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Relative Humidity Problem

Physics Asked on April 12, 2021

I have been assigned to solve the problem: the ongoing temperature of the room $T_1 = 7$; RH is 85% and I am asked to figure out how much the temperature must be increased in order that RH would drop to 70% or less.

I would like to know what I did wrong – why I could not find this second temperature. I can infinitely use the definition of RH and Antoine equation, but never find $T_2$ in the way. Why? Because I do not see any connection between parameters $p_1, p_{sat_1}, v_1, v_{sat_1}, p_2, p_{sat_2}, v_2, v_{sat_2}$. Knowing $T_1$ I can find $p_1$ and then $v_1$; and also the temperature at which the air shall be saturated with water vapour: $T_{d_1}$. Knowing them does not solve the risen problem. That means that I do not understand something about how these parameters are connected between themselves.

The only possibility for me is to look into what happens in the room upon the rising the temperature up to $T_2$. Will the RH drop? Not necessarily, because the more environment’s temperature, the more moisture the air can store, so boosting RH. But okay we did somehow rise $T$ up to $T_2$ and $varphi_2 = 0.7$. Definitely, $T_{d_2} ne T_{d_1}$; maybe $p_{sat_2} = p_{sat_1}$? I think no as well, because both parameters depend on the specific temperature.

What do I misunderstand? Maybe a quantity gets preserved? And by it I should solve the problem. I am confused. Volumes are not given. Only the concurrent temperature of the room and RH; then they boost temperature and tell that RH is 0.7.

NOTATION

$T$ is temperature,

$varphi$ is RH

$v$ is the water vapor contained in the air

$p$ is the partial water vapour pressure

$p_{sat}$ when the air gets saturated with the water vapour

$v_{sat}$ is the amount of the water vapour contained in the saturated air.

2 Answers

Comment turned answer as requested by Bill N.

This question seems to assume that the amount of water vapor in the air stays constant. Then this statement

[...] Will the RH drop? Not necessarily, because the more environment's temperature, the more moisture the air can store, so boosting RH.

makes no sense. If the air can store more moisture, but the amount of water in the air stays constant, RH must drop.


Answering the follow-up question in the comments:

You can see this as taking some volume of air from the outside and heating it up. You know the new temperature and RH, but the absolute amount of water will stay constant. Therefore, if the air outside was saturated with water vapor, the temperature there must have been the dew point of the air inside (which is also the dew point of the air outside, since this only depends on the absolute amount of water in the air, which stays constant).

Correct answer by noah on April 12, 2021

The relative humidity of air is a ratio between the maximal amount of water vapour the air can hold (before fog forms or, in technical terms, the dew point) and the amount it actually holds.

The maximum carrying capacity of water vapour of air depends on the temperature. You can just look up this value (your psat). This means you can calculate the amount of vapour in the air from the RH and the psat at that temperature. The amount of water vapour is the conserved quantity. The problem boils down to working out which temperature is needed to give an RH of 70% with that amount of vapour.

The carrying capacity of air does change with temperature but, unless you add extra water vapour from some other source, the RH will drop as temperature increases. In your logic you have confused the psat with the RH.

Answered by matt_black on April 12, 2021

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