# Q & A: boiling point vs. pressure

Q:
There is a lower boiling of water in vacuum. If moisture sensitive devices would be baked in the vacuum chamber instead of 100 - 125 C temperature we can use 75 C (0.1 bar vacuum). Is there anywhere a precision table or graph with water boiling temeperature (evaporation) in ICs according 0.1 - 1 bar vacuum.
- Zygimantas (age 25)
Kaunas, Lithuania
A:

I haven't found a table, although I'm sure that there must be one available somewhere. There is a basic exact equation (Clausius-Clapeyron) relating the boiling point to the pressure. In simple cases where the latent heat of vaporization doesn't depend much on the pressure and where the vapor is an ideal gas with volume much bigger than that of the liquid, it takes on the following form

T(p)= T0/(1+T0*ln(p0/p)/(L/k))
where T0 is the boiling point in absolute temperature (e.g. Kelvin) at some standard pressure p0. p is the pressure of interest, k is Boltzmann's constant, and L is the latent heat of vaporization, per molecule, and ln stands for natural log.

So here's the numbers for water, at least approximately:
T0 =373 K (same as 100 °C)

L/k = 4800 K

At 0.1 bar, you should be able to reduce the temperature some 50 or 60 °C below what you need at an atmosphere, more or less consistent with your experience. The formula should let you interpolate.

Mike W.

(published on 10/22/2007)

## Follow-Up #1: units

Q:
"T(p)= T_0/(1+T_0*ln(p_0/p)/(L/k)) where T_0 is the boiling point in absolute temperature (e.g. Kelvin) at some standard pressure p_0. p is the pressure of interest, k is Boltzmanns constant, and L is the latent heat of vaporization, per molecule, and ln stands for natural log." k is not Boltzmann's constant but the Gas constant R: 8.3145 J K^-1 mol^-1, (otherwise the dimensions aren't correct!), i.e., L/R should be 40660/8.3145 = 4890 K Regards Steve Battison
- Steve Battison (age old)
Guildford, England, UK
A:
Steve- No, you use the gas constant R if you use the latent heat per mole for L. If by L you mean per molecule, you use Boltzmann's constant.
Think of the units. Your expression would change if we decided to use a different number of particle as the number per mole. Nature can't care what names we use.

Mike W.

(published on 01/23/2010)