4
$\begingroup$

In my textbook, it's written that vapour pressure of a liquid does not depend on shape and size of a container but won't a container that provides less surface area to the liquid will have lower vapour pressure?

for example, consider 2 containers give in the diagram below, the volume of container and liquid is same in both the diagram just the shape is changed. enter image description here

wouldn't the vapor pressure be lower in the diagram on the left side? and how does changing size not affect vapor pressure?

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ Consider though that "vapor pressure" is really "equilibrium vapor pressure" thus ignoring dynamics. Due to smaller surface area the left system would be slower to come to equilibrium than the right system. $\endgroup$ – MaxW Sep 23 '19 at 16:24
  • 3
    $\begingroup$ For a sufficiently tall container, vapor pressure does vary with height, in a gravitational field, so it is dependent on shape. $\endgroup$ – DrMoishe Pippik Sep 23 '19 at 17:56
  • 2
    $\begingroup$ At equilibrium, both forward and reverse rate increase with increasing surface area, so the equilibrium constant is not affected. Your diagram is lacking blue molecules in the gas phase. $\endgroup$ – Karsten Theis Sep 23 '19 at 19:38
  • 1
    $\begingroup$ @DrMoishePippik I am afraid your comment is confusing OP. $\endgroup$ – Alchimista Sep 24 '19 at 8:49
  • 1
    $\begingroup$ @DrMoishePippik. Yes and of course. But this remark about tall containers in a gravitational field can be done all the times if the gas isn't perfect. Thus, for basically every question regarding gases. What is important is the P at the condensed ph./ gas interface, or, by definition, vapour P is function of T only. The P atop has nothing to do with vapour P has for that region is not an interface. And again the same thinking applies to P in general. Atm P at ground and at K2 are of course different. But vapour P is the same. It might be a different definition in earth science and meteo... $\endgroup$ – Alchimista Sep 27 '19 at 10:00
3
$\begingroup$

Vapor pressure does not depend on surface area because it is derived from the thermodynamic equilibrium between the liquid and gas phases of a substance. The closer the vapor pressure is to atmospheric pressure, the closer that substance is to boiling. At its boiling point, vapor pressure equals atmospheric pressure and the molecules that comprise the liquid now have enough thermal energy to overcome the intermolecular forces that are maintaining it in a condensed phase. So it might help to think about how container geometry does not affect the boiling point of a liquid; the same reasons are true for why it does not affect the vapor pressure.

The evaporation rate in terms of mass/time is absolutely a function of the container geometry, but equilibrium vapor pressure is an intrinsic molecular property.

Improve this answer
$\endgroup$
3
$\begingroup$

Evaporation of a liquid (and condensation of vapour) are physical processes. We can write the equations for a liquid material $\text{M}$ as follows:

$$\ce{M_{(l)}} \rightleftharpoons \ce{M_{(g)}}$$

By the law of mass action, we can write the expression for the equilibrium constant as follows:

$$K = \dfrac{[\ce{M_{(g)}}]}{[\ce{M_{(l)}]}}$$

We typically use partial pressures for the activity of gases and unity for the activity of liquids. Thus,

$$K_{p} = \dfrac{p_{M}}{1} = p_{M}$$

Suppose we start with the liquid $\ce{M}$ in a container with an arbitrary amount of headspace. It will evaporate till the vapour exerts a pressure equivalent to $p_{M}$ (of course, $p_{M}$ depends on the temperature).

Thus, with a change in the size of a container, there will be a difference in the amount of $\ce{M}$ in the vapour phase, but not the partial pressure exerted by the vapours of $\ce{M}$ at equilibrium (a.k.a. the vapour pressure).

The shape might affect the flux at which the dynamic exchange between vapour and liquid phase $\ce{M}$ is taking place, but nothing more.

Improve this answer
$\endgroup$

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.