Why does relative humidity drop if I open the window in winter? What about the temperature drop?

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

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

$e_s$ 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? $$e=ftimes e_s(T)$$ $f$ is relative humidity, $e$ is the water vapour pressure

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

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

The next step is to combine these two masses:

$$M=m(293K)+m(273K)=0.0089kg$$

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

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?

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