# Clausius-Clapeyron equation: what it is for, examples, exercises

This equation is derived from the thermodynamic laws and from the PT diagrams for a substance, where the line of coexistence is observed; one that separates two different phases (liquid-vapor, solid-liquid, etc.). To cross this line, the gain or loss of heat is necessary, such as the enthalpy of vaporization, ΔH _{vap} .

The upper image shows the Clausius-Clapeyron equation before it is integrated. It is usually applied for liquid-vapor systems, where ΔH _{vap} is used and you want to calculate what the vapor pressure of the liquid will be at a certain temperature. It is also used to calculate the ΔH _{vap} of a certain liquid in a range of temperatures.

_{sub}, is considered .

__What is the Clausisu-Clapeyron equation for?__

__What is the Clausisu-Clapeyron equation for?__

**Calculation of pressure changes and enthalpies of vaporization**

From the Clausius-Clapeyron equation above, we proceed to have some mathematical considerations to finally carry out an integration. For example, for a liquid-vapor system, it is assumed that ΔH _{vap} does not vary with temperature, and that ΔV corresponds exclusively to the volume of the vapor, neglecting the volume of the liquid (V _{vapor} -V _{liquid} = V _{vapor} ).

Assuming that the steam behaves as an ideal gas and integrating, the integrated Clausius-Clapeyron equation is obtained:

This equation corresponds to that of a line that is graphed as:

Ln P vs 1 / T

And whose negative slope is (ΔH / R). For this equation to hold, therefore, ΔH must be constant over the temperature interval (T _{2} -T _{1} ) in which the vapor pressures in equilibrium with the liquid are measured.

In this way, if it is assumed that ΔH varies little within small temperature intervals, it is possible to use the equation of this line to predict changes in the vapor pressure of a liquid; and even more, its ΔH of vaporization can be determined.

The larger the temperature ranges considered, the larger the deviation of this equation from the experimental data, and the less it will be true.

**Determination of phase changes**

Thus, the Clausius-Clapeyron equation becomes the development of a tangent line to the line of coexistence between two physical phases, which is observed in any PT diagram for a substance.

__Usage examples__

__Usage examples__

– The Clausius-Clapeyron equation has been used in meteorology to study the behavior of clouds, even those present on other planets or moons with atmospheres.

– It has been used to determine the enthalpy of fusion of various metals such as sodium and gallium, and to extrapolate their vapor pressures at very high temperatures.

– It has also been used to determine the enthalpy of vaporization of substances such as chlorine gas, carbon tetrachloride, liquid water, ice and iodine.

– It has also served to study phase changes in crystalline structures. In this last example, the integrated Clausius-Clapeyron equation looks remarkably different, since the same considerations as for the liquid-vapor system cannot be made for ΔV. The volume variations from one phase to the other this time are small.

__Solved exercises__

__Solved exercises__

**– Exercise 1**

The vapor pressure of ice is 4.58 torr at 0 ° C and 1.95 torr at -10 ° C. What is its enthalpy of sublimation in that temperature range?

Note that we have two pressures and two temperatures:

P _{1} = 4.58 torr

P _{2} = 1.95 torr

T _{1} = 0 ° C + 273 = 273 K

T _{2} = -10 ° C + 273 = 263 K

We convert the temperature units from ° C to K, since the gas constant R has K in its units:

R = 8.314 J / K

Thus, we use the integrated Clausius-Clapeyron equation and solve for ΔH _{sub} , which would look like:

ΔH _{sub} = -RLn (P _{2} / P _{1} ) / (1 / T _{2} – 1 / T _{1} )

For greater comfort, we will proceed to replace only with the numbers, but knowing that the final unit will be the Joule:

ΔH _{sub} = – (8.314) Ln (1.95 / 4.58) / (1/263 – 1/273)

= 50.97 J

Or 51.07 J considering few decimals. This value will fluctuate depending on the T _{2} -T _{1} intervals and the determined vapor pressures.

**– Exercise 2**

The boiling point of ethanol at a pressure of 760 torr (1 atm) is 78.3 ° C, and its enthalpy of vaporization is 39.3 kJ. What will its vapor pressure be at a temperature of 46 ° C?

We identify the data:

P _{1} = 760 torr

P _{2} =?

T _{1} = 78.3 ° C + 273 = 351.3 K

T _{2} = 46 ° C + 273 = 319 K

ΔH _{vap} = 39.3 kJ or 39300 J

Thus, we must solve for P _{2} from the integrated Clausius-Clapeyron equation. Again, the units will be omitted for convenience and the calculations will be developed step by step:

Ln (P _{2} / P _{1} ) = – (ΔH _{vap} / R) (1 / T _{2} – 1 / T _{1} )

Ln (P _{2} /760) = – (39,300 / 8314) (1/319 – 1 / 351.3)

Ln (P _{2} /760) = -1.36

Applying the exponential function to both sides of the equation to be able to solve for P _{2} we will have:

and (ln P _{2} /760) = e ^{(-1.36)}

P _{2} /760 = 0.256

P _{2} = 0.256 (760)

= 195 torr

The lower the temperature (46 ° C), the lower the vapor pressure (195 torr). In fact, since ethanol has a pressure of 760 torr at 78.3 ° C, we are talking about its normal boiling point . This is the behavior that is expected for all liquids

In general, Clausius-Clapeyron exercises of this type consist of clearing P _{2} , T _{2,} or ΔH from vaporization or sublimation. The calculations change notably when ΔV must also be considered, especially when it comes to systems or solid-liquid equilibria.