Vapor Pressure Independent of Container?

[drXanthine]

This is from TBR. Consider a liquid in a closed, cylindrical container that is half full. The molecules at the surface of the container and along the edges have the fewest neighbors; therefore, they will evaporate more quickly than other molecules. This intuitively makes sense.

The text also says that vapor pressure is independent of the shape and volume of a container. Now consider two cylindrical tubes of the same volume. The first has a much smaller radius than the second. Since the smaller radius cylinder has a greater proportion of its molecules along the edges, I would think that it would have a slightly higher vapor pressure. Is this correct? Perhaps I'm just over-complicating things. Any help is appreciated!

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[sps27]

This is from TBR. Consider a liquid in a closed, cylindrical container that is half full. The molecules at the surface of the container and along the edges have the fewest neighbors; therefore, they will evaporate more quickly than other molecules. This intuitively makes sense.

The text also says that vapor pressure is independent of the shape and volume of a container. Now consider two cylindrical tubes of the same volume. The first has a much smaller radius than the second. Since the smaller radius cylinder has a greater proportion of its molecules along the edges, I would think that it would have a slightly higher vapor pressure. Is this correct? Perhaps I'm just over-complicating things. Any help is appreciated!

if you decrease the radius, you also essentially reduce the exposed surface area. So the vapor pressure goes down due to exposed surface area and goes up due to longer edge contact. So the total value essentially remains the same, I guess.

 

[Schenker]

Rate of evaporation is to vapor pressure as reaction rate is to equilibrium concentration. Vapor pressure is a description of an equilibrium value, whereas rate of evaporation is a dynamic value.

It will help if you imagine the following two containers as starting from no liquid having evaporated yet:

A closed container that is shaped such that the liquid inside evaporates more quickly will reach equilibrium more quickly, but it will reach the same equilibrium as if the liquid were in another closed container with a slower rate of evaporation. The pressure of the liquid's vapor in the former container will reach vapor pressure before the latter container, but they will both reach the same vapor pressure.

(I'm not sure if you directly quoted TBR, but I don't buy that liquid molecules along the sides of a container generally evaporate more quickly. What if the adhesive forces between the liquid and the container were greater than the cohesive forces between molecules of the liquid?)

 

[sps27]

Rate of evaporation is to vapor pressure as reaction rate is to equilibrium concentration. Vapor pressure is a description of an equilibrium value, whereas rate of evaporation is a dynamic value.

It will help if you imagine the following two containers as starting from no liquid having evaporated yet:

A closed container that is shaped such that the liquid inside evaporates more quickly will reach equilibrium more quickly, but it will reach the same equilibrium as if the liquid were in another closed container with a slower rate of evaporation. The pressure of the liquid's vapor in the former container will reach vapor pressure before the latter container, but they will both reach the same vapor pressure.

(I'm not sure if you directly quoted TBR, but I don't buy that liquid molecules along the sides of a container generally evaporate more quickly. What if the adhesive forces between the liquid and the container were greater than the cohesive forces between molecules of the liquid?)

That was a good explanation, that vapor pressure is essentially a matter of equilibrium. It just depends how quickly or how late the equilibrium is reached. Earlier I thought that the larger exposed surface will bring about greater vapor pressure but after reading your comments I realized that it will reach equilibrium sooner.

 

[drXanthine]

I'm not sure that is the case. In a chemical reaction, increasing reaction rate by, say, increasing temperature would increase the rates of both the forward and reverse reactions. Equilibrium is reached when the forward and reverse rates are equal. This is why a temperature change affects the rate at which equilibrium is reach but not the equilibrium itself.

In the situation posed, the particles along the edge of the container would be less attracted to the liquid than are the particles fully bounded by neighboring particles. This would essentially increase the rate of the "forward reaction." The reverse reaction would then be particles returning to the liquid. There is no corresponding increase in incentive for the particles to return to the liquid. They are only more likely to leave.

Alternatively, the energetics of such a situation can be considered. Say that the particles along the edge of the container have more or less intrinsic energy than do those in the center. Increasing the proportion of molecules along the edge would therefore change the energetics of the situation--thereby affecting equilibrium as opposed to just kinetics.

 

[Schenker]

I'm not sure that is the case.

Are you referring to my statement on vapor pressure and equilibria or on my comment in parentheses?

In a chemical reaction, increasing reaction rate by, say, increasing temperature would increase the rates of both the forward and reverse reactions. Equilibrium is reached when the forward and reverse rates are equal. This is why a temperature change affects the rate at which equilibrium is reach but not the equilibrium itself.

Increasing temperature increases only the rate of either the forward reaction or the reverse reaction, not both. This is because one reaction will be exothermic, while the other will be endothermic, assuming that the deltaH of the reaction is non-zero. Temperature changes will indeed shift the equilibrium of a chemical reaction.

The equilibrium between a liquid and its vapor is affected in the same way by temperature. Increasing the temperature would increase the rate at which molecules leave the liquid phase and decrease the rate at which molecules leave the gas phase. Thus, increasing the temperature shifts the liquid-vapor equilibrium more towards the gas phase, i.e. increasing the temperature raises the vapor pressure.

Alternatively, the energetics of such a situation can be considered. Say that the particles along the edge of the container have more or less intrinsic energy than do those in the center. Increasing the proportion of molecules along the edge would therefore change the energetics of the situation--thereby affecting equilibrium as opposed to just kinetics.

This would imply that vapor pressure is affected by the shape of a liquid's container, which we know not to be true.

(I'm still unconvinced that liquid molecules on the sides of a container are more likely to leave the liquid phase. While this may be true for boiling due to increased nucleation, it's not for evaporation. If this was something mentioned in one line of your TBR text and then never mentioned again, I would be inclined to disregard it, unless you can find another source.)

 

[drXanthine]

Sorry, I misspoke (mis-typed?) about the temperature and equilibrium. What I wrote is obviously not true. I was simply trying to give an example in which the kinetics was affected, but the thermodynamics of the situation was not. I should have said that adding a catalyst would have this affect.

I am almost convinced that vapor pressure is affected by the shape of a liquid's container. Since the molecules around the container sides will almost assuredly be energetically distinct than those neighboring only similar molecules.

When discussing how the vapor pressure is constant, TBR was concerned with the force per unit area side of things. Going to a large surface area would increase the amount of vapor above the surface (increasing the "force"), but this value would remain the same per unit area. I now think my reasoning is overkill and would probably be negligible in almost any case.

I think this is similar to your statement concerning the particles around the sides not evaporating more quickly. In general, the cohesive forces are probably stronger, but there are exceptions.

When I get home later I will post what TBR said about this specifically.

 

[drXanthine]

"Molecules can evaporate only from the surface, and tend to evaporate from the corners. Corner molecules have the fewest neighbors, so they have the fewest intermolecular forces. Because of the minimal forces, molecules evaporate most readily from the corners and next most readily from the surface."

There is an accompanying picture with a liquid-filled cylinder that has been sliced vertically down the middle. The corner molecules in this case are the molecules along the sides of the container as described above.

 

[Schenker]

"Molecules can evaporate only from the surface, and tend to evaporate from the corners. Corner molecules have the fewest neighbors, so they have the fewest intermolecular forces. Because of the minimal forces, molecules evaporate most readily from the corners and next most readily from the surface."

There is an accompanying picture with a liquid-filled cylinder that has been sliced vertically down the middle. The corner molecules in this case are the molecules along the sides of the container as described above.

Oh ok, sorry, I thought you meant side surfaces of the container rather than the edges of the surface of the liquid. My bad.

Even if the molecules at the edges of the surface evaporated more quickly, the shape of a container would not affect vapor pressure. However, the shape of a container may hasten or delay the attainment of that vapor pressure, like a catalyst that speeds up a reaction without affecting equilibrium.