<p>Let's do this with accurate equations. We know how to calculate the saturated water vapour pressure for temperature <span class="math-container">$T$</span>:</p>
<p><span class="math-container">$$e_s(T)=e_{s0}e^{\frac{h_i}{R_w}(\frac{1}{T_{s0}}-\frac{1}{T})}=610\text{ Pa} \times e^{5423(\frac{1}{273\text{ K}}-\frac{1}{T})}$$</span></p>
<p><span class="math-container">$e_s$</span> is saturated water vapour pressure as function of <span class="math-container">$T$</span> in Pascals<br />
<span class="math-container">$T$</span> is the temperature in Kelvins</p>
<p>So, the saturated water vapour pressure at <span class="math-container">$T=0°C=273K$</span> is equal to <span class="math-container">$e_s(273K)=610 Pa$</span>. Similarly, <span class="math-container">$e_s(293K)=2360Pa$</span> and <span class="math-container">$e_s(283K)=1230Pa$</span>.</p>
<p>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?
<span class="math-container">$$e=f\times e_s(T)$$</span>
<span class="math-container">$f$</span> is relative humidity, <span class="math-container">$e$</span> is the water vapour pressure</p>
<p><span class="math-container">$$m(T)=\frac{eV}{R_w T}=\frac{f\times e_s(T)\times V}{R_w\times T}=\frac{fe_s(T)V}{461\frac{J}{kg}T}$$</span></p>
<p><span class="math-container">$$m(293K)=0.0070kg$$</span>
<span class="math-container">$$m(273K)=0.0019kg$$</span></p>
<p>The next step is to combine these two masses:</p>
<p><span class="math-container">$$M=m(293K)+m(273K)=0.0089kg$$</span></p>
<p>So the density is <span class="math-container">$\rho=\frac{M}{V}=0.0045\frac{kg}{m^3}$</span></p>
<p>The water vapour pressure is given by:
<span class="math-container">$$e=\rho RT=587Pa$$</span>
The relative humidity is then:
<span class="math-container">$$f=\frac{e}{e_s(283K)}=0.477$$</span></p>
<p>So this is around 48% of the relative humidity.</p>
<p>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?</p>
<hr />
<p>Also, for Skew-T Log-P lovers:
<a href="https://i.sstatic.net/xqtpj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xqtpj.png" alt="Skew-T Log-P solution" /></a>
This solution is even more concise and less error prone.</p>

CC BY-SA 4.0
2021-02-27T17:55:38.587Z
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.
Joscha Fregin
CC BY-SA 4.0
2021-02-27T19:04:30.767Z
@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.
User123
CC BY-SA 4.0
2021-03-24T22:28:25.397Z
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.
user250614
CC BY-SA 4.0
2021-03-25T18:56:56.183Z
Which hygrometer do you use? Maybe it measures the humidity inaccurately because of the wind coming from the outside.
User123
CC BY-SA 4.0
2021-04-01T12:01:09.260Z
I purchased this one: https://www.amazon.de/gp/product/B0814NRMQ2/ref=ppx_yo_dt_b_asin_title_o03_s01?ie=UTF8&psc=1

Do you get different readings on yours?
user250614
CC BY-SA 4.0
2021-04-01T18:32:36.717Z
@user250614 I don't have any : ) Otherwise, it seems to me pretty cheap and I heard somewhere that the humidity is hard to read on 5 percents even with expensive hygrometers. I think that your hygrometer is based on price prone to other factors, like wind speed and the direction.
User123
CC BY-SA 4.0
2021-02-27T11:33:04.020Z
2021-04-01T18:56:03.400Z
2021-04-01T18:56:03.400Z
User123
<p>I understand that 40% rel. hum. at 0 °C is less absolute humidity than 40% at 20 °C. However, assume that we have 2<span class="math-container">$~m^3$</span> of air. Both are 40% humid at 20 °C. Now we exchange 1<span class="math-container">$~m^3$</span> with air that is 40% humid at 0°C. Then according to <a href="https://wiki.unece.org/download/attachments/23101992/image2014-5-19%2010%3A38%3A41.png?version=1&amp;modificationDate=1400488842518&amp;api=v2" rel="nofollow noreferrer">this picture</a> we should get 7<span class="math-container">$~g\, m^{-3}$</span> and 2<span class="math-container">$~g\, m^{-3}$</span> of absolute humidity for each part. So that's 4.5<span class="math-container">$~g\, m^{-3}$</span> on average. Now the temperature should be 10 °C (right?). Then, 4.5<span class="math-container">$~g\, m^{-3}$</span> 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?</p>

<temperature><humidity>
Why does relative humidity drop if I open the window in winter? What about the temperature drop?
user250614
# Why does relative humidity drop if I open the window in winter? What about the temperature drop?

(Asked by: username_0 on 2021-02-26)

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](https://wiki.unece.org/download/attachments/23101992/image2014-5-19%2010%3A38%3A41.png?version=1&modificationDate=1400488842518&api=v2) 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?

**Tags:** temperature, humidity


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## 1 answers

### Answer by username_1 on 2021-02-27. Score: 4

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:

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

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](https://i.sstatic.net/xqtpj.png)](https://i.sstatic.net/xqtpj.png)
This solution is even more concise and less error prone.

> **Comments:**
> **username_2**: 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.  
> **username_1**: @J.username_2 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.  
> **username_0**: 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.  
> **username_1**: Which hygrometer do you use? Maybe it measures the humidity inaccurately because of the wind coming from the outside.  
> **username_0**: I purchased this one: https://www.amazon.de/gp/product/B0814NRMQ2/ref=ppx_yo_dt_b_asin_title_o03_s01?ie=UTF8&psc=1  Do you get different readings on yours?  
> **username_1**: @username_0 I don't have any : ) Otherwise, it seems to me pretty cheap and I heard somewhere that the humidity is hard to read on 5 percents even with expensive hygrometers. I think that your hygrometer is based on price prone to other factors, like wind speed and the direction.  

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