(I am self-studying as can be verified from my post history and, additionally, I believe this question has more general applicability.)

Callen asks us to consider the following in his famous textbook:

Question: Discuss the equilibrium that eventually results if a solid is placed in an initially evacuated closed container and is maintained at a given temperature. Explain why the solid-gas coexistence curve is said to define the " vapor pressure of the solid" at the given temperature.

Essentially, the goal of the problem seems to be to define vapor pressure.

I have answered as follows:

When the solid is initially placed in the evacuated chamber (we suppose it is maintained at a temperature $T$ throughout), it will experience zero pressure. In general, the solid phase is not the minimum $G$ phase at zero pressure, so the solid will begin to sublimate (i.e off-gas) to form the gas phase (which is generally the minimum $G$ phase at zero pressure). As $N_g$ (the moles of gas phase) increases, we will begin to observe a pressure increase. This process will tend to continue until we reach the point $(T,P)$ at which the solid and vapor can coexist. At this point, any "parcel" of one mole of substance has equal $g$, whether it's in the gas or solid phase. It's therefore no more likely to be in one or the other and, unless there is some further driving force, the system will now stay where it is (with solid in equilibrium with said amount of gas). This amount of gas will have a certain pressure -- the vapor pressure!

But I have two questions here:

  1. I wrote that "unless there is some further driving force, the system will now stay where it is", but why should this be? Surely fluctuations will occur?
  2. Given 1), if there are fluctuations, then some fluctuations will be such as to produce more gas phase. But, if this is so, then how can the vapor pressure be well-defined given that (presumably) more $N_g$ means higher pressure?

Edit: I think some of my problem may be in not appreciating that, from a kinetic theory perspective, solids can also exert pressure, but I am not sure.

  • 2
    $\begingroup$ Fluctuations are ignored in macroscopic thermodynamics, because observable fluctuation in state variables is very unlikely and is not easily observed. More detailed view takes into account fluctuations, and in this view the instantaneous properties can change in time. However, pressure is unlikely to be the same everywhere, so its fluctuation depends on position. Macroscopic vapor pressure is understood as the average value of the instantaneous pressure. $\endgroup$ Jul 14, 2023 at 0:20
  • $\begingroup$ Fluctuations perhaps scale as the square root of the number of particles in the volume. For a mole of particles that is a very small number. $\endgroup$
    – Jon Custer
    Jul 14, 2023 at 1:09
  • $\begingroup$ Is my assessment otherwise correct @JánLalinský ? That is, would the equilibrium state in this case be just at the “start/onset” of sublimation (rather than proceeding to some intermediate amount of sublimation)? I am also curious in terms of physical mechanism as to how the (vapor) pressure remains constant throughout the sublimation. Surely as we add many more particles which were previously in the solid phase, we crank up the pressure? Or is the equilibrium condition in terms of equal $\mu$ such that there are very few (relatively) particles in the solid, so that further sublimation does… $\endgroup$
    – EE18
    Jul 14, 2023 at 2:29
  • $\begingroup$ …not add relatively more particles and thus the pressure is negligible changed? $\endgroup$
    – EE18
    Jul 14, 2023 at 2:30
  • 1
    $\begingroup$ Sublimation is loss of mass of the solid, due to evaporation. This means the system is not in equilibrium. This is possible when the gas concentration (partial pressure) is low enough. When equilibrium is reached, in any given time interval, the number of particles that evaporate is the same (up to a fluctuation) as the number of particles that condense back on the solid. There are always two opposing processes - evaporation and condensation - and equilibrium is when these processes balance each other. $\endgroup$ Jul 14, 2023 at 14:58

2 Answers 2


In thermodynamics an equilibrium is generally dynamic.

If we look at the solid phase in the system we would see the particles are vibrating due to the thermal energy of the solid, and as the thermal energy randomly fluctuates occasionally a particle will get enough energy to be ejected from the surface into the gas phase. The exact timings of the particle ejections will be random, but since we have huge numbers of particles (of order of Avagadro's number) the average number of particles leaving the surface per second will be well defined.

At the same time particles in the gas phase are continually colliding with the solid surface. In general these collisions will be inelastic as some of the kinetic energy of the particle is exchanged with vibrational energy of the particles in the solid. So sometimes a gas particle will lose enough energy to be bound to the solid. The energy exchange is random, so the exact times particles from the gas phase condense onto the solid will be random, but again we have huge numbers of gas particles so the average number of particles condensing onto the surface per second will be well defined.

The equilibrium state is the one in which the two rates are the same. We expect that the rate of sublimation is controlled mainly by the temperature because the greater the thermal energy of the solid the more likely it is that the particles in the solid will randomly get enough energy to be ejected.

The condensation rate will also be affected by temperature since the higher the thermal energy of the particles the less likely they are to randomly lose enough energy to stick to the solid surface. However the rate will also be related to the number of gas particles per second hitting the solid surface and this will be proportional to the pressure. Hence increasing pressure increases the rate of condensation.

And we can now see why an equilibrium exists. If we keep the temperature and volume constant we expect the rate of sublimation, $R_s$, to be constant but the rate of condensation will be proportional to the pressure:

$$ R_c = kP $$

for some constant $k$ that depends on the temperature. Then we get equilibrium when $R_s = R_c$ i.e. at the pressure given by:

$$ R_s = k P $$

And this pressure is the vapour pressure.

All very well, but your question is whether we can usefully define a vapour pressure or whether it would randomly fluctuate with time. After all, both the ejection of solid particles from the surface and the binding of gas particles back onto the surface are random events.

And the answer is that yes the pressure will fluctuate. However the law of large numbers comes to our rescue. Assuming we have around a mole of the solid we are considering around Avagadro's number of particles, and this is such a huge number that the random events average out to give well defined average rates of sublimation and condensation. Hence we get a well defined average vapour pressure.

All of thermodynamics is based upon the assumption that the large number of particles produces well defined average values for properties. Your case of a vapour pressure is just one example of this.

  • $\begingroup$ Thank you for your answer! If it's OK, I just want to recapitulate my question slightly because I think this answer isn't quite getting at my question 2) in the OP. As written in a comment above, my question 2) can be restated as follows: let $x_g$ be the mole fraction of gas. If we are in equilibrium for any $x \in (0,1)$ then how is it that we have the same pressure in all cases? That is, when the subliming solid finally achieves equilibrium with the gas (which it has been off-gassing), it will do so at some $x_0 \in (0,1)$, right? But then, based on my understanding of phase transitions,... $\endgroup$
    – EE18
    Jul 14, 2023 at 18:31
  • $\begingroup$ ...we will have a solid-vapor equilibrium for any $x \in [x_0,1)$. Take some $x_1 \in (x_0,1)$ (i.e. $x_1>x_0$). Surely there are (macroscopically) more particles in the gas in this case, and one would correspondingly expect a higher pressure (this is a kinetic theory argument, I guess)? $\endgroup$
    – EE18
    Jul 14, 2023 at 18:32
  • $\begingroup$ May I also get you to comment on my question 1) in the OP? When the solid in this scenario first achieves the equilibrium at gas mole fraction $x_0$, will it in general stay there? That is, simply left to its own devices (no external changes), will it simply remain at $x_0$ rather than proceeding to some $x_1>x_0$? Is this a correct understanding? $\endgroup$
    – EE18
    Jul 14, 2023 at 18:34
  • $\begingroup$ That is, I think my emphasis on fluctuations may be misplaced, but I think my question still stands. $\endgroup$
    – EE18
    Jul 14, 2023 at 18:55
  • $\begingroup$ @EE18 Are you asking why the vapour pressure is independent of the amount of solid that is present? $\endgroup$ Jul 15, 2023 at 12:05

I wrote that "unless there is some further driving force, the system will now stay where it is", but why should this be? Surely fluctuations will occur?

This question is about equilibrium in general, not about vapor pressure specifically. We always have fluctuations around the equilibrium state. If the state is stable, fluctuations always produce a restoring "force" that brings the system back to the same equilibrium point. Thermodynamics describes this situation through appropriate potentials, in the case of constant $T$, $P$, $N$, that potential is $G$.

If you are looking for a physical explanation, this will come not from thermodynamics but from empirical observation, specifically, the observation that gradients of pressure, temperature and chemical potential produce flows (convective flow, heat, or molecular flow) in a direction that decreases the gradient. This produces a stable state: if temperature fluctuations heat up a spot at the expense of a neighboring spot, it will induce a flow of heat from the hot spot to the colder one and in the process the temperature gradient will diminish.

  • $\begingroup$ Thank you for your answer Themis. I've added some commentary to John Rennie's answer below which may help clarify my confusion. It seems your answer here is directed at my question 1) in the OP. To clarify, my issue is that any $x \in (x_0,1)$ also represents an equilibrium, and these would represent macroscopic differences in pressure surely? These are not fluctuations I don't think, but rather macroscopic differences as we proceed "fully from one phase to another". I am thinking somewhat analogously to what the lever rule says here. $\endgroup$
    – EE18
    Jul 14, 2023 at 18:36
  • $\begingroup$ You say that "any $x\in(0,1)$ represents equilibrium" but this this is not true. The $x^*$ at equilibrium must satisfy $\mu^\text{gas}(T,P^*,x^*) = \mu^\text{solid}(T,P^*)$: solve this equation to obtain $x^*$. This is an easier problem to solve analytically in vapor/liquid equilibrium using the van der Waals equation. I suggest that you do that to convince yourself of how things works at the phase boundary. $\endgroup$
    – Themis
    Jul 15, 2023 at 6:23
  • $\begingroup$ Not any $x$, but multiple $x$. Do you not agree that as the solid turns into vapor, there are various $x$ ranging from some $x_0$ at which it initially achieved equilibrium, all the way to $x = 1$? That is, the amount of solid changes/can change. I am asking why this does not change the pressure as you add more to the gas phase. $\endgroup$
    – EE18
    Jul 15, 2023 at 14:55
  • $\begingroup$ No. Only one $x$ satisfies the equilibrium condition. That’s what Gibbs’s rule says for one component. $\endgroup$
    – Themis
    Jul 15, 2023 at 15:04
  • $\begingroup$ What do you mean? The lever rule explicitly gives us a rule for determining $x$ and clearly it can vary? $\endgroup$
    – EE18
    Jul 15, 2023 at 17:23

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