Clausius Clapeyron Equation

Core Concepts: In this tutorial, you will learn what the Clausius Clapeyron Equation is and how to apply it to problems. At the end, there are several problems with step-by-step solutions.

What is the Clausius Clapeyron Equation

The Clausius-Clapeyron equation relates the enthalpy of vaporization and vapor pressure. The Clausius Clapeyron equation is shown below in a form similar to a linear equation ( $y=mx+b$ ).

$\begin{gather*} {lnP_{vap} = \left(\frac{-\Delta H_{vap}}{R}\right) \frac{1}{T} + C} \end{gather*}$

Vapor pressure rises non-linearly as temperature rises since the y-term in this case is a natural log of the vapor pressure. $\Delta H_{vap}$ is the heat of vaporization. $R$ is the gas constant. Be careful of which value for the gas constant is used, depending on the units of the other variables. $T$ is the temperature. And $C$ is a constant that is specific to the liquid being examined.

Below is a plot of the vapor pressure of water verse the temperature.

To make the graph linear, the natural log of the vapor pressure against the inverse temperature is plotted. This plot is shown below.

vapor pressure of water formatted with Clausius Clapeyron Equation

Other names for the equation include the Clapeyron equation or the Clapeyron-Clausius equation.

Using the Clausius-Clapeyron Equation

The equation can be used to solve for the heat of vaporization or the vapor pressure at any temperature.

To determine the heat of vaporization, measure the vapor pressure at several different temperatures. The vapor pressure and temperature can then be plotted. The slope of the line will be the vapor pressure.

Another technique is to determine algebraically what the heat of vaporization is. This is done by plugging in two known vapor pressures and temperatures to either of the equations below.

$\begin{align*} {ln\frac{P_{1}}{P_{2}} &= \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)} \\ {ln\frac{P_{1}}{P_{2}} &= \frac{-\Delta H_{vap}}{R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)} \end{align*}$

In these equations, $P_{1}$ and $P_{2}$ are the vapor pressures at $T_{1}$ and $T_{2}$ . $\Delta H_{vap}$ is the enthalpy of vaporization for the substance. The equation comes from the fact that at both temperatures, the constant $C$ (from the very first equation) must be equal. That is:

$\begin{gather*} {lnP_{1} + \left( \frac{\Delta H_{vap}}{R} \right) \frac{1}{T_{1}} = C = -lnP_{2} + \left( \frac{\Delta H_{vap}}{R} \right) \frac{1}{T_{2}}} \end{gather*}$

The equation can then be rearranged into the other forms to easily solve for whatever variable you need.

Clausius Clapeyron Equation Example Problems

Calculating Heat of Vaporization

Given a liquid that has a normal boiling point of $41.0\degree \text{C}$ and a $P_{vap} = 400 \text{mmHg at }22 \degree \text{C}$ . What is the heat of vaporization in $\text{kJ/mol}$ ?

Worked Answer:

We could solve the problem either graphically or algebraically. We have the equation from above that we can easily plug our two known points into and solve for.

$\begin{gather*} {ln\frac{P_{1}}{P_{2}} = \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)} \end{gather*}$

Our $P_{1}$ and $T_{1}$ will be $400 \text{mmHg and }295 \text{K }(22\degree \text{C})$ . The temperature needs to be in kelvin because our $R$ constant has units of Kelvin. $R$ is equal to $8.3145 \text{J/Kmol}$ . Our second $P_{2}$ and $T_{2}$ are $760\text{mmHg and }314 \text{K }(41\degree [text{C})$ . We know $P_{2}$ because the boiling point will be at the standard pressure, which is $760 \text{mmHg}$ .

Plug these values into our equation above:

$\begin{gather*} {ln\frac{400\text{mmHg}}{760\text{mmHg}} = \frac{\Delta H_{vap}}{8.3145\text{J/Kmol}} \left( \frac{1}{314\text{K}} - \frac{1}{295\text{K}} \right)} \end{gather*}$

The simplify and rearrange:

$\begin{gather*} {\frac{-0.698 \cdot 8.3145 \text{J/Kmol}}{\frac{1}{314\text{K}} - \frac{1}{295\text{K}}} = \Delta H_{vap} = 26017 \text{J/mol} = 26.017 \text{J/mol}} \end{gather*}$

Make sure to always check your units! For example, here R has units of joules but the question asked for units of kilojoules. So, at the end, we need to convert units.

Calculating the Vapor Pressure

Suppose you have a liquid that you heat to $40\degree \text{C}$ . You know at $23\degree \text{C}$ it had a vapor pressure of $120 \text{mmHg}$ . The heat of vaporization of the liquid is $48.2 \text{kJ/mol}$ . What is the vapor pressure at elevated temperature for the liquid?

Worked Answer:

First, let’s layout the variables we know. We know one the $\Delta H_{vap}$ is $48.2 \text{kJ/mol}$ . One temperature and pressure point is $296 \text{K }(23\degree \text{C})$ and $120 \text{mmHg}$ . We only have the temperature for the other point, $313 \text{K }(40\degree \text{C})$ . Looking at our equations, all but one variable is known. Therefore, our strategy will be to rearrange and then plug into the equation to solve. We want to solve for $P_{2}$ .

$\begin{gather*} {ln\frac{P_{1}}{P_{2}} = \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)} \end{gather*}$

We need to rearrange to solve for $P_{2}$ .

Using logarithm rules we rearrange to:

$\begin{align*} {lnP_{1} - lnP_{2} &= \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)} \\ {-lnP_{2} &= \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right) -lnP_{1}} \end{align*}$

Then, to make it easier we multiply both sides by -1 to get rid of the negative sign on $P_{2}$ .

$\begin{align*} {lnP_{2} &= \frac{-\Delta H_{vap}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right) +lnP_{1}} \end{align*}$

Now we plug in our known values and solve. Keep an eye on your units!

$\begin{align*} {lnP_{2} = \frac{-48.2\text{kJ/mol}}{8.3145\text{J/Kmol}} \left( \frac{1}{313\text{K}} - \frac{1}{296\text{K}} \right) +ln(120\text{mmHg})} \end{align*}$

As we solve, the heat of vaporization and the $R$ constant have different units. Also, don’t forget the negative sign in front of the heat of vaporization!

$\begin{align*} {lnP_{2} &= -\frac{48.2\text{kJ/mol}}{8.3145\text{J/Kmol}} \left( -0.00018 \text{K}^{-1} \right) +4.787 \text{mmHg}} \\ {P_{2} &= 347 \text{mmHg}} \end{align*}$

We get our final answer of $347 \text{mmHg}$ . This answer seems reasonable. An example of an unreasonable answer would be if you got a negative number since negative pressures don’t make sense.

Clausius Clapeyron Equation Practice Problems

Problem 1

At $300\text{K}$ , a sample of gas has a pressure of $150\text{mmHg}$ , while at $350\text{K}$ , the gas has a pressure of $400\text{mmHg}$ . What is the heat of vaporization of this unknown gas?

Problem 2

A certain liquid has an enthalpy of vaporization of $14.4 \text{kJ/mol}$ at $180\text{K}$ , its normal boiling point. The molar volume the vapour at the boiling point is $14.5 \text{L/mol}$ . What is the $C$ of this liquid?