Two vessels of the same volume, in the same location and open to the atmosphere. One is of a small diameter, the other is of a larger diameter. The question is will the vessel with the larger surface area evaporate the liquid faster than the smaller one? If not! I need answer to other question ,if i am not wrong. Why does vapour pressure decreases with decrease in surface area?
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$\begingroup$ this question was discussed there: physics.stackexchange.com/questions/55205/… $\endgroup$– Jaroslav KotowskiCommented Jan 28, 2014 at 20:17
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3 Answers
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Why does vapour pressure decreases with decrease in surface area.
Well, it does not. Vapour pressure is a function of temperature only. It is in fact the equilibrium constant for the reaction $\ce{H2O_{(l)}->H2O_{(g)}}$, and like all other equilibrium constants, is independent of surface area, amount of the species but depends only on the temperature. At a given T, only for one specific value of the partial pressure is water and its vapour in equilibrium.
That said, water is not actually in equilibrium when in an open vessel evaporating into the atmosphere. The vapour pressure above the water filled vessel is far less than what is required for equilibrium. Water constantly evaporates as a desperate attempt to increase the vapour pressure to its equilibrium value, but since the vessel is open and the vapours not confined, the equilibrium value is seldom reached. If the surface area is larger, it means that the liquid phase can evaporate a larger amount of water while attempting to attain equilibrium. Therefore, larger surface area increases the amount of water that is lost as vapour while attaining equilibrium, since it is that part of the liquid exposed to the vapour phase which tries to maintain equilibrium by evaporating or condensing. Evaporation is essentially a surface phenomenon, as opposed to boiling which becomes considerably a bulk phenomenon.
Since it is practically the surface layer which participates actively in the phase equilibrium, larger the surface, greater the attempt to attain equilibrium and quicker is the evaporation.
answered Feb 2, 2014 at 2:33
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1$\begingroup$ Nicely done! Specifically, I liked the visual part of this answer- water is trying hard to equate its fugacity, but it's always coming short. That's why after few days we will see empty vessel. $\endgroup$ Commented Apr 14, 2017 at 15:15
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It depends what you mean by "rate". If the question is in which vessel does the liquid disappear fastest, then the answer is yes. But some people might assume that rate means rate per unit area in which case the answer would be no as that just depends on the vapour pressure.
It is pretty easy to observe the effect with any volatile liquid.
answered Jan 28, 2014 at 19:41
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Evaporation means the change from liquid to gas. That's why water in the larger surface will evaporate more than smaller surface. The water surface in a larger vessel will have more direct contact with the air. Another thing is that evaporation will be influenced by pressure and temperature and both will influence the boiling point. In this point the liquid evaporation will occur.
The pressure inside the vessel which different surface, should be different and the pressure difference also found in between inside and outside the vessel.
