Estimating vapor pressure from boiling point

Originally posted 2019-10-30


Did you know that you can estimate the volatility of a substance at room temperature if you know its boiling point? This can be a useful calculation if you’d like to estimate, e.g. ppm of some volatile organic compound in the air.

The equations

The Clausius Clapyeron equation relates several quantities: two temperatures \(T_1\) and \(T_2\), the vapor pressures \(P_1\) and \(P_2\) at those temperatures, the ideal gas constant \(R\), and the enthalpy of vaporization.

\[ \ln{\frac{P_2}{P_1}} = \frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]

We know most of these quantities. \(T_1\), \(P_1\) are (boiling point, 1 bar), by definition of the boiling point. \(T_2\) is room temperature. \(P_2\) is the quantity we want to compute. \(R\), the ideal gas constant, is known.

The only missing quantity is \(\Delta H_{vap}\), the enthalpy of vaporization. This is an empirically measured value which is more difficult to measure, and therefore less commonly measured than boiling point. However, Trouton’s Rule is an observation that empirically, almost all organic molecules have near-identical entropy of vaporization \(\Delta S_{vap}\) = 85 J/K, or about 10.5 times \(R\). (Note that entropy != enthalpy). If we could link entropy back to enthalpy somehow, we’d have all the quantities we’d need to estimate volatility at room temperature.

Luckily, the Gibbs equation says that at the boiling point, we have an equilibrium between vapor and liquid phases given as

\[ \Delta G = 0 = \Delta H - T\Delta S \]

i.e. \(\Delta H_{vap} = T_{BP}\Delta S_{vap} = 10.5T_{BP}R\).

All together, this yields the following equation for estimating vapor pressure at room temperature of a substance:

\[ \ln{\frac{P_2}{\textrm{[1 bar]}}} = \frac{10.5T_{BP}R}{R}\left(\frac{1}{T_{BP}} - \frac{1}{T_2}\right) \]

\[ P_2 = \textrm{[1 bar]} \cdot e^{10.5 \left(1 - \frac{T_{BP}}{T_2} \right)} \]

Sanity check

When \(T_2 = T_{BP}\), the exponential turns into a factor of 1, as desired. At a temperature that’s half of the boiling point, we have vapor pressure is \(e^{-10.5}\) bars. At a temperature that’s twice the boiling point, we have vapor pressure = \(e^{10.5}\) bars. At a temperature that’s 80% the boiling point (i.e. water boils at 373 K, room temperature is 300 K), we have vapor pressure of \(e^{-10.5 \cdot 0.25}\) = 0.07 bar. Water at room temperature has a saturation of about 4%, or 0.04 bar, so this is about right.

A fun calculation

When you take a sniff of an essential oil, how many grams of material did you just inhale?

Let’s assume the essential oil has a boiling point of 200C, or 470K. Room temperature is 300K. Plugging these numbers into our equation yields a vapor pressure of 0.003 bar. A sniff is maybe 100 mL of air. Using the ideal gas law, we end up with \(10^{-5}\) moles of material in that whiff. Assuming that our essential oil has a molecular weight of about 200 daltons, this yields 2 milligrams of material. If the oil is diluted to 10% in some carrier medium, then the vapor pressure drops accordingly and we’d have inhaled 0.2 milligrams of material.