1. Dec 8, 2016

### Frank Einstein

1. The problem statement, all variables and given/known data
Hi everybody. I am currently trinyg to solve the first exercice of the fifth chapter of R.R. Rogers book about cloud physics.

"Show that the vapor pressure in equilibrium over a pure water drop of radius r decreases with T if r<2σ/LρL.

2. Relevant equations

es(r)=es(r=infinite)*(2σ/rRvρLT).

This equation relates the vapor pressure of saturation of a drop of radius r and the one from a plain surface.
es is the vapor pressure, r is radius, σ is surface tension, Rv is the gas constant, ρL is density and T is temperature.

3. The attempt at a solution

As far as I understand, I have to derivate es(r) respect to T and see if the result is similar to the crtical radius which value is 2σ/RvρLTlnS.

If I have understood correctly, none of the values in the expression of es depends on T, therefore d/dT=∂/∂T. If I derivate exp(a/T) I obtain -a*exp(a/T)/T2. This expression doesn't look like the one expressed in the upper paragraph.
Note: a=2σ/rRvρL

If someone could give me some guidance I would be extremley grateful.

2. Dec 8, 2016

### haruspex

What is L there (not the subscript, the other one)?
Are you saying svp does not depend on temperature? That would be a controversial claim.
For the question to make sense es(∞) must depend on temperature. You need to get two terms from the differentiation, one positive, one negative. Which wins will depend on whether r is above or below a threshold.

3. Dec 9, 2016

### Frank Einstein

L is the latent heat of evaporation.

In the book I can't find the value of es(r=infinite) so I just assumed that it didn't change.

4. Dec 9, 2016

### haruspex

Well, it must change, but I don't understand this equation:
Where did you get that from? It looks extremely unlikely to me. As r tends to infinity, the factor on the right after the asterisk should tend to 1.

5. Dec 9, 2016

### Frank Einstein

Is the equation 6.1 from the book A Short Course in Cloud Physics. It is in page 84. If you wish, you can find it here:
https://es.scribd.com/doc/293198948/A-Short-Course-in-Cloud-Physics-pdf

6. Dec 9, 2016

7. Dec 9, 2016

8. Dec 9, 2016

### haruspex

Last edited by a moderator: May 8, 2017
9. Dec 9, 2016

### Frank Einstein

I'm sorry. I didn't realize. I posted it in a hurry and I didn't realize. I apollogize for the inconvenience I have caused you.

But still I have the same problem that I started with. In the attempted solution, I derivate the propper expression and I don't find the anwser I seek.

Then If I have understood you correctly, I have to seek the expression of es(r=infinite) and try to add it to the derivate to try to obtain the expression of the crytical radius?

10. Dec 9, 2016

### haruspex

You would have to apply the product rule.
I tried working backwards from the thing to be proved and arrived at $e_s(\infty)=Ae^{-\frac L{R_vT}}$ for some constant A. I must say the curves I've seen for svp do not look much like that.

Edit: but it seems like it might follow by combining Clausius-Clapeyron with PV=nRT.
In the C-C equation, I guess you can take Δv as the V of the gas, since the liquid phase has almost no volume in comparison.

Last edited: Dec 9, 2016
11. Dec 10, 2016

### Frank Einstein

After some more thoght I think I can use Classius Clapeyro equation to solve this problem.

Thak you very much for the help

12. Dec 10, 2016

### haruspex

You are welcome.
I wrote earlier that the shape of y=e-1/x seemed wrong for the svp curve, but later I looked more carefully and realised it looks right in the small x region. The larger x regions are probably beyond the range of interest for engineers, so do not appear in published svp curves.
It is quite an interesting curve. Zero gradient at the origin, curving swiftly upwards, in a roughly circular arc, through a point of inflection, then levelling off to a horizontal asymptote.