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?

  • $\begingroup$ What is the model of your instrument? Is it an electronic/scientific instrument? $\endgroup$ – marsisalie May 30 '19 at 19:13
  • $\begingroup$ @EtienneGodin believe it is this one or a similar model, I can double check tomorrow: omega.com/en-us/sensors-and-sensing-equipment/humidity/… $\endgroup$ – Curtis May 30 '19 at 19:31
  • $\begingroup$ ok for the model of your sensor - also can you explain how the pressure can be 140 kPa in your setting ? $\endgroup$ – marsisalie May 30 '19 at 22:38
  • 1
    $\begingroup$ @EtienneGodin The sensor is basically in a container that's under pressure with humid air flowing through it. $\endgroup$ – Curtis May 30 '19 at 23:19
  • $\begingroup$ I live a mile above sea level (pressure about 20% lower than at sea level) and it never occurred to me to adjust the humidity reading. I'd start with the Ideal Gas Law and the Clausius-Clapeyron equation (en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation). My wild guess: since relative humidity is a ratio, and both water vapor and nitrogen/oxygen change in the same way, pressure doesn't affect relative humidity. $\endgroup$ – user967 Jun 5 '19 at 2:55

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.