1
$\begingroup$

I found this problem in a practice sheet. The original source doesn't indicate any authorship or whatsoever.

It states the following:

A sealed container of $\pu{6.24 L}$ in capacity has humid air at $\pu{30 ^\circ C}$ with a $RH=90\%$. The container temperature is then lower to $\pu{27 ^\circ C}$. Find the volume of condensed water. The vapor pressure of water is $\pu{32 mmHg}$ at $\pu{30 ^\circ C}$ and $\pu{27 mmHg}$ at $\pu{27 ^\circ C}$. You can consider that the volume of condensed water is insignificant.

The problem doesn't specifically mentions any other assumptions but I'm assuming that the gas is behaving ideally and since it is referring to water. I will need the density of water, which is $\pu{1 \frac{g}{mL}}$ and the formula weight of water, which is $\pu{18 g mol-1}$.

The strategy which I thought was that the moles of air will remain the same when the vessel is cooled but the total moles will change before and after, as the moles of water initially will be different when the vessel it is cooled.

Then to find the volume what I need to find is that difference in moles and use the formula weight of water and the density to get volume.

However this is a problem because, how can I characterize the vessel if I'm not given an initial pressure?

I'm aware that:

$$RH=\frac{\textrm{partial pressure}}{\textrm{vapor pressure}}\times 100$$

Therefore I can say that the partial pressure for water is:

$$0.9\times \pu{32 mmHg} = \pu{28.8 mmHg}$$

But what I need is the total pressure in the vessel. Is it possible to find it?

If I could find the total moles, what I could do is find the moles of air by difference as follows:

$$n_{air}=n_{t}-n_{water}$$

These moles will not change.

The moles of water could be found by using the ratio of the partial pressure of water to that of the total is equal to the moles of water divided by the total moles.

$$\frac{n_{water}}{n_{total}}=\frac{p_{water}}{p_{total}}$$

But as it can be seen, I'm stuck. I did not included numerical calculations because if I were to use the equation of the ideal gas:

$$PV=nRT$$

What would I had use as the total pressure? Therefore I'm trapped. Is there any way to find what it is being asked?

Edit:

Following the indications by the answer I did the following:

The number of moles of water vapor at $\pu{30 ^\circ C}$ can be found from using the equation of ideal gas:

$\left(0.9\times 32\,mmHg\right)(6.24\,L)=n_{water}(62.4\frac{mmHg\cdot L}{mol\cdot K})(30+273)K$

$n_{water}=0.009505\,mol$

Then the number of moles of water vapor at $\pu{27 ^\circ C}$:

$\left(27\,mmHg\right)(6.24\,L)=n_{water}(62.4\frac{mmHg\cdot L}{mol\cdot K})(27+273)K$

$n_{water}=0.009\,mol$

Then:

$\Delta n_{water}=0.009505-0.009=0.000505\,mol$

From this quantity the grams of water and volume can be obtained using density of water and the formula weight.

$V=0.000505\,mol\times\frac{18\,g}{1\,mol\,\pu{H_{2}O}}\times\frac{1\,mL}{1g}=0.009090\,mL$

Which would be the volume of the condensed water. Judging from the ammount obtained it makes sense that it is a very tiny volume.

$\endgroup$
  • 2
    $\begingroup$ Why do you need to mention air at all? Air is quite irrelevant. It could just as well have not been there. You have the pressure of water vapor before and after, you don't need anything else. $\endgroup$ – Ivan Neretin Dec 21 '19 at 20:03
  • $\begingroup$ Is your answer $\pu{8.99 mg}$? $\endgroup$ – Mathew Mahindaratne Dec 22 '19 at 5:46
  • $\begingroup$ @MathewMahindaratne I have obtained $0.009\,mL$ as the volume since the density of water is $1\frac{g}{mL}$ that would be what you obtained. I'll show it as an edit in my question. $\endgroup$ – Chris Steinbeck Bell Dec 22 '19 at 5:50
  • $\begingroup$ Your calculations are seemingly okay to me. But you must recheck the numerical value of $R$ (gas constant). Also, you must state that your assumption of saturated vapor pressure at $\pu{27 ^\circ C}$, because some water is condensed. $\endgroup$ – Mathew Mahindaratne Dec 22 '19 at 6:40
  • 1
    $\begingroup$ Chris, I get essentially the same answer as you, though you should express it to only two sig figs (based on what was given in the problem): 0.0091 mL. Expressing it to four sig figs, as you did, misrepresents the precision of the answer. $\endgroup$ – theorist Dec 23 '19 at 2:19
1
$\begingroup$

This problem is sloppily worded and, technically, it is not possible to give a definitive answer based on the information provided. But the missing information is not what you think -- you don't need to know the air pressure in the flask. What's missing are two key pieces: The degree of humidity ("humid" doesn't mean "saturated") and the volume of water in the flask. If the flask contains no water, and air that, while humid, is sufficiently below 100% $RH$ at $\pu{30 ^\circ C}$, there might be no water condensation when cooling to $\pu{27 ^\circ C}$ (i.e. you'd just be increasing the $RH$ from some number <100% to a value that, while larger, is still <100%). Alternately, if the flask is initially half liquid water, then the volume of water that condenses would be about half that if the flask instead initially contained a very small volume of liquid water. Thus here is what I think the author meant to ask:

A sealed container of $\pu{6.24 L}$ contains an equilibrated mixture of air and liquid water at $\pu{30 ^\circ C}$. The container temperature is then lowered to $\pu{27 ^\circ C}$. Find the volume of water that condenses. Assume the volume of liquid water is insigificant relative to the volume of the container. The vapor pressure of water is $\pu{32 mmHg}$ at $\pu{30 ^\circ C}$ and $\pu{27 mmHg}$ at $\pu{27 ^\circ C}$.

As to how to approach the question, that's straightforward: You know the partial pressure of the water vapor at each temperature. You also know the total volume of the air. Given this, you can determine the number of moles of water vapor at each temperature (using the ideal gas law). The difference is the number of moles of water that condenses, from which you can then determine the volume of the water that condenses.

$\endgroup$
  • $\begingroup$ I'm sorry, during my transcription of the problem I forgot to give the relative humidity which was 90 percent. The other piece which I omitted in a rush was that the problem did say to assume that the volume of liquid water is insignificant. But it used the term condensed and because of such I found it misleading because it said that I have to find the volume of condensed water and the it mentions that the volume of condensed water is negligible. I did not understood this part. How does this changes the way to approach this question? $\endgroup$ – Chris Steinbeck Bell Dec 22 '19 at 2:18
  • $\begingroup$ In my question I did mentioned that density of water and the formula weight can be used to find that volume. Am I right with this part? $\endgroup$ – Chris Steinbeck Bell Dec 22 '19 at 2:20
  • 2
    $\begingroup$ @ChrisSteinbeckBell (1) Since the RH is 90% (and assuming equil. conditions), there can be no water in the container to start (if there were water, the RH would be 100%). So the final V of condensed water equals the V that condenses. What they meant by "negligible" is prob. that the V of condensed water is negligible relative to the V of the container (such that you can use V_container = V_gas even after the water condenses). (2) Yes, you would use the molecular weight and density to convert from moles to volume: molecular weight converts moles to mass, and density converts mass to volume $\endgroup$ – theorist Dec 22 '19 at 3:22
  • 1
    $\begingroup$ @ChrisSteinbeckBell (1) Image you're cooling the air from Th to Tc. Assume at Th and Tc a certain V of air can hold 1 mol and 0.9 mols H2O vapor, respectively. Further assume the container has no water, so the air is not saturated and that, in addition, the RH at Th is 80%, meaning it is actually holding 0.8 moles H2O vapor. Now we cool it down to Tc. At Tc the air is capable of holding 0.9 mols vapor, so none condenses. Further, the RH at Tc is 0.8/0.9 x 100 = 89%. Hence the RH has changed from some number <100% (80%) to a number that, while larger, is still <100% (89%). $\endgroup$ – theorist Dec 23 '19 at 1:36
  • 1
    $\begingroup$ @ChrisSteinbeckBell (2) Suppose instead the flask has water, in which case the air is saturated. If the flask is half-filled with water, the volume of air is half what it would be if the volume of water were negligible. Thus half as many moles condense, because the number of moles of water that condense when you cool the flask is proportional to the volume of the air. Also: If you found my answer above helpful, you should up-vote it and/or accept it $\endgroup$ – theorist Dec 23 '19 at 1:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.