<p>The maximum vapor pressure in air is a function of the temperature and not of the total pressure. So when you humidify air at 12°C you get a maximum vapor pressure of approximately 1.4 kPa. However, when the air is humidified at 140 kPa the vapor pressure is still only 1.4 kPa. Therefore, you have a water/air ratio of 1.4/140= 1%. </p>

<p>Therefore, for every mol of water, you will have 100 mol of air. So when the total pressure drops to 100 kPa, the mol/volume fractions will not change. This is also valid for the vapor pressure ratio. Therefore, in this case the vapor pressure will still be 1% of the total pressure. Thus, the vapor pressure will only be 1.0 kPa at a total pressure of 100 kPa.   </p>

<p>Therefore, to "correct" an "under-pressure" reading you will need to multiply your "under pressure" reading by (101.3 kPa)/(P_actual), where P_actual is your actual total pressure in kPa.</p>

CC BY-SA 4.0
2019-06-21T17:27:55.833Z
2019-06-21T17:27:55.833Z
Curtis
<p>Obviously pressurized air can hold more water than regular air, so an increase in overall pressure would result in an inversely proportional decrease in relative humidity. However, I can't seem to find any equations which reflect this.</p>

<p>Basically my issue is that my thermohygrometer calculates the relative humidity assuming that atmospheric pressure is 101.3kPa but where I am measuring it the pressure is about 140kPa. This can be fixed by multiplying the reading by (101.3kPa/140kPa), but I am having a hard time expressing that scientifically and mathematically.</p>

<p>So, what is the mathematical relationship between pressure and relative humidity?</p>

CC BY-SA 4.0
2019-05-30T19:13:58.780Z
What is the model of your instrument? Is it an electronic/scientific instrument?
marsisalie
CC BY-SA 4.0
2019-05-30T19:31:15.840Z
@EtienneGodin  believe it is this one or a similar model, I can double check tomorrow:

https://www.omega.com/en-us/sensors-and-sensing-equipment/humidity/humidity-meters/p/HX15
Curtis
CC BY-SA 4.0
2019-05-30T22:38:25.310Z
ok for the model of your sensor - also can you explain how the pressure can be 140 kPa in your setting ?
marsisalie
CC BY-SA 4.0
2019-05-30T23:19:23.050Z
@EtienneGodin The sensor is basically in a container that's under pressure with humid air flowing through it.
Curtis
CC BY-SA 4.0
2019-06-07T19:54:28.433Z
I think it may be calculating the saturation pressure of water vapour from a temperature-based correlation like the one on page 653 of this appendix: https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470496251.app2
and just gets the partial pressure from the voltage change, in which case multiplying by my ratio (P/P') would remedy this.
Curtis
<atmosphere><temperature><humidity>
How does total pressure affect relative humidity and how do I correct my hygrometer for pressure?
Curtis
# How does total pressure affect relative humidity and how do I correct my hygrometer for pressure?

(Asked by: username_0 on 2019-05-30)

Obviously pressurized air can hold more water than regular air, so an increase in overall pressure would result in an inversely proportional decrease in relative humidity. However, I can't seem to find any equations which reflect this.

Basically my issue is that my thermohygrometer calculates the relative humidity assuming that atmospheric pressure is 101.3kPa but where I am measuring it the pressure is about 140kPa. This can be fixed by multiplying the reading by (101.3kPa/140kPa), but I am having a hard time expressing that scientifically and mathematically.

So, what is the mathematical relationship between pressure and relative humidity?

**Tags:** atmosphere, temperature, humidity

> **Comments:**
> **username_1**: What is the model of your instrument? Is it an electronic/scientific instrument?  
> **username_0**: @EtienneGodin  believe it is this one or a similar model, I can double check tomorrow:  https://www.omega.com/en-us/sensors-and-sensing-equipment/humidity/humidity-meters/p/HX15  
> **username_1**: ok for the model of your sensor - also can you explain how the pressure can be 140 kPa in your setting ?  
> **username_0**: @EtienneGodin The sensor is basically in a container that's under pressure with humid air flowing through it.  
> **username_0**: I think it may be calculating the saturation pressure of water vapour from a temperature-based correlation like the one on page 653 of this appendix: https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470496251.app2 and just gets the partial pressure from the voltage change, in which case multiplying by my ratio (P/P') would remedy this.  

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## 1 answers

### Answer by username_0 on 2019-06-21. Score: 2 | Accepted answer

The maximum vapor pressure in air is a function of the temperature and not of the total pressure. So when you humidify air at 12°C you get a maximum vapor pressure of approximately 1.4 kPa. However, when the air is humidified at 140 kPa the vapor pressure is still only 1.4 kPa. Therefore, you have a water/air ratio of 1.4/140= 1%.

Therefore, for every mol of water, you will have 100 mol of air. So when the total pressure drops to 100 kPa, the mol/volume fractions will not change. This is also valid for the vapor pressure ratio. Therefore, in this case the vapor pressure will still be 1% of the total pressure. Thus, the vapor pressure will only be 1.0 kPa at a total pressure of 100 kPa.

Therefore, to "correct" an "under-pressure" reading you will need to multiply your "under pressure" reading by (101.3 kPa)/(P\_actual), where P\_actual is your actual total pressure in kPa.


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