Vapour pressure equation

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 5.35 – 5.36.

The Clausius-Clapeyron relation is a differential equation relating pressure and temperature on a phase boundary

\displaystyle \frac{dP}{dT}=\frac{L}{T\Delta V} \ \ \ \ \ (1)

 

where {L} is the latent heat for the phase transition and {\Delta V} is the volume change. In order to solve this as a differential equation, we need to know {L} and {\Delta V} as functions of {P} and {T}, which ordinarily we don’t (or if we do, the relation is too complicated to solve analytically). However, if one of the phases is an ideal gas and we can take {L} constant over the region of interest, then usually the volume of the condensed phase (liquid or solid) is negligible compared to the gas phase. In this case, {\Delta V} becomes essentially just {V}, the volume of the ideal gas, at temperature {T} and pressure {P}, that is, {\Delta V=V=nRT/P} and if we consider 1 mole so that {n=1}, we have

\displaystyle \frac{dP}{dT}=\frac{L}{TV}=\frac{LP}{RT^{2}} \ \ \ \ \ (2)

 

We can then do the integration:

\displaystyle \frac{dP}{P} \displaystyle = \displaystyle \frac{L}{RT^{2}}dT\ \ \ \ \ (3)
\displaystyle \ln P \displaystyle = \displaystyle -\frac{L}{RT}+\ln K\ \ \ \ \ (4)
\displaystyle P \displaystyle = \displaystyle Ke^{-L/RT} \ \ \ \ \ (5)

where {K} is a constant. This is known as the vapour pressure equation, and gives a measure of how the pressure at which the phase transition occurs varies as the temperature is raised.

As an example, we can consider the steam-water phase transition, for which data are given in Schroeder’s Figure 5.11. Looking at the temperature range between {50^{\circ}\mbox{ C}} and {100^{\circ}\mbox{ C}}, we see that the latent heat isn’t quite constant over this range, so we can use the average

\displaystyle L \displaystyle = \displaystyle \frac{1}{2}\left(42.92+40.66\right)\times10^{3}\mbox{J mol}^{-1}\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle 41.79\times10^{3}\mbox{J mol}^{-1} \ \ \ \ \ (7)

The gas constant is

\displaystyle R=8.314\mbox{ J mol}^{-1}\mbox{K}^{-1} \ \ \ \ \ (8)

Using the given value of {P=0.1234\times10^{5}\mbox{ Pa}} at {T=50^{\circ}\mbox{ C}=323\mbox{ K}} we find

\displaystyle K=0.1234\times10^{5}e^{41.79\times10^{3}/8.314\times323}=7.07\times10^{10}\mbox{ Pa} \ \ \ \ \ (9)

With this value of {K}, we can use 5 to predict the pressure at {T=100^{\circ}\mbox{ C}=373\mbox{ K}}:

\displaystyle P\left(373\right) \displaystyle = \displaystyle 7.07\times10^{10}e^{-41.79\times10^{3}/8.314\times373}\ \ \ \ \ (10)
\displaystyle \displaystyle = \displaystyle 9.93\times10^{4}\mbox{ Pa} \ \ \ \ \ (11)

The value given in Schroeder’s Figure 5.11 is {P=1.013\mbox{ bar}=10.13\times10^{4}\mbox{ Pa}} (that is, 1 atmosphere, since water boils at {100^{\circ}\mbox{ C}} at sea level) so we’re not too far off.

A plot of {P} versus {V} using 5 looks like this:

It has the same general shape as the steam-water phase boundary shown in Schroeder’s Figure 5.11.

We can invert 5 to find the boiling temperature for various altitudes. We get

\displaystyle T=\frac{L}{R\ln\frac{K}{P}} \ \ \ \ \ (12)

Or, in centrigrade:

\displaystyle T_{c}=\frac{L}{R\ln\frac{K}{P}}-273 \ \ \ \ \ (13)

Using the barometric equation, we found the atmospheric pressures at various altitudes. We can use this equation to add a column for the boiling point.

Altitude {z} (m) {P\left(z\right)} (atm) {T\mbox{ (}^{\circ}\mbox{C)}}
1430 0.844 {95.9}
3090 0.693 {90.6}
4420 0.592 {86.5}
8850 0.350 {73.5}

The last line is the height of Mt Everest.

We can get an estimate of the dependence of {T} on {z}, the height above sea level, as follows. We need to be careful to distinguish the boiling temperature {T_{w}} of the water from the surrounding air temperature {T_{a}}. From 2, the change {\Delta T_{w}} as a result of a change {\Delta P} in atmospheric pressure is

\displaystyle \Delta T_{w}=\frac{RT_{w}^{2}}{LP}\Delta P \ \ \ \ \ (14)

 

From the barometric equation, the change in pressure due to an increase in height {\Delta z} is

\displaystyle \Delta P=-\frac{MgP}{RT_{a}}\Delta z \ \ \ \ \ (15)

 

where {M} is the mass of a mole of air molecules. Inserting 15 into 14, we get

\displaystyle \Delta T_{w}=-\frac{Mg}{L}\frac{T_{w}^{2}}{T_{a}}\Delta z \ \ \ \ \ (16)

In the limit, we get the differential equation

\displaystyle \frac{dT_{w}}{dz}=-\frac{Mg}{LT_{a}}T_{w}^{2} \ \ \ \ \ (17)

which we can integrate to get

\displaystyle \frac{1}{T_{w}} \displaystyle = \displaystyle \frac{Mg}{LT_{a}}z+A\ \ \ \ \ (18)
\displaystyle T_{w} \displaystyle = \displaystyle \frac{1}{A+\frac{Mg}{LT_{a}}z} \ \ \ \ \ (19)

for a constant of integration {A}. At {z=0}, {T_{w}=373} so {A=1/373=2.68\times10^{-3}\mbox{ K}^{-1}}. For a molar mass of {M=0.029\mbox{ kg}} and an air temperature of {T_{a}=10^{\circ}\mbox{C}=283\mbox{ K}}, we have

\displaystyle \frac{Mg}{LT_{a}} \displaystyle = \displaystyle \frac{\left(0.029\right)\left(9.8\right)}{\left(41.79\times10^{3}\right)\left(283\right)}=2.4\times10^{-8}\mbox{m}^{-1}\mbox{K}^{-1}\ \ \ \ \ (20)
\displaystyle \frac{Mg}{LT_{a}A} \displaystyle = \displaystyle \frac{2.4\times10^{-8}}{2.68\times10^{-3}}=8.97\times10^{-6}\mbox{m}^{-1} \ \ \ \ \ (21)

Thus we can write 19 as

\displaystyle T_{w} \displaystyle = \displaystyle \frac{1}{A}\left(\frac{1}{1+\frac{Mg}{LT_{a}A}z}\right)\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle \frac{373}{1+8.97\times10^{-6}z} \ \ \ \ \ (23)

where {z} is in metres. Even if {z} is several thousand metres, the {z} term in the denominator is quite small, so we can approximate it by

\displaystyle T_{w}\approx373\left(1-8.97\times10^{-6}z\right) \ \ \ \ \ (24)

which shows that {T_{w}} is roughly linear in {z}. The slope is

\displaystyle \frac{dT_{w}}{dz}\approx-3.34\times10^{-3}\mbox{K m}^{-1} \ \ \ \ \ (25)

which is a decrease in the boiling point of about 3.34 K per kilometre.

4 thoughts on “Vapour pressure equation

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