[callagga]

Hi,

Is water evaporation rate for different temperatures of water proportional to the difference (ratio) of the Water Vapour Pressure (mmHg)? 

For example would my swimming pool's heater was on such that the average temperature was 30 degress C, as opposed to 25 degress C, would the increased evaporation rate be equal to P(28)/P(25)?  i.e. 31.8 / 23.76 (about 34% more) based on the tables at http://hyperphysics.phy-astr.gsu.edu/Hbase/kinetic/watvap.html

Thanks
Greg

[Valdorod]

I do not believe that they are.  Evaporation is part of the liquid/vapor pressure equilibrium which follows the Classius-Clapeyron equation, which is not linear in its original form, the relationship can be made linear by relating ln P with 1/T.  Also the relationship of the rate of a reaction with respect to temperature is also not linear (Arrhenius equation).


http://www.epcc.edu/Portals/259/PowerPoint/1412/chapter_11_powerpoint_l.ppt#290,42,Slide 42

http://www.epcc.edu/Portals/259/PowerPoint/1412/chapter_13_powerpoint_l.ppt#278,24,Slide 24

Valdo

[callagga]

Valdo - I'm not sure how to relate the "Rate Constant" equation from your 2nd link to what I did.  So my approach was:

a) Assuming:  "The evaporation rate of water from a swimming pool is fairly easy to calculate; given you have access to a pschyometric chart or a water vapor tables. W=(A (69.4+30.8 (V)) / Y )(Pw-Pa) W = lb / hr of evaporation. A = surface area of the swimming pool. V = mean wind velocity (mph). Y= Latent heat (approx. 1000). Pw - Sat. Press at Room Air temp (in Hg). Pa - Sat. Vapor press at water temp (in Hg)".  Reference: http://wiki.answers.com/Q/How_much_water_evaporates_from_a_pool.

b) So evaporation rate is proportional to Pw-Pa (i.e. difference in water vapour pressure at Air Temperation and Water Temperature).   [I don't fully understand this - this seems to imply that if the pool water was at the same temperature as the air there would be no evaporation???]

c) Looked up the Water Vapour Pression tables from http://hyperphysics.phy-astr.gsu.edu/Hbase/kinetic/watvap.html, i.e. so this would have taken into account your 1st point.

d) Formed the ratio between the two values, e.g. 31.8 / 23.76 (about 34% more) for my 30 degress C and 25 degress C example.

Q1 - Is my thinking incorrect somewhere?  I had seen a post somewhere where someone suggested the increased evaporation at a higher temperature for water could be calculated on this basis (i.e. post, not in a text or anything).

Q2 - Re Arrhenius equation - I'm not sure this would be valid here no?  It seems to deal with reaction between 2 substances together at the same temperature.  Here we have the pool water temperature changing (i.e. the question is the difference in evaporation rate in the case one turns on the pool heater) and outside air which isn't changing in temperature.

Signed
Still a bit confused 

[Valdorod]

Callaga,

You are heading in the right track, I'll try to clarify a few points, from what I understand.  First, to clarify there is always the theory and then the actual application of the theory.

A)  For part A everything you found is correct, you need all that information to actualy calculate the rate of evaporation for a particular system taking into account pressure, temperature and surface area.  Nothing incorrect there.

B) For part B, what it is referring to is how much water needs to evaporate to achieve full vapor pressure.  Think of it as this: Pw is the vapor pressure of water at ambient temperature, while Pa is the actual vapor pressure.
For example at 25 Centigrade the vapor pressure of water is supposed to be 23.756 mmHg.  If the water temperature is at 20 Celcius the vapor pressure is 17.535.  Thus enough water needs to evaporate to make up the difference.  Now in relative terms the bigger the difference the higher the rate of evaporation (the ratio you obtained).  However, as the two vapor pressures approach each other the rate decreases (the ratio gets smaller).

Also at the same time you are evaporating water, water vapor is condensing, just at a lower rate.  Eventually the two pressures are equal and the rate of evaporation remains constant, and it equals the rate of condensation.

Q1)  If you increase the temperature of the water, you are just providing more kinetic energy, thus providing more molecules with enough enthalpy to evaporate.  Thus your thinking is correct, however, this is where you might run into problems with the equation being used in part a) (theory vs practice).

If the temperature of the water is higher than the air temperature, the ratio is now negative because Pa is bigger than Pw, thus that equation can only be used for normal evaporation of water systems, which are not being heated artificially.

Q2) "Technically" both fusion and vaporization are thermodynamic reactions since they involve the transfer of heat and the breaking of intermolecular bonds.  Thus

H₂O(l) --->  H₂O(g)    ΔH_vap = 40.8 kJ/mol at 1 atm

You can very easily derive the equation for a system at two different temperatures and arrive at this form of the equation

Ln (k₁/k₂) = -Ea/R (1/t₁ - 1/t₂)

in this form it becomes obious that as temp increases so does the rate.

The important part of the the Arrhenius equation is that the rate of a reaction increases with temperature, however the increase is not linear but exponential.  If the temperature difference between the two systems is small enough a ratio can approximate the change, however, over a wider range then the ratio is not valid anymore. ( kind of like the slope in calculus)

As you can see in this equation there is a term Ea (activation energy); increasing the temperature of the water allows more water molecules to obtain enough energy to go over this barrier and thus evaporate.

Valdo

[callagga]

thanks Valdo,

So if I've substituted things correctly the evaporation rate difference between 25C and 30C is ~31%, and not ~34% like to calculated previously?


Ln (k1/k2) = -Ea/R (1/t1 - 1/t2)            
            
Ea   40800   40.8 kJ/mol at 1 atm      
R   8.314472       R = 8.314472(15) J · K-1 · mol-1       
Ea/R   4907.10654867802         
T1(C)   25         
            
T2(C)   T2(K)   ln(k1/k2)   k1/k2(%)   
25   298.15   0   0.00%   
26   299.15   0.055017603245846   5.66%   
27   300.15   0.10966860577041   11.59%   
28   301.15   0.16395665958155   17.82%   
29   302.15   0.21788536834016   24.34%   
30   303.15   0.27145828815759   31.19%   
31   304.15   0.32467892837734   38.36%   
32   305.15   0.37755075234135   45.87%   
33   306.15   0.43007717814129   53.74%   

Tks
Greg