Calculating indoor humidity -- do I need to account for pressure?

Question

I'm developing a calculation that converts an outdoor temperature, pressure and relative humidity into an indoor relative humidity (for a given indoor temperature).

E.g. if it's 39°F outdoors at 87% relative humidity, then indoors (with the same air, the same absolute humidity) at 72°F it might be 35% relative humidity.

So far this has been relatively straightforward -- I calculate the vapor pressure of water for the original temperature (39°F), use that together with the original relative humidity to calculate the absolute humidity (mass of water divided by volume, i.e. volumetric humidity), figure out the vapor pressure of water for the new temperature (72°F), and then use the absolute humidity relative to that to calculate the new relative humidity. (Which gives the result here of 35%.)

However, I'm worried that I'm not accounting for the effects of pressure. My calculation would seem to assume that the mass of water vapor remains fixed indoors. But the increase in temperature means that, if the indoors were sealed shut, the pressure would increase because of the rise in temperature. Obviously indoors aren't sealed shut, so as temperature rises due to indoor heating, both air and water vapor flow outwards (to the outside environment) until the indoor pressure is equal to the outdoor.

Does this mean my indoor relative humidity calculation also has to account for a reduction in mass of indoor water vapor, i.e. an additional reduction in absolute humidity due to raising the temperature?

Or am I overthinking this somehow -- is there some reason why the increase in temperature won't result in a loss of air and vapor, or why vapor would migrate back in to remain at equal pressure?

Also is there any other factor I ought to be taking into account that I'm missing? (I'm obviously intentionally ignoring indoor sources of humidity like shower, humidifier, etc.)

(Experimentally here in my home, my indoor humidity readings do seem to be ~5 percentage points lower than what my current calculation is producing, but I'm not sure how accurate the devices are.)

Answer 1

Houses are never perfectly air tight. Pressure inside and outside are the same.

Temperature is different, so the density of air is different. But the temperature is about $300$ K, give or take. A $10$ K difference is about $3$ %. Small, but present. You are right to include it.

There are lots of reasons why you might find a difference from what your formula predicts.

Relative humidity in the bathroom is often higher than one might expect from outside conditions because of all the hot water in the room. In other rooms, people give off moisture. It can have an effect. You have heard of fogging up the windows of a car? That might make your readings higher than you would expect.

Moisture in the air migrates into and out of the walls, carpets, and such. If you take a shower and air out the bathroom, moisture in the towel will keep air more humid than you might expect.

If doors and windows are closed and the air conditioner/heater fan is off, it takes a while for outside air to replace inside air. A dry day yesterday might have caused moisture in the walls to flow out into the air, leaving the walls drier than usual. As moister air from today moves in, moisture might flow back into the walls, leaving the air drier than you might expect.

An air conditioner can cool air a lot, making moisture condense. Then that air mixes with the room, warming back up. But it is drier air than you expect.