# Water equilibrium when relative humidity is high

Suppose a puddle has water and another component whose composition is known.

The water in it would be in equilibrium with the water vapour following Raoult and Dalton's law (also here):

$$P_T y=P^0_{H_2O} x$$

Where:

$$P_T$$: Total ambient pressure

$$P^0_{H_2O}$$: Vapor pressure of pure water

$$x$$: Molar composition of water in liquid phase

$$y$$: Molar composition of water on gaseous phase

Considering a day of $$35°C$$ and $$1 \ atm$$, according to Antoine's equation $$P^0_{H_2O}= 42.08 \ mm\ Hg$$.

Let's say that $$x_{H_2O}=0.097$$, then

$$y=\frac{P^0_{H_2O} x}{P_T}=\frac{42.08 \ mm \ Hg \ 0.097}{760 \ mm \ Hg}=0.0054$$

It happens that the day's relative humidity is $$80 \%$$, so

$$\frac{\bar{P_{H_2O}}}{P^0_{H_2O}}=0.8$$

$$\bar{P_{H_2O}}=0.8 \times 42.08 \ mm \ Hg=33.664 \ mm \ Hg$$

$$\bar{P_{H_2O}}=P_T y$$

$$y=\frac{\bar{P_{H_2O}}}{P_T}=\frac{33,664 \ mm \ Hg}{760 \ mm \ Hg}=0.044$$

Here's my doubt: How comes the composition of water in air is higher than the composition in equilibrium on the surface of the puddle? Is water condensing over the puddle?

To second this question: What happens to the other component in the puddle? Does the humidity affects it's evaporation rate?

I am assuming the following process: a puddle of water contains $$x_W$$ mol fraction of water and $$1-x_w$$ fraction of a non volatile component. The air is humid with relative humidity RH that corresponds to mol fraction of moisture $$y^*_W$$. If we let the puddle equilibrate what will be the mol fraction of water in the paddle?

To establish equilibrium either some water will evaporate or some water vapor will condense into the puddle. I either case $$y^*_W$$ will not change because the air above the puddle is so much bigger than the puddle.

Suppose the final mol fraction of water in the puddle is $$x^*_W$$. We must have $$y^*_W P = P_W^\text{sat} x^*_W \Rightarrow \boxed{ x^*_W = \frac{y^*_W P}{P_W^\text{\sat}} }$$ This value does not depend on the mol fraction of water initially.

Using your numbers,$$y^*_W=0.044$$,$$P=760$$ mmHg, $$P_W^\text{sat}=42.08$$ mmHg we find $$x^*_W=0.795$$. If this is more than your starting value, some humidity will condense into the puddle; if it is less, some of the water in the puddle will evaporate.

You say that the initial mol fraction of water in the paddle is 0.097. That's very small and doesn't sound like a puddle of water, rather a paddle of something that contains a tiny bit of water. But if that's what you have, then some moisture will condense into the puddle.

Comment 1 if the amount of water in the puddle is so small, Raoult's law may not be accurate. You will likely need activity coefficients unless the system happens to be ideal.

Comment 2 If the rest of the stuff in the paddle is volatile, then all of the water will evaporate (why?)