# Does the term "vapor pressure" even mean anything in an open system?

I know this type of question has been addressed before, but I believe mine is slightly different and hints at a potential misunderstanding I've had for a long time.

I understand the concept of equilibrium vapor pressure, that molecules with high enough energies escape the surface of a liquid until the rate of evaporation = rate of condensation. I also understand that for a given temperature, a substance has a characteristic saturation vapor pressure. But one thing bothers me: when considering a system open to the atmosphere, does it even make sense to talk about a vapor pressure developing? Consider, for example, water vapor at 95 C. Obviously, evaporation still occurs at the surface, but since its saturation vapor pressure is less than the surrounding atmospheric pressure, can we even consider a "vapor pressure" to develop, or is this a term that holds meaning only within the context of closed systems? (I suppose, however, that it does make sense to speak of the vapor pressure in an open system once the boiling point is reached since there is quite literally a vapor pressure building up in bubbles that can push back the atmosphere...)

• Umm, that's perhaps the most obvious one. Simply how much pressure something creates in a certain point and time. If system is open or not is irrelevant - it's not like pressure truly averages out either way. Commented Sep 25, 2023 at 17:57
• How about this: in an open system, a substance with a higher vapor pressure will evaporate faster than an substance with a lower vapor pressure. Commented Sep 25, 2023 at 18:32
• @user123462 With higher equilibrium vapor pressure, I think OP gets this one. Commented Sep 25, 2023 at 19:05

In an open system like you describe, the partial pressure at the liquid surface is observed to be at the equilibrium vapor pressure. This is typically higher than the partial pressure in the free stream of air, so there is a concentration driving force for diffusion of vapor molecules away from the liquid surface and into the air. Usually, all the action takes place in a thin concentration boundary layer close to the surface where the partial pressure varies rapidly with spatial position from the equilibrium vapor pressure at the surface to the free stream value at the edge of the thin boundary layer. The evaporation rate (mass transfer rate of vapor) is usually quantified by $$\phi=\frac{k}{RT}(p^*-p_{\infty})$$where $$\phi$$ is the water vapor mass flux (moles/cm^2), p* is the equilibrium vapor pressure at the liquid surface, $$p_{\infty}$$ is the partial pressure at the outer edge of the boundary layer (free stream value in air), and k (cm/sec) is the "mass transfer coefficient." The mass transfer coefficient is related to the diffusion coefficient of vapor molecules in air D and the boundary layer thickness $$\delta$$ by $$k=\frac{D}{\delta}$$There are experimental (dimensionless) correlations which have been developed to calculate the mass transfer coefficient from the air flow parameters.

Imagine the open system, but the evaporation is faster than mass transfer. Then the open system about reaches saturation. The partial pressure of vapor is approximately equal to (saturated) vapor pressure and the pressure difference to external pressure is compensated by partial pressure of the system air.

If the mass transfer is faster than evaporation, the open system does not reach saturation and vapor partial pressure is smaller than (saturated) vapor pressure.

Still, the vapor pressure is good criteria to evaluate volatility of liquids at given temperature. Liquids with somewhat higher boiling point and smaller molar enthalpy of evaporation than water, like toluene, have at room temperature higher vapor pressure than water. And therefore toluene is more volatile.

It sounds from the question as if you may not be taking into account the concept of partial pressure. In the example of water vapor at 95C, the relevant vapor pressure is the partial pressure, which is the pressure the water vapor would have if there were no other gases present. So it does not matter that the vapor pressure is less than the ambient atmospheric pressure, since you can treat it separately from the rest of the atmosphere.

In a well-mixed sample of ideal gases, the partial pressures can be added together to provide the total pressure. But the other gases cannot be completely ignored, since they slow the rate at which the water vapor leaves the neighborhood of the liquid surface (mass transfer). This allows there to be an approximately equilibrium vapor pressure near the surface, with a much lower pressure further away.

I apologize if this answer is telling you things you already knew, but I thought it might be helpful if you didn't.