Estimate of condensation rate

We want to estimate how quickly a layer of condensation would form on our sample, and how much of the incoming light will be absorbed by this layer. The estimate below is very crude, but serves as a starting point.

Known quantities

Pressure p = 1e-3 Pa (or 1e-5mbar)

Temperature T = 293 K

Volume V = 0.014 m³ (estimated volume of vacuum chamber only)

Gas constant R = 8.314 Jmol^-1K^-1

Avogadro's constant N_A = 6.022e23 mol^-1

Boltzmann constant k_B = 1.380649e-23 JK^-1

With the ideal gas law, we calculate the amount of air in the vacuum chamber to be 5.75e-9 mol. At sea level we will assume the atmosphere contains 2% water vapour. So, we get n = 0.02*5.75e-9 = 1.15e-10 mol water in our system. The molar mass M of H₂O is 18*1.67e-27 = 3.0e-26 kg mol^-1.

For the velocity v of the water molecules, we use v = sqrt(3*k_B*T/M) and calculate the collision rate per second per square meter as r = 0.5*n*N_A*v/V = 1.57e18 s^-1m^-2.

Now let's calculate how many molecules there will be in a monolayer of water covering our 3cm x 3cm sample. A cm³ of water has N_A/18 water molecules, taking the cube root of that gives us 32212404 layers of molecules, with each 32212404*32212404 = 9.3e15 molecules.

Using the collision rate and assuming that all molecules that hit the sample will neatly arrange themselves into a single layer as long as there is space, we calculate that it takes 9.3e15/(r*0.03*0.03) = 6.6 s to form a single layer of water on our sample.

Now, a single layer has a thickness of 1/32212404 cm with the same argument as before. So, after 60 seconds we will have a 2.82e-7 cm thick layer of water on our sample. Now, with the Lambert-Beer law, we calculate that at 125 nm, where the absorption coefficient is 1.978e5 cm^-1, we have I=100*exp(-1.978e5*2.82e-7) = 94% of our light remains and so about 6% is absorbed. At 175 nm the absorption coefficient is 1.705e1 and therefore we keep 99.999% of our light remains.