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A balloon is half filled with water at sea level ($P = \pu{1 atm}$, $T = \pu{298K}$) such that the partial pressure of water is $\pu{24 mmHg}$. It is then submerged to a point where $P = \pu{2 atm}, T = \pu{298K}$. What will be the vapor pressure of water at that point?

My first instinct is to find the mole fraction of water in the balloon. Using Dalton's law,

$P_{total} = P_{\ce{H2O}} + P_{\mathrm{atm}} = 24 + 760 = \pu{784 mmHg}$

$P_{\ce{H2O}} = X_{\ce{H2O}}\,P_{total}$

Solving for $X_{\ce{H2O}}$, I get $24/784$. Then I use that to find the new partial pressure at the new depth, $(24/784)\cdot 1520$. But the answer seems to be that the partial pressure of water remains constant. Can someone explain why the partial pressure will be constant?

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Excellent question. On the one hand, the vapor pressure of water (or any other volatile liquid) is a function of the temperature alone, but on the other, if you assume that the mole fraction is constant you find that the vapor pressure of the water changes even though the temperature is constant.

The short answer

The water vapor pressure stays constant because as the pressure in the balloon increases, the mole fraction of the water reduces proportionally to maintain the equilibrium it has established.


The longer answer

Let's dig into the former approach I mentioned in the introduction first. We can begin with the Clausius-Clapeyron relationship.

The Clausius-Clapeyron relationship (often is simplified to):

$$\ln \left(P_{vapor}\right) = -\frac{L}{R}\left(\frac{1}{T}\right) + C$$

Where $L$ and $C$ are constants and $R$ is the ideal gas constant.

Using this model, we would expect the vapor pressure of the water to remain constant and therefore the mole fraction of water in the gas phase to decrease.

Dalton's law tells us: $$X_{water}=\frac{n_{water}}{n_{tot}} = \frac{P_{water}}{P_{tot}}$$

Using this expression we can see that, as the total pressure of the gas increases and the vapor pressure of the water stays constant, the mole fraction of water in the gas phase must go down. If we assume, as a first approximation, that the total number of gas molecules in the gas phase is constant, then a reduction in mole fraction can only be accomplished by some of the molecules of water in the gas phase being reabsorbed into the liquid water phase. This consequence is in agreement with the Le Chatelier principle prediction. A decrease in volume available to the water molecules will shift the equilibrium below to the left:

$$H_2O\,(l) \rightleftharpoons H_2O\,(g)$$

Thank you for such a rich question! How interesting that there are so many topics it relates too! Excellent!


Other thoughts

  • When we calculate the initial pressure of the other gases in the volume, we should use:

$$P_{tot} = P_{water} + P_{all\,other\,gas}$$

$$P_{tot} = P_{external} = 760\,mm\,Hg$$

$$P_{all\,other\,gas} = P_{tot} - P_{water} = 760\,mm\,Hg - 24\,mm\,Hg = 736\,mm \,Hg$$

  • The dissolved gas in the water in the balloon will also increase at higher pressures. How will this affect our calculations?
  • We have not considered energy concepts explicitly in this argument about the pressures being equal (in the first bullet above), but we could have considered the elastic energy stored in the bonds of the balloon. How could we include this to get a better estimate of the pressure of the other gases in the balloon?
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  • $\begingroup$ +1 for this terrific answer! And welcome to chemistry.SE :) $\endgroup$ – Michiel May 19 '13 at 16:33
  • $\begingroup$ Thanks michielm! I am so excited that this resource is available to the chemistry community! :) $\endgroup$ – Harvey Ryan Johnson May 19 '13 at 17:52

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