# Calculating the amount of water condensed when cooling below the dew point

I've been given the following question:

Calculate the dew point temperature of a 60% humid air at $$30^oC$$. How much water mass wil condensate out of 1 $$m^3$$ of such air when the temperature is lowered to $$5^oC$$?

I tried to find the dew point temperature corresponding to those conditions, which is $$T_d=21.55^oC$$ and then I took the density of the partial water at that temperature and at $$5^oC$$ from thermodynamic stables. That would be $$\rho=0.01943[ \frac{kg}{m^3}] ~@~ 22^oC , \rho=0.006797[ \frac{kg}{m^3}] ~@~ 5^oC$$

And I calculated the mass condensated by $$m=(\rho_{22} - \rho_5)*[m^3]=0.01264 [kg]$$

But according to our course professor, he says that all of the water vapor will condense. So $$m=\rho_{22}*[m^3]=0.01943[kg]$$

Why is that? Does always all the water condense when cooling in constant pressure below the dew point temperature? Will it ever depend on the specific temperature I'm cooling to (in our case, $$5^oC$$)? Are there scenarios where my method would be correct?

Edit: It wasn't given in the question, but i think he assumed that we are cooling the air in constant pressure. First I calculated the mass of water vapor at $$30^oC$$ and I used the ideal gas equation with the pressure of $$P(30,\nu)=RH*P_{sat}(30^oC)$$, where RH stands for relative humidity.

If I look at the following $$T-\nu$$ diagram,

the pressure line of the dew point is $$P_{sat}(22^oC)$$, and if the temperature is further lowered to $$5^oC$$ I get $$P(5^oC,v)=Psat(22^oC)=> 5^oC$$ which is far below the saturation line. I know that cooling below the dew point will condensate the excess water vapour. In that case, will all of the mass that I calculated in the start of the process condence (0.0126 kg)?

• Your professor has decided to neglect the vapor pressure of water at $5°$. It is valid as a first approximation. Your result is better, if your numerical values are correct, which I have not checked. . Jun 21, 2022 at 12:57
• As a rule of thumb,maximum absolute humidity in $\pu{g m-3}$ goes approximately to 1/2 by cooling by 10 °C. So saying cooling from 22 °C to 5 °C condenses all water is very bad first approximation. Jun 21, 2022 at 13:21
• @Maurice (and also Poutnik), I've added further clarification. Is my way still the correct one? Jun 21, 2022 at 14:29
• For precise calculation, one has to involve thermal volume contraction and even condensation based contraction, as absolute humidity relates to the volume (with or without air). Jun 21, 2022 at 14:45

The maximum vapor pressure of water is $$\pu{4243 Pa}$$ at $$\pu{30°C}$$, and $$\pu{872 Pa}$$ at $$\pu{5°C}$$. If the original humidity is $$\pu{60}$$%, it means that the initial vapor pressure is $$\pu{0.600 · 4243 Pa = 2546 Pa}$$.
The initial amount of water in $$\pu{1 m3}$$ humid ($$\pu{60}$$%) air at $$\ce{30°C}$$ is : $$\ce{n_{H2O} = pV/RT = \frac{2546 Pa·1 m^3}{8.314 J K^{-1}mol^{-1}·303 K} = 1.0106 mol. This amount of water weighs 18.19 g}$$.
If cooled down to $$\pu{5°C}$$, $$\pu{1 m^3}$$ air has a volume equal to $$\pu{\frac{278}{303} 1 m^3 = 0.917 m^3}$$. At $$\pu{5°C}$$, the amount of water remaining in $$\pu{0.917 m3}$$ air (saturated in water) is :
$$\ce{n'_{H2O} = pV/RT = \frac{872 Pa · 0.917 m^3}{8.314 J K^{-1} mol^{-1} 278 K} = 0.345 mol}$$. This amount of water weighs $$\pu{6.22 g}$$
The water condensed during the cooling operation weighs : $$\pu{18.19 g - 6.22 g = 11.97 g}$$.
So the proportion of water condensed during the cooling to $$\pu{5°C}$$ of $$\pu{1 m^3}$$ air taken at $$\pu{30°C}$$ with $$60$$% humidity is $$\pu{\frac{11.97 g}{18.19 g} = 65.80}$$%. It is much smaller than $$\pu{100}$$%. This result is far from your teacher's statement.