Water vapor equilibrium calculations

The first part said:

Calculate the amount of water required to saturate a vessel of volume 82 L at 300 K given that the vapor pressure of water is 38 torr at 300 K.

The solution went like this:

Since we have to saturate the vessel, we have to find the minimum number of moles that will occupy the given volume (that is what I understood).

Now, applying the gas equation

$$PV=nRT$$

$$\frac{1}{20}*82 = n * 0.82 * 300$$

$$n = \frac{1}{6}; w=n * M = 3 gm$$

This part is fine.

Now the second part:

If 5 gm of water is introduced in the vessel (initially evacuated), then, calculate the amount of water remaining.

So, my teacher said that, 3 gm will evaporate and 2 gm will remain. But, 3 gm will evaporate for 82 L. So, when I pour 5 gm of water, shouldn't the volume decrease?

• You are correct that there will initially be a volume of 82L - 5cc. Then 3cc evaporates, so the volume of the container is 82 L - 0.002 L. Will this make a difference in your calculation that will show up in a 2 significant figure answer? Apr 28 '17 at 1:47
• @airhuff then, in the first question also, the volume should be 82-0.003. right? Apr 28 '17 at 1:54
• But the point for the second question, is would it matter if you just ignored the effect of the volume of the liquid water? In the first question there is no liquid water left because it all evaporated and you have nothing but water vapor. Back to the second question: 82L - 0.002L = 81.998L, which rounds to 82L (note that you are only given 2 significant figures in the problem), and for that matter it rounds to 82.00L! So, is the volume of the small amount of water insignificant compared to the size of the container in this question (hint, hint), such that you can just ignore it? Apr 28 '17 at 2:07
• You are not supposed to write "Thanks" or "Help is appreciated" or any statements of weaknesses in your post. And fix that Vap0r in the title. [I edited his post, apparently he rejected it] Apr 28 '17 at 12:18