Partial Pressures of an Ideal Gas Mixture Containing Water Vapor

In summary: I am still not understanding something. The book and instructor are clueless as to why this is in here since it is a first for the class.Water vapor behaves like an ideal gas at pressures below 75 mmHg - this is the saturation pressure of water at this temperature. In your case, water vapor behaves as an ideal gas, because its partial pressure is 14.56 mmHg - it is less than 75 mmHg.In summary, a gas mixture of 0.13 mol NH3, 1.27 mol N2, and 0.025 mol H2O vapor at a total pressure of 830 mm Hg and 323 K was analyzed. The mole fraction of each component
  • #1
kilgorethecat
5
0

Homework Statement



A gas mixture of 0.13 mol NH3, 1.27 mol N2, and 0.025 mol H2O vapor is contained at a total pressure of 830 mm Hg and 323 K. Calculate the following:
(a) Mole fraction of each component.
(b) Partial pressure of each component in mm Hg.
(c) Total volume of mixture in m3 and ft3.

Homework Equations



Xi = ni/ntotal
Ptotal = P1 + P2 + P3 ... + Pn
Pi = XiPtotal
PV = nRT
perhaps others...

The Attempt at a Solution



OK, so I have been working on this for a while now and can't seem to find the answer. Since water vapor only behaves like an ideal gas at pressures below 75 mm Hg, it cannot be treated as so in this problem.

I calculated the mole fractions no problem:
(0.13 mol NH3)/1.425 mol total = 0.0912
(1.27 mol N2)/1.425 mol total = 0.891
(0.025 mol H2O)/1.425 mol total = 0.0175

The problem comes when I try to calculate partial pressures of each gas and the water vapor. Since vapor pressure is a function of temperature alone, I was able to reference a table of values which says at 323 K (50 C), water vapor has a saturation pressure of 92.5 mm Hg. Subtracting this from 830 mm Hg gives 737.5 mm Hg, which should be the remaining combined pressures of ammonia and nitrogen gas. Next, I multiplied each gas's mole fraction by 737.5 mm Hg to obtain their partial pressures. However, according to the ideal gas law, these partial pressures should add up to the total pressure in the container, but they don't.

Can someone please help me understand how to approach and complete this problem? Thank you so much!
 
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  • #2
Why do you assume water vapor is saturated?
 
  • #3
V=ntot*R*T/Ptot

Pi=niR*T/V

Am I missing something?
 
  • #4
RTW69 said:
V=ntot*R*T/Ptot

Pi=niR*T/V

Am I missing something?

No need to calculate volume - molar fractions are enough.
 
  • #5
I'm sorry, I didn't mean saturation pressure. I meant vapor pressure. So, should I recalculate mole fractions of the two ideal gases excluding water vapor and use that to calculate their partial pressures?
 
  • #6
kilgorethecat said:
I'm sorry, I didn't mean saturation pressure.

I am afraid you meant it - you have read saturated vapor pressure from the tables and used it in calculations. This is equivalent to assume water vapor was saturated.

I meant vapor pressure. So, should I recalculate mole fractions of the two ideal gases excluding water vapor and use that to calculate their partial pressures?

Use all gases in exactly the same way - that is, ignore water vapor pressure tables. THEN, after you know partial pressure of water, check if it is saturated - IF it is, you have to check how much water would condense and become a liquid. If not, water vapor behaves exactly as every other gas.
 
  • #7
Doesn't NH3 dissolve rapidly in water to give NH4OH? Can we really consider these components (phases?) as not reacting with each other?
 
  • #8
Borek said:
I am afraid you meant it - you have read saturated vapor pressure from the tables and used it in calculations. This is equivalent to assume water vapor was saturated.

Crap, you're right! Sorry, my brain gets jumbled from the numbers flying around in my head all day with these engineering classes... and I didn't do so great in psychrometrics or thermodynamics. But I'm in biothermodynamics now, so I need to understand this or it's fail for me.


Borek said:
Use all gases in exactly the same way - that is, ignore water vapor pressure tables. THEN, after you know partial pressure of water, check if it is saturated - IF it is, you have to check how much water would condense and become a liquid. If not, water vapor behaves exactly as every other gas.

So, I calculated the partial pressure of water vapor from the original mole fraction, and I got 14.56 mm Hg. The saturation pressure of water at 50 C from the table is 92.5 mm Hg. Since the value I calculated is less than the value of saturation pressure at that temperature, this means that it is NOT saturated, correct? In that case, if it behaves like an ideal gas, why does almost every source I come across say that the pressure conditions must be below 75 mm Hg? This is the part that confused me in the first place, because I didn't think my given conditions allowed water vapor to behave as the other gases...

On a side note, this is a 300 level class in Biosystems Engineering and I didn't think they would give us that simple of a problem.
 
  • #9
rude man said:
Doesn't NH3 dissolve rapidly in water to give NH4OH? Can we really consider these components (phases?) as not reacting with each other?

You don't have a liquid water here. No idea if they react in gaseous phase - but I strongly doubt.
 
  • #10
kilgorethecat said:
So, I calculated the partial pressure of water vapor from the original mole fraction, and I got 14.56 mm Hg.

Looks OK to me.

The saturation pressure of water at 50 C from the table is 92.5 mm Hg. Since the value I calculated is less than the value of saturation pressure at that temperature, this means that it is NOT saturated, correct?

Yes.

In that case, if it behaves like an ideal gas, why does almost every source I come across say that the pressure conditions must be below 75 mm Hg?

Must be below 75 mmHg for what?
 
  • #11
Borek said:
Must be below 75 mmHg for what?

From my original post:

kilgorethecat said:
Since water vapor only behaves like an ideal gas at pressures below 75 mm Hg, it cannot be treated as so in this problem.

In my last post when I said I didn't understand how I could treat water vapor as an ideal gas, I was referring to the total pressure (given) of 830 mm Hg. Perhaps to be more clear: I am confused as to whether or not this rule is referring to the TOTAL pressure of a gaseous mixture or the partial pressure of the water vapor itself.

I cannot seem to find a better explanation from a reputable source in my text nor on the web...Either we still don't understand water's role in gaseous mixtures or I am a moron.

Also, thank you SO MUCH for your help so far.
 
  • #12
OK, I see what it is about - whether ideal gas approximation is correct, or not.

To be honest with you I don't know what are limitations. A lot depends on the required accuracy. However, I don't think we have much better approximation, especially for the random mixtures of gases. I seem to remember seeing elaborate tables for some gas mixtures - note that they contain experimental data, not calculated

You can try to ask about ideal gas approximation limitations in the Engineering forum - link to this thread so that people there can check what we have already discussed here.
 
  • #13
borek said:
you don't have a liquid water here. No idea if they react in gaseous phase - but i strongly doubt.

10-4.
 
  • #14
rude man said:
10-4.

Makes 6, but I fail to see your point.
 
  • #15
That's 'Highway Patrol' talk for "OK, understood". This miserable TV show was popular back around 50-60 yrs ago.
 
  • #16
Ah, OK.
 
  • #17
Dang, I was hoping to come to a solid conclusion on this one today since it's due, but you know, I do appreciate your help and will certainly repost for curiosity's sake. You'll hear from me again, thanks again!
 

1. What is the importance of understanding partial pressures in an ideal gas mixture containing water vapor?

The partial pressures of an ideal gas mixture containing water vapor is important because it determines the amount of water vapor present in the mixture, which can have significant effects on the behavior and properties of the gas mixture. This is particularly important in applications such as air conditioning and combustion processes.

2. How is the partial pressure of water vapor calculated in an ideal gas mixture?

The partial pressure of water vapor in an ideal gas mixture can be calculated using Dalton's Law of Partial Pressures, which states that the total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of each individual gas. The partial pressure of water vapor can be determined by multiplying the total pressure of the mixture by the mole fraction of water vapor.

3. What factors can affect the partial pressure of water vapor in an ideal gas mixture?

The partial pressure of water vapor in an ideal gas mixture can be affected by factors such as temperature, pressure, and the amount of water vapor present in the mixture. As temperature and pressure increase, the partial pressure of water vapor also increases. Additionally, the amount of water vapor present in the mixture can change as a result of evaporation, condensation, and other processes.

4. How does the partial pressure of water vapor impact the properties of an ideal gas mixture?

The partial pressure of water vapor can have a significant impact on the properties of an ideal gas mixture. It can affect the density, specific heat, and thermal conductivity of the mixture, as well as its ability to undergo chemical reactions. In addition, the partial pressure of water vapor can also impact the equilibrium composition of the mixture, particularly in cases where the mixture contains multiple components.

5. Can the partial pressure of water vapor in an ideal gas mixture ever exceed the total pressure?

No, the partial pressure of water vapor in an ideal gas mixture cannot exceed the total pressure. This is because Dalton's Law of Partial Pressures states that the total pressure of a mixture of ideal gases is always equal to the sum of the partial pressures of each individual gas. Therefore, the partial pressure of water vapor can never be greater than the total pressure of the mixture.

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