The Clausius-Clapeyron equation describes the transition between the phases of two states of matter of the same substance. This is the case of water and the transitions between its different states, as shown in the phase diagram in the figure. The Clausius-Clapeyron equation can be used to determine vapor pressure as a function of temperature, or also to calculate the heat of phase transition, which involves given vapor pressures at two different temperatures. Vapor pressure and temperature do not usually have a linear relationship; In the case of water, the vapor pressure increases faster than the temperature. The Clausius-Clapeyron equation allows us to calculate the slope of the tangent line at each point of the curve that represents the variation of vapor pressure as a function of temperature.
Let’s see an application of the equation proposed by Rudolf Clausius and Benoit Emile Clapeyron. The vapor pressure of 1-propanol is 10 torr at 14.7 °C, and the heat of vaporization of 1-propanol = 47.2 kJ/mol; what is the vapor pressure at 52.8°C?
The expression of the Clausius-Clapeyron equation is as follows
ln[P T1,vap / P T2,vap ] = (ΔH vap / R)[1/T 2 – 1/T 1 ]
This equation relates the vapor pressures and the temperature in two states, 1 and 2, and the heat of vaporization, expressed by the enthalpy of vaporization ΔH vap . In our problem, state 1 will correspond to temperature T 1 = 14.7 °C and vapor pressure P T1,vap = 10 torr, while state 2 will be the one with temperature T 2 = 52.8 °C, being the pressure P T2,vap the value that we want to determine. R is the ideal gas constant; R = 0.008314 kJ/K mol.
In the Clausius-Clapeyron equation, temperature is expressed in Kelvin scale values, so the first step is to convert the temperatures from our Celsius degree problem to the Kelvin scale. To do this, we must add to 273.15, and then T 1 = 287.85 K and T2 = 325.95 K
Now we can plug the values from our problem into the Clausius-Clapeyron equation.
ln[10 / P T2,vap ] = (47.2 / 0.008314)[1/325.95 – 1/287.85]
If we carry out the operations indicated in the right-hand term of the equality, we obtain
ln[10 / P T2,vap ] = -2,305
In order to isolate the value of P T2,vap that is affected by the logarithm, we apply the antilogarithm to both sides of the equality, or what is equivalent, we apply the power of both terms of the equality to the number e (2.718), and the following equality is obtained:
10 / P T2,vap = 0.09972
Calculating the inverse value of both sides of the equality and passing the value 10, it is obtained that
P T2,vap = 100.3
Therefore, the vapor pressure of 1-propanol at 52.8 °C is 100.3 torr.
Sources
Goldberg, David. 3000 Solved problems in chemistry . McGraw-Hill Education 2011.
Haynes, William. CRC Handbook of Chemistry and Physics . CRC Press Book, 2012.
