I understand that 40% rel. hum. at 0 °C is less absolute humidity than 40% at 20 °C. However, assume that we have 2$~m^3$ of air. Both are 40% humid at 20 °C. Now we exchange 1$~m^3$ with air that is 40% humid at 0°C. Then according to this picture we should get 7$~g\, m^{-3}$ and 2$~g\, m^{-3}$ of absolute humidity for each part. So that's 4.5$~g\, m^{-3}$ on average. Now the temperature should be 10 °C (right?). Then, 4.5$~g\, m^{-3}$ means a relative humidity of around 50%! So why does relative humidity drop in reality? Is it because the temperature would actually be higher somehow?


Let's do this with accurate equations. We know how to calculate the saturated water vapour pressure for temperature $T$:

$$e_s(T)=e_{s0}e^{\frac{h_i}{R_w}(\frac{1}{T_{s0}}-\frac{1}{T})}=610\text{ Pa} \times e^{5423(\frac{1}{273\text{ K}}-\frac{1}{T})}$$

$e_s$ is saturated water vapour pressure as function of $T$ in Pascals
$T$ is the temperature in Kelvins

So, the saturated water vapour pressure at $T=0°C=273K$ is equal to $e_s(273K)=610 Pa$. Similarly, $e_s(293K)=2360Pa$ and $e_s(283K)=1230Pa$.

So we have two parcels of air: 1. 20°C, 40% and 2. 0°C, 40%. What are the masses of the water vapour in both parcels? $$e=f\times e_s(T)$$ $f$ is relative humidity, $e$ is the water vapour pressure

$$m(T)=\frac{eV}{R_w T}=\frac{f\times e_s(T)\times V}{R_w\times T}=\frac{fe_s(T)V}{461\frac{J}{kg}T}$$

$$m(293K)=0.0070kg$$ $$m(273K)=0.0019kg$$

The next step is to combine these two masses:


So the density is $\rho=\frac{M}{V}=0.0045\frac{kg}{m^3}$

The water vapour pressure is given by: $$e=\rho RT=587Pa$$ The relative humidity is then: $$f=\frac{e}{e_s(283K)}=0.477$$

So this is around 48% of the relative humidity.

Why did I put here so many equations? Simple! These equations hold on high degree of accuracy, so they can't be wrong. Why would the relative humidity drop in reality?

Also, for Skew-T Log-P lovers: Skew-T Log-P solution This solution is even more concise and less error prone.

  • $\begingroup$ You can't just combine the water masses due to the volume change. You start with 2 $m^3$ but you will not end up with 2 $m^3$ after mixing the air masses. $\endgroup$ – J. Fregin Feb 27 at 17:55
  • $\begingroup$ @J.Fregin I feel like this is negligible because of small temperature change and it doesn't that affect the solution. If we would be picky, we could also use virtual temperature instead of temperature. It affects the answer pretty much the same as the volume adding inconsistency. $\endgroup$ – User123 Feb 27 at 19:04
  • $\begingroup$ Thank you for your answer. You seem to basically support what I have written in the OP. My problem is that my actual measurements with my hygrometer indicate a drop in relative humidity when I open the window. That's what I don't get. $\endgroup$ – user250614 Mar 24 at 22:28
  • $\begingroup$ Which hygrometer do you use? Maybe it measures the humidity inaccurately because of the wind coming from the outside. $\endgroup$ – User123 Mar 25 at 18:56
  • $\begingroup$ I purchased this one: amazon.de/gp/product/B0814NRMQ2/… Do you get different readings on yours? $\endgroup$ – user250614 Apr 1 at 12:01

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