**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 heat of vaporization and vapor pressure. The Clausius Clapeyron equation is shown below in a form similar to a linear equation (y=mx+b).

Vapor pressure rises non-linearly as temperature rises since they-term in this case is a natural log of the vapor pressure. ΔH_{vap} is the heat of vaporization. R is the gas constant. 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 verse temperature is plotted. This plot is shown below.

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.

In these equations, P_{1} and P_{2} are the vapor pressures at T_{1} and T_{2}. ΔH_{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:

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^{o}C and a P_{vap}= 400 mmHg at 22^{o}C. What is the heat of vaporization in 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.

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

Plug these values into our equation above:

The simplify and rearrange:

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^{o}C. You know at 23^{o}C it had a vapor pressure of 120 mmHg. The heat of vaporization of the liquid is 48.2 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 ΔH_{vap} is 48.2 KJ/mol. One temperature and pressure point is 296 K (23^{o}C) and 120 mmHg. We only have the temperature for the other point, 313 K (40^{o}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}.

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

Using logarithm rules we rearrange to:

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

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

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!

We get our final answer of 347 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.