Understanding Vapour Pressure & Temperature Difference

[Jacob White]

Homework Statement

https://www.ioc.ee/~kalda/ipho/Thermodyn.pdf

Problem 22 (It contains a drawing so it's better to see it in a booklet than copying here)

Relevant Equations

pv = ps(T)

I have a problem at the very beginning. I don't know how to relate this vapour pressure to the temperature difference. I have read the hint:

Recall the idea 7: for dynamical processes, at first, a mechanical equilibrium is reached, which means the equality
of pressures; the other equilibria (e.g. thermal) will be
reached later (if ever within a reasonable time frame).

In particular, this means that if there is evaporation from
a water surface, and because of that, close to the water
surface, there is an higher concentration of water vapours,
then there must be a lower concentration of air molecules.

Indeed, while due to mechanical equilibrium, the total
pressure must remain equal to the atmospheric one, it also
equals to the sum of the vapour pressure and air pressure
13. Equivalently can be said that the air pressure equals
to the atmospheric pressure minus the vapour pressure.
If the saturation pressure ps(T) becomes larger than the
atmospheric pressure patm then mechanical equilibrium is
no longer possible: as we learned earlier, very close to
the water surface, there is thermal quasi-equilibrium and
r = 100% and hence, in that layer, the vapour pressure
pv = ps(T).

The total pressure p in that layer is sum of
the vapour pressure and air pressure, hence p ≥ pv > patm.
Therefore, the vapours at the liquid surface will have larger
pressure than the atmospheric one, and the surrounding
air will be pushed away. Furthermore, if there were a small
bubble inside the liquid, it would also have higher pressure
of vapours inside than the pressure of the surroundings,
hence the bubble would start growing.

It should be noted
that there are always either microscopic bubbles or other
impurities inside the liquid which can serve as evaporation
centres.

But I don't know what is happening the saturation pressure is about 100 times smaller than atmospheric. How could it be bigger than it? Or something else is the case? Could someone explain it to me? I would be very grateful.

 

[Chestermiller]

I assume you are struggling with part (a).

What is your understanding of the definition of "relative humidity?" Based on this definition, what is the partial pressure of the water vapor in the bulk air?

 

[Jacob White]

It's the ratio of the current vapour pressure to the saturation pressure at this temperature. Outside, in the bulk air has relative humidity 90% so using the graph vapour has pressure p ≈ 0,9 * 2,3kPa = 2,07kPa

 

[Chestermiller]

It's the ratio of the current vapour pressure to the saturation pressure at this temperature. Outside, in the bulk air has relative humidity 90% so using the graph vapour has pressure p ≈ 0,9 * 2,3kPa = 2,07kPa

So, from the graph, what temperature can the water in the cloth cool down to before no more can evaporate?

 

[Jacob White]

Now I have done the problem assuming as you have written that the water will cool to the temperature at which the saturated pressure would equal to the vapour pressure outside(all the result are correct). But why the water stops evaporating now? Does the vapour that is very close to the cloth has to have the same pressure as vapour outside? I was thinking that the sum of vapour's and air's pressure has to be equal to that outside.

 

[Chestermiller]

Now I have done the problem assuming as you have written that the water will cool to the temperature at which the saturated pressure would equal to the vapour pressure outside(all the result are correct). But why the water stops evaporating now? Does the vapour that is very close to the cloth has to have the same pressure as vapour outside?

Yes. Once the cloth has cooled to the temperature you determined, no more vapor can evaporate.

I was thinking that the sum of vapour's and air's pressure has to be equal to that outside.

No way. It has nothing to do with the air pressure. And, at the new temperature of the wet cloth, the vapor pressure is only 0.0207 bars.[/quote][/QUOTE]

 

[Jacob White]

But why the water vapour pressure couldn't be different next to the cloth?

 

[Chestermiller]

But why the water vapour pressure couldn't be different next to the cloth?

The water vapor partial pressure immediately at the wet cloth surface is equal to the equilibrium vapor pressure at the cloth temperature. Equilibrium is always present immediately at a liquid surface.

 

[Jacob White]

So at a liquid surface there is a saturation pressure at given temperature but why this pressure couldn't be lower than this outside?

If we suppose that cloth has temperature corresponding to lower saturation pressure this would mean that just near its surface the vapour has lower pressure than outside and so this vapour would be shrinked to water again? But why the outside partial vapour pressure determines this process and not all the outside pressure?

 

[Chestermiller]

So at a liquid surface there is a saturation pressure at given temperature but why this pressure couldn't be lower than this outside?

It only has to match the saturation pressure at the boundary. But if the saturation pressure at the boundary is lower than outside, water vapor will condense at the boundary, and if it is lower, water will evaporate at the boundary.

If we suppose that cloth has temperature corresponding to lower saturation pressure this would mean that just near its surface the vapour has lower pressure than outside and so this vapour would be shrinked to water again? But why the outside partial vapour pressure determines this process and not all the outside pressure?

Because an ideal gas mixture behaves as if each component is totally independent of the other components present. So the water vapor doesn't know that there is air in the mixture, and the air doesn't know that there is water vapor in the mixture.

 

[Jacob White]

I know Dalton's law for partial pressure but I didn't know that we could treat these gases completely separately. Could you explain why is it possible? Or say where I can read about it? When I was searching on the internet I have found only something about Dalton's law. I don't see why Dalton's law would imply that.
And it is valid only for ideal gases?

 

[Chestermiller]

I know Dalton's law for partial pressure but I didn't know that we could treat these gases completely separately. Could you explain why is it possible? Or say where I can read about it? When I was searching on the internet I have found only something about Dalton's law. I don't see why Dalton's law would imply that.
And it is valid only for ideal gases?

This is in every thermodynamics textbook. Two thermo textbooks that I highly recommend are:

Fundamentals of Engineering Thermodynamics by Moran, et al
Introduction to Chemical Engineering Thermodynamics by Smith and van Ness

For non-ideal gas mixtures, the interactions of the various species molecules begins to come into play.

 

[Jacob White]

I have just read a fragment about Dalton model and moist air(from Fundamentals of Engineering Thermodynamics by Moran ). And still I have my previous doubts. We are getting the vapour pressure treating it as it doesn't know about the air(Dalton model). But why this outside partial pressure has impact on the vapour at boundary?

I'am imagining this situation as the vapour at boundary has its saturation pressure and outside vapour has its partial pressure. And when saturation pressure is lower, this vapour couldn't further escape to the surrounding and evaporation is no longer possible. Is it correct thinking? And now I don't know why only the partial pressure of vapour determines whenether molecules of water are escaping from this boundary thin layer of saturated vapour.

 

[Chestermiller]

I have just read a fragment about Dalton model and moist air(from Fundamentals of Engineering Thermodynamics by Moran ). And still I have my previous doubts. We are getting the vapour pressure treating it as it doesn't know about the air(Dalton model). But why this outside partial pressure has impact on the vapour at boundary?

I'am imagining this situation as the vapour at boundary has its saturation pressure and outside vapour has its partial pressure. And when saturation pressure is lower, this vapour couldn't further escape to the surrounding and evaporation is no longer possible. Is it correct thinking? And now I don't know why only the partial pressure of vapour determines whenether molecules of water are escaping from this boundary thin layer of saturated vapour.

Sorry. I don't understand these questions. Maybe someone else can help.

Another book that may help answer all your questions is Mass Transfer Operations by Treybel

 

[Jacob White]

So maybe I will reformulate my problem. If we have two adjacent volumes of air(pure air and water vapour). And in the first vapour has lower partial pressure, then vapour would star moving from the second to the first until they would have equal partial pressures?

 

[Chestermiller]

So maybe I will reformulate my problem. If we have two adjacent volumes of air(pure air and water vapour). And in the first vapour has lower partial pressure, then vapour would star moving from the second to the first until they would have equal partial pressures?

This is different from the example that you have been considering. In that example, you have a small amount of liquid water in contact with a large gaseous mixture of water vapor in air. The mixture is blown past the liquid water surface, and the concentration of water vapor in the bulk of the air doesn't change (because the volume is so large and the amount of water that evaporates is insignificant). A thin boundary layer develops adjacent to the liquid surface in which the water vapor partial pressure decreases from the saturation vapor pressure at the liquid surface to the water vapor partial pressure in the bulk of the gas mixture. This partial pressure difference provides the driving force for water vapor to diffuse through the boundary layer from the liquid surface to the bulk. This enables water vapor to evaporate at the liquid surface. The evaporation robs heat from the liquid water, and the liquid water cools. This continues until the temperature of the liquid water is lowered to the point where its equilibrium vapor pressure has dropped to the partial pressure of the water vapor in the bulk air outside the boundary layer. At that point, the pressure driving force has dropped to zero, and no further evaporation takes place.

 

[Jacob White]

Thank you, this clarifies a lot!
But I have one another little question about the driving force you have written.
Why only the partial pressure makes that driving force for vapour? Why vapour molecules react only to the pressure made up by the vapour?

 

[Chestermiller]

Thank you, this clarifies a lot!
But I have one another little question about the driving force you have written.
Why only the partial pressure makes that driving force for vapour? Why vapour molecules react only to the pressure made up by the vapour?

It follows from Fick's law of diffusion. The molar flux is related to the mole fraction variation by:
$$J=-\frac{P}{RT}D\frac{dx}{dy}$$ where P is the total pressure, D is the diffusion coefficient, x is the mole fraction (in our case, water), and y is the distance from the liquid water surface. For constant total pressure over the boundary layer thickness $\delta$, this yields: $$J=\frac{D}{RT}\frac{(p_{sat}-p_{\infty})}{\delta}$$where p is the partial pressure.

 

[Jacob White]

Thank you a lot! Now I think that I understand it. I have also found an interesting derivation of this law:
http://web.mit.edu/biophysics/sbio/PDFs/L15_notes.pdf