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$48~\mathrm{l}$ of dry $\ce{N2}$ passes through $36~\mathrm{g}$ $\ce{H2O}$ at $300~\mathrm{K}$, and this results in a water loss of $1.20~\mathrm{g}$. The vapor pressure of water is?

The solution says, that water loss is observed because of escape of water molecules with $\ce{N2}$ gas. Then it says that water vapours occupy the volume of $\ce{N2}$, i.e $48~\mathrm{L}$, this part I am not able to understand. Why would water occupy the same volume as that of $\ce{N2}$, it isn't mentioned that these two are in some container of volume of $48~\mathrm{L}$. What is the logic behind this? Then the solution proceeds, by applying $pV=\frac{m\mathcal{R}T}{M}$.

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It doesn't matter if there is a container. I'm picturing bubbling the nitrogen through the water, for example the water is in a flask, and there is a hose going down into the water, maybe a frit at the end of the hose.

The important assumption is that nitrogen reaches equillibrium with the water, which is a big assumption.

But adding 1.2 grams of water would increase the total volume by about 1 L. I would use the total volume to answer the question, not 48L, but it will only change the answer by about 2%.

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  • $\begingroup$ True, they should have asked for partial pressure, not vapor pressure so that assumption wouldn't be necessary. $\endgroup$ – Fred Senese Jan 15 '15 at 19:49
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This is an example of using partial pressures for a mixture of ideal gases.

We are making a big assumption - that water is an ideal gas - but once we make that assumption, the way to proceed is to treat each gas in the mixture as if it were by itself in the container. As you pointed out, it isn't very obvious from the way the question is worded that we would assume that both would occupy 48 L, but, since the volume of a gas depends either on the container volume or on temperature and pressure when there is no container, it does make sense. The idea is that whatever is constraining the $\ce{N2}$ to 48 L would also constrain the $\ce{H2O}$ to the same volume, since it is a mixture of gases.

Since the partial pressure equals the vapor pressure, the solution is given by calculating the pressure of this mass of water under these conditions.

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  • $\begingroup$ Thanks again , i did get what you are saying , but i think , we need to make another assumption that , the weight of N2 gas taken= weight of water displaced , only then will they occupy equal volume . $\endgroup$ – user47024 Jan 16 '15 at 10:51
  • $\begingroup$ No, for ideal gases you don't need to make that assumption. And I am pretty sure that that assumption would never hold - I can't think of a situation where a gas would displace an equal mass of water while bubbling through. Maybe you are trying to bring in buoyancy? That would make sense for incompressible fluids and for solids, but not for gases. $\endgroup$ – thomij Jan 16 '15 at 14:14

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