Question about water drop's critical radius

[Frank Einstein]

Homework Statement

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/Lρ_L.

Homework Equations

e_s(r)=e_s(r=infinite)*(2σ/rR_vρ_LT).

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

3. The Attempt at a Solution

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

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/rR_vρ_L

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

 

[haruspex]

r<2σ/Lρ_L

What is L there (not the subscript, the other one)?

none of the values in the expression of es depends on T

Are you saying svp does not depend on temperature? That would be a controversial claim.
For the question to make sense e_s(∞) 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.

 

[Frank Einstein]

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 e_s(∞) 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.

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 e_s(∞) 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.

L is the latent heat of evaporation.

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

 

[haruspex]

L is the latent heat of evaporation.

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

Well, it must change, but I don't understand this equation:

es(r)=es(r=infinite)*(2σ/rRvρLT).

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.

 

[Frank Einstein]

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.

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

 

[haruspex]

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

That seems to be behind a paywall.
Can you just post an image of that page?

 

[Frank Einstein]

It seems I can't post it directly, I will post the link.
https://postimg.org/image/4rzbx74u5/https://postimg.org/image/4rzbx74u5/

 

[haruspex]

It seems I can't post it directly, I will post the link.
https://postimg.org/image/4rzbx74u5/[PLAIN]https://postimg.org/image/4rzbx74u5/[/QUOTE]
That makes more sense. You left out the exp. It says

e_s(r)=e_s(∞)e^{2σ/rR_vρ_LT}.

 

[Frank Einstein]

That makes more sense. You left out the exp. It says

e_s(r)=e_s(∞)e^{2σ/rR_vρ_LT}.

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 e_s(r=infinite) and try to add it to the derivate to try to obtain the expression of the crytical radius?

 

[haruspex]

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?

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.

 

[Frank Einstein]

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.

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

Thak you very much for the help

 

[haruspex]

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

Thak you very much for the help

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 realized 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.