How is it even possible that vapour pressure of liquid and vapour of solid are equal at freezing point? [duplicate]

Question

My text book states The freezing point is defined as "the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase" Also the vapour pressure is the pressure at phase equilibrium by definition. By looking at a typical phase diagram

Its clear that theres no single temp or even different temp corresponding to same pressure at equilibrium for solid with gas and liquid with gas (except the triple point) . The how can the vapour pressure be assumed equal at freezing points. Many sources explain this in term of chemical potential but doesn’t the phase diagram contradict the statement .Please correct if anything i stated is incorrect. Thanks

Answer 1

The liquid-solid line in the phase diagram consists of the two phases with no gas phase present so the question of having the same vapor pressure is moot. If the physical constraint is removed, the common example is the freezing point of water in the open air, there are the additional equilibria involving water vapor. The vapor pressures of the liquid and solid must be equal. In essence a new triple point is defined for every inert gas pressure. This is why the freezing point of water for the initial Celsius scale had to be specified at a specific pressure [this pressure was the sum of the inert atmosphere and the vapor pressure]. Water in an inert atmosphere at a temperature has a definite vapor pressure. When both water and ice are present at a given T and P a triple point is established. Each inert gas pressure establishes a unique triple point, a distinct vapor pressure; solid and liquid must have the same VP at equilibrium.

Answer 2

Below the melting point, the vapor pressure over the solid phase is lower than over the respective liquid phase. And vice versa above the melting points. At the melting point, they are equal.

Look at it from the different angle. At the freezing point, the solid and liquid phase are at equilibrium. How could they be at equilibrium if each had different vapor pressure? The one with higher pressure would have tendency to transform to the one with the lower pressure. Similarly as if jars with water and saturated salt solution were place in the common, closed container.

Answer 3

To understand what you are asking, you need to remember two things:

1. Each point in a phase diagram is not related to a certain state function, but it represents the phase state (solid, liquid, vapor) that is thermodynamically stable at a fixed pair of values $(p, T)$. For a single component (i.e., a simple pure substance), this means that if the point lies in a certain region of the phase diagram (not specifically on any single curve or the triple point), then it is in that physical state (single phase).
2. Each point in the phase diagram lying on a curve represents an equilibrium between two different phases, such that if you artificially maintain $(p, T)$ constant, the phase transition between these two states will be perpetually ongoing. To be clear, for a freezing (melting) transition, you would observe that in the bulk material certain volume elements melt while others freeze until you let the system evolve to a new pair $(p', T')$, thus allowing for the phase transition to end.

Now, if you understand that each point on the separating curve between the two regions associated with compressible liquid and solid phases is equal to a pair $(p_{\text{melt.}}, T_{\text{melt.}})$ of parameters at which solid and liquid phases coexist, and that for the coexistence of these two in an isolated system, you need them to be in equilibrium, this means that their chemical potential is actually the same. So, at different fixed freezing temperatures, there exist a set of freezing (melting) vapor pressures that define the separation curve between the phases, meaning that for each point on the curve $p_{\text{solid}} = p_{\text{liquid}} = p_{\text{on the curve}}$. The key point is to note that in the phase diagram, $p$ is not associated with any of the phases but it is a generic parameter (so on the curve it must be equal for the two coexisting phases at a specific melting temperature $T$).

EDIT: I got the topic wrong. Sorry, but I was checking out my notes to try and understand if I was totally on the wrong track and as @jimchmst correctly said that there is just one point at which 3 phases do coexist for a pure substance. However, considering the atmosphere, you should look at a different graph, namely the one depicting a two phase mixture of an inert atmosphere and the pure substance. Honestly I tried to find one that depicts it correctly but I did not succeed. However, consider that the phase diagram you depict considers just the pure substance.

There is however also the possibility that your text is just using an old convenction: to consider the vapor pressure as the pressure of the pure component in an inert atmosphere in equilibrium with the pure substance (for example at the melting point for your substance in an atmosphere the vapour pressure of the liquid and the vapour pressure of the solid must be equal to allow for the transition to happen).