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. ΔHvap 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, P1 and P2 are the vapor pressures at T1 and T2. ΔHvap 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 41.0oC and a Pvap= 400 mmHg at 22oC. What is the heat of vaporization in KJ/mol?
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 P1 and T1 will be 400 mmHg and 295 K (22oC). 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 P2 and T2 are 760mmHg and 314 K (41oC). We know P2 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 40oC. You know at 23oC 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?
First, let’s layout the variables we know. We know one the ΔHvap is 48.2 KJ/mol. One temperature and pressure point is 296 K (23oC) and 120 mmHg. We only have the temperature for the other point, 313 K (40oC). 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 P2.
We need to rearrange to solve for P2.
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 P2.
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.