homework and exercises

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Closed 9 months ago.

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)?