2
$\begingroup$

I know this has been asked before but i could not understand the replies. If someone has a simpler answer please do so.
My simple Geog lesson just said that warm air molecules are further apart leaving bigger gaps for water vapour. Is this correct somewhat is it more than this.

$\endgroup$
  • 1
    $\begingroup$ It would be better if you linked the question you found here but didn't understand. That'll help the answerers to better gauge the level of their answers. $\endgroup$ – user258881 May 29 at 10:43
  • $\begingroup$ Warm air being hotter can carry more number of water molecules in vapour state , I believe $\endgroup$ – Noah J. Standerson May 29 at 10:50
3
$\begingroup$

That explanation is wrong. Cold water has less vapor pressure than warm water, and so we find less water vapor in air that is cold, and (if the water is available) more of it in air that is warm. This has only to do with the temperature dependence of vapor pressure.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Between freezing and boiling point, the thermodinamical stable phase is liquid water. So, at first no water molecules should be present in air.

But in order to join together, the water molecules must form a boundary, and it requires energy. The balance equation is:

$$\Delta G .V + \gamma A < 0$$

where $\Delta G$ is the free energy of the bulk of water, and is negative in temperatures and pressures where liquid water is the stable phase. $V$ is the volume of the water droplet

$\gamma$ is the energy per unit of area of water at the temperature, and $A$ is the surface area of the droplet. This term is positive.

A spherical droplet has the minimum area per volume, and for a radius R the inequality becomes:

$$\Delta G \frac{4}{3}\pi R^3 + \gamma 4\pi R^2 < 0$$

$$R > \frac{3 \gamma}{-\Delta G}$$

Any random meeting of molecules that form a smaller radius are not stable and returns to water vapour.

$\Delta G$ is negative just below the boiling point, and becomes more and more negative as the temperature decreases.

So, in colder air the critical radius is smaller, and it is easier for random meetings form stable water drops, decreasing the amount in vapor phase.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ if there would be an extra about the free energy itself and how it is temperature dependent, then this would be a perfect explanation! $\endgroup$ – Ben Jul 10 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.