I need clarity on saturated vapour pressure in a closed system at equilibrium. How does saturated vapour pressure relate to vapour pressure?

  • $\begingroup$ What exactly are you confused about? Are you just asking about the meaning of 'saturated'? $\endgroup$ – bon Aug 4 '15 at 18:11
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    $\begingroup$ The saturated vapor pressure will be the vapor pressure of a closed system containing liquid and vapor in equilibrium. It will change with the temperature of the system. $\endgroup$ – Jon Custer Aug 4 '15 at 19:15
  • $\begingroup$ Perhaps chemguide.co.uk/physical/phaseeqia/vapourpress.html might help? $\endgroup$ – chipbuster Aug 4 '15 at 21:15

If you're at equilibrium, you're saturated. If you're saturated, you're at equilibrium. The end.

If things aren't at equilibrium, the situation is different. Consider a dish of water in the middle of the Sahara desert. The amount of water in the atomsphere in the Sahara desert is generally nonzero, say in the range of 1 $\mathrm g~{kg}^{-1}$. Even though the water vapor pressure (even in the Sahara!) is nonzero, it is way less than "saturated". Therefore it isn't at equilibrium, and the pan of water quickly evaporates.


How does saturated vapour pressure relate to vapour pressure?

There are four terms you need to understand.

  • pressure (i.e. the total pressure)
  • vapor pressure
  • partial vapor pressure
  • saturated vapor pressure (identical with equilibrium vapor pressure.)

Let's assume that water has a equilibrium vapor pressure of 18 torr at STP (760 toor, 20 degrees C).

Case 1

I have a flask sealed with some air and some water at my STP. The total pressure (i.e. the pressure) is 760 torr. Let's say that there is 750 torr of dry air. That would mean that the partial pressure of water is 10 torr.

Case 2

So I have a flask with water and some air is sealed above. The air/water are at STP and are at equilibrium. So the total pressure of the gas phase (water + dry air) is 760 torr. 742 torr is dry air and 18 torr is the partial vapor pressure of water. The air/water mixture has 100% humidity which means that it is saturated with water, hence the term saturated vapor pressure. (In other words if the temperature falls then dew would form, meaning that the air loses water.)

Case 3

I put a sample of water in a flask and try to pull a "vacuum." So long as the water stays at the standard temperature then the pressure will stay at 18 torr. The water keeps "evaporating" into the gas phase. So 18 torr is the equilibrium vapor pressure as well.

Case 4

Now water has a density of 1g/cm^3, and 18 g/mole. So 18 g of water at the boiling point (100 degrees C) occupies 22.4 liters. If the 18 grams of water was at 100 degrees C, then I need a volume of 946 liters to get all of the water to evaporate. So if I had a 1000 liter tank, then 18 grams of water wouldn't be enough to fill it to saturation at 100 degrees C. In other words all the 18 grams would be in the gas phase with no liquid. (If I dropped the temperature to 99 degrees C, water still wouldn't condense out.) So in this situation the water would have a vapor pressure less than the saturated vapor pressure.

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    $\begingroup$ Why don't you use SI units? $\endgroup$ – mcocdawc Nov 3 '15 at 9:50
  • $\begingroup$ @MaxW Case 1 is not correct. Since it is at STP, the water would have a vapor pressure of 18 Torr. Since you said there is 750 Torr of dry air, the total pressure inside the flask would be 768 Torr. $\endgroup$ – Nova Apr 7 '17 at 21:29
  • $\begingroup$ @Nova - No, the point I was trying to make was that there wasn't enough water to get to equilibrium pressure. $\endgroup$ – MaxW Apr 7 '17 at 23:34
  • $\begingroup$ @MaxW Are you saying that there was so little water that it all evaporated and only produced 10 Torr of pressure? $\endgroup$ – Nova Apr 8 '17 at 9:21
  • $\begingroup$ @Nova - Yes, that was what I meant. $\endgroup$ – MaxW Apr 8 '17 at 15:59

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