My teacher said that vapour pressure is independent of the surface area and volume of a liquid. But I think it should be dependent, because surface area is directly proportional to the evaporation rate and if I increase surface area of a liquid then more liquid will be evaporated and more vapours will form and hence vapour pressure will be increased. And similar approach for volume of liquid also. Where am I wrong?
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4$\begingroup$ I think that you forget about molecules re-entering the liquid from the vapour. Eventually the vapour will come into equilibrium with its liquid and so the rate of leaving the liquid and the rate of colliding with the liquid surface and so re-entering, become equal. When this happens the vapour pressure becomes constant. $\endgroup$– porphyrinCommented Apr 10, 2023 at 13:15
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1$\begingroup$ Sir, I kept thinking on it and got what you were trying to say, thank you sir for helping. $\endgroup$– Rohan SinghCommented Apr 10, 2023 at 14:33
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4$\begingroup$ It's equilibrium vapor pressure. What actual partial pressure of H2O vapor is, in one place or another, depends on, like, everything. $\endgroup$– MithoronCommented Apr 10, 2023 at 14:54
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2$\begingroup$ Also I wouldn't be surprised if this is a dupe of half a dozen old q. $\endgroup$– MithoronCommented Apr 10, 2023 at 14:56
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2 Answers
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Vapour pressure is intensive property.
It depends for pure liquids (if we neglect minor secondary effects) only on temperature. Water molecules are not aware of the surface size nor the liquid volume to modify their behaviour.
Larger liquid surface areas accelerate the evaporation and condensation the same way. The system with larger surface area converges to equilibrium faster, but the equilibrium is the same regardless of the surface area or liquid volume if measured at the same temperature.
You can try an experiment at home. Fill 2 small containers with water. One wide and shallow, the other tall and narrow. Put them in an air tight container like the common clip boxes for food.
If your hypothesis were correct, water from shallow and wide container would progressively evaporate, condensing in the tall and narrow one.
If you observe it, let us know. It could lead to the effective way how to distill water without providing thermal energy.
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$\begingroup$ @Poutnik Hello Poutnik. This has left me thinking. If you have a system where area effects cannot be neglected, you cannot postulate that free energy is only a function pressure, temperature, and composition of the system. In this case, the saturation pressure will not be a function of temperature, but also of surface energy, and thus will be different like, e.g., what you find in water tables. Do you agree with me? $\endgroup$ Commented Apr 11, 2023 at 12:43
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$\begingroup$ @MetalStorm I have intentionally avoided minor effects like for non-flat surfaces $p = p_0 + \frac{2\gamma}{r}$. // Like it does not make much sense to show students learning about impulse p=m.v, what it is like as 4-vector in the Theory of relativity. $\endgroup$– PoutnikCommented Apr 11, 2023 at 13:29
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$\begingroup$ @Poutnik But, isn't it the case that the OP is precisely asking for? $\endgroup$ Commented Apr 11, 2023 at 13:42
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$\begingroup$ @MetalStorm If you read what he wrote here and in the related question about liquid volume, it is clear he is confused about basics of evaporation kinetics and thermodynamics. $\endgroup$– PoutnikCommented Apr 11, 2023 at 13:46
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$\begingroup$ @Poutnik Yes, he is confusing kinetics and thermodynamics, but well... $\endgroup$ Commented Apr 11, 2023 at 13:57
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Your intuition confuses the long term equilibrium level with how fast it is reached
Your intuition that a larger surface area should increase evaporation rate is correct.
But we normally define "vapour pressure" as the equilibrium pressure. And this means the pressure that would be reached given infinite time. So, while a larger surface area in a closed vessel would reach that equilibrium pressure far faster, this doesn't matter as we have as much time as we need to reach equilibrium even if the surface area were tiny. The rate of evaporation is irrelevant in both cases.
We don't often observe things that have reached thermodynamic equilibrium, which explains why intuition doesn't often match theory. But the idea of equilibrium if often important in chemistry and it is worth learning why it is independent on how fast things happen.
answered Apr 10, 2023 at 16:19
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1$\begingroup$ In which area or subject is this true? We normally define the vapour pressure (or tension) as the actual vapour partial pressure, which may or may not be equal to the saturation vapour pressure (tension of saturated vapour) or the equilibrium vapour pressure which are the pressures in some equilibria. $\endgroup$ Commented Apr 11, 2023 at 6:38
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1$\begingroup$ @VladimirFГероямслава The context where the question was asked. The context where the vapor pressure is constant. Had the teacher said the actual vapour pressure, they you would be right but the question would be meaningless as the actual vapour pressure is not something constant. We measure, for example, the humidity of air relative to the fixed, known capacity of air to hold water vapour which is the only constant relevant to an elementary chemistry class. $\endgroup$ Commented Apr 11, 2023 at 8:44