this topic:

The Clausius-Clapeyron equation relates which three values to each other?

The Clausius-Clapeyron equation provides a way to quantify the relationship between the pressure, temperature, and the enthalpy of vaporization (or sublimation) during a phase transition between two phases of matter, such as from liquid to gas. Specifically, it relates the change in vapor pressure with the change in temperature to the latent heat of vaporization (or sublimation) of the substance. In a more technical sense, the equation gives the slope of the curve on a pressure-temperature (P-T) diagram that represents the phase boundaries. The equation is usually expressed in the form:

$\frac{\mathrm{dP}}{\mathrm{dT}}=\frac{L}{T(V_g-V_l)}$

where $\frac{\mathrm{dP}}{\mathrm{dT}}$ is the rate of change of pressure with respect to temperature, $L$ is the latent heat of vaporization or sublimation, $T$ is the absolute temperature, and

What does the Clausius-Clapeyron equation represent?

The Clausius-Clapeyron equation is a way to characterize the relationship between the pressure and temperature of a substance during a phase change, such as from liquid to gas (evaporation) or from solid to liquid (melting). Specifically, it describes how the pressure of a substance's vapor changes in response to its temperature, assuming the phase change occurs at equilibrium.

The equation is usually written as:

$\frac{dp}{dT}=\frac{L}{T∆v}$

Here, $\frac{dp}{dT}$ represents the rate of change of pressure with respect to temperature, $L$ is the latent heat of the phase transition (the amount of heat required to change the phase of a unit mass of the substance), $T$ is the temperature in Kelvin, and $∆v$ is the change in specific volume between the two phases.

In essence, the Clausius-Clapeyron equation allows us to predict how the equilibrium pressure for a phase transition varies with temperature,

What is "r" in the Clausius-Clapeyron equation?

In the Clausius-Clapeyron equation, "r" stands for the latent heat of vaporization or sublimation. This is the amount of heat energy required to change a substance from a liquid to a gas or from a solid to a gas, without changing its temperature. It's important to note that this value is specific to the substance in question and can vary depending on the pressure and temperature at which the phase change occurs. In the context of the equation, "r" is used to describe the relationship between the pressure and temperature during a phase change, providing a way to predict how the equilibrium vapor pressure of a substance changes with temperature.

What is "r" in the Clausius-Clapeyron equation?

In the Clausius-Clapeyron equation, "r" stands for the latent heat of vaporization or sublimation. This is the amount of heat energy required to change a substance from a liquid to a gas or from a solid to a gas, without changing its temperature. It's important to note that this value is specific to the substance in question and can vary depending on the pressure and temperature at which the phase change occurs. In the context of the equation, "r" is used to describe the relationship between the pressure and temperature during a phase change, providing a way to predict how the equilibrium vapor pressure of a substance changes with temperature.