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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?

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  • $\begingroup$ What is the model of your instrument? Is it an electronic/scientific instrument? $\endgroup$ – Etienne Godin 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 ? $\endgroup$ – Etienne Godin May 30 '19 at 22:38
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    $\begingroup$ @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. $\endgroup$ – Barry Carter Jun 5 '19 at 2:55
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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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