Why water evaporates below 100 ºC

  • Thread starter Fernsanz
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  • #26
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Thank you very very much for your support Marty.

You have summed up the problem perfectly well and I am glad that you have fully understood my question.

Once again, thank you very much.
 
  • #27
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The post from Marty has encouraged me to better restate my question which I have been chewing over.

Lets suppose the partial pressure of water in the air is 0.03 atm and we are at 30ºC. If I asked you how much have I to cool the air to get liquid water, most of you would say 25 ºC. If I asked you why, your answer would be "because if look at the phase diagram of water at 25ºC and a vapor partial pressure of 0.03 atm the water just start to being in equilibrium with liquid water".

Well, but there is a little point here you have passed over: if look at the phase diagram of water at 25ºC and a vapor partial pressure of 0.03 atm the water just start to being in equilibrium with liquid water at those same 0.03 atm, not at 1 atm!!!!!! So, how could be your reasoning right? In fact it is not generally right.

That was what didn't fit in my mind; and I recognize that perhpas I didn't know to state it clearly, but the post from Marty has made clear to me that I could have been understood.

Thanks and thanks Marty.
 
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  • #28
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So, the dew point is, strictly talking, less than 25ºC.

But how much less? Well, we have to make liquid and vapor chemical potentials equal to find the dew point:

[tex]\mu_v(T_{dew}, 0.03 \mbox{ atm})=\mu_l(T_{dew}, 1 \mbox{ atm})[/tex]​

where l stands for liquid and v for vapor. As I have pointed out in a previous reply, at 1 atm and 25ºC the liquid water chemical potential is greater than that of the vapor at 0.03 atm and 25ºC by an amount - molar volume times pressure difference in right units- [tex]v \cdot (\Delta P) = 18 \cdot 10^{-6} \cdot (101125-101125 \cdot 0.03)=1.766 J/mol[/tex], where [tex]18 \cdot 10^{-6} m^3/mol[/tex] is the molar volume of liquid water; i.e.

[tex]\mu_l(25 \mbox{C}, 1 \mbox{ atm}) = \mu_v(25 \mbox{C}, 0.03 \mbox{ atm}) + 1.766[/tex]​

If we want to compensate for this difference we can lower the temperature. The contribution of a difference in temperature to the chemical potential is, if the temperature difference is not to big, [tex]\mu(T_2, P) = \mu(T_1, P) -s \cdot (\Delta T)[/tex], where s is the molar entropy and [tex]\Delta T=T_2-T_1[/tex] is the temperature difference. The molar entropy for water vapor is 188.83 and for liquid water is 69.91. So, the balance in order for liquid water at 1 atm and water vapor at 0.03 atm to have equals chemical potential starting from the temperature [tex]T_1[/tex]=25ºC is

[tex]\mu_v(T_{dew}, 0.03 \mbox{ atm})=\mu_l(T_{dew}, 1 \mbox{ atm}) \hspace{20} \Rightarrow \hspace{20} 1.766 - 69.91·(T_{dew}-T_1) +188.83 \cdot (T_{dew}-T_1) = 0 [/tex]​

Solving for [tex]T_{dew}-T_1[/tex] we have [tex]T_{dew}-T_1=-0.015[/tex] ºC. That is, the dew point is practically 25ºC.

So, numerical calculations show us that the true dew point is practically the same that the dew point you would arrive by the wrong way. However this is just a consequence of the numbers. If you apply the wrong reasoning you will obtain wrong results in general. And, what is more important by far, you will show a lack of understanding of the physics which is going on and will transmit misconceptions to those who receive ur answer.

______________________
"Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don't understand it, but by that time you are so used to it, so it doesn't bother you any more". Arnold Sommerfeld

Anyone is in that second stage?

Regards.
 
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