Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
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.
wouldn't the vapor pressure be lower in the diagram on the left side? and how does changing size not affect vapor pressure?
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.
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.
