Volume change when cooling a saturated volume of air

  • #1
Gstoettenmayr
7
4
TL;DR Summary: How much does a bag with 100 % water saturated air shrink in volume when it is cooled?

A bag filled with 100% water saturated air at 85 degrees C (inside and outside) at a pressure of 738 torr inside and outside is cooled to 50 degrees C (inside and outside bag) and the pressure (inside and outside of bag) remains at 738 torr. How much will the bag shrink?

Mentor note: Moved from a technical forum.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Welcome to PF.

Are you aware that some water will condense, to liquid in the bag, when the bag is cooled?
 
  • #3
Thank you for your reply. Yes, I am aware of this. What shrinkage did you calculate for the question I posted?
 
  • #4
Gstoettenmayr said:
Thank you for your reply. Yes, I am aware of this. What shrinkage did you calculate for the question I posted?
You first. Let's see your analysis.
 
  • #5
@Gstoettenmayr
Why do you ask such a specific question?
What accuracy does the answer require?
What is your application?
 
  • #6
Here are my calculations, what are your thoughts?
 

Attachments

  • 02988FAA-496E-4C38-9C10-DA0E5ED7C865.jpeg
    02988FAA-496E-4C38-9C10-DA0E5ED7C865.jpeg
    45.4 KB · Views: 43
  • CCF6AFD9-5D21-4D90-8919-A72F06F08279.jpeg
    CCF6AFD9-5D21-4D90-8919-A72F06F08279.jpeg
    49.8 KB · Views: 41
  • #7
Baluncore said:
@Gstoettenmayr
Why do you ask such a specific question?
What accuracy does the answer require?
What is your application?
It is just a thought experiment and ideal gas laws will suffice for the accuracy required.
 
  • #8
I do not have confidence in your 61% volume solution.

A first approximation.
Charles' Law. https://en.wikipedia.org/wiki/Charles's_law
T1 = 273 K + 85°C = 358 K.
T2 = 273 K + 50°C = 323 K.
If we wrongly assume there is no condensation,
T2 / T1 = 323 / 358 = 0.90 = 90%

A second approximation.
CalcTool https://www.calctool.org/atmospheric-thermodynamics/air-density
Air Density Calculator. 738 Torr. RH = 100%.
The density of saturated air at 85°C = 0.82513 kg/m³
The density of saturated air at 50°C = 1.01042 kg/m³
The mass in the bag remains constant, so the density ratio is also the volume ratio.
0.82513 / 1.01042 = 0.817 = 81.7%
That ignores the volume of liquid condensate that remains in the bag.

A third approximation.
Requires an absolute humidity calculator to work out the mass of liquid water that will condense during the cooling process. Assume water density is 1g/cc, and use that to correct the previous 50°C density.

An accurate computation.
You must correct for the density of the liquid water at 50°C.
 
  • #9
Thank you for very much for your detailed answer. Regarding the second approximation, the mass in the bag does remain constant. However, the mass in the gas phase does decrease due to the water condensing. Therefore, I would think that in this case the density ratios cannot be used to calculate the volume ratios. How far off the mark would I be with my reasoning?
(Or perhaps that is what you already pointed out by saying that this ignores the volume of condensate in the bag?)
 
  • #10
I can't really follow what you did, but here is my approach:

$$P_{w,85}=433.6\ torr$$$$P_{w,50}=92.5\ torr$$Assume an initial number of moles of gas = 1 mole.

Initial moles water vapor = 433.6/738=0.5875
Initial moles air in gas = 1 - 0.5875 = 0.4125

At 50 C, mole fraction water in vapor = 92.5/738= 0.1253
mole fraction air in gas phase = 1 - 0.1253 = 0.8747

Since none of the air condenses, the number of moles of air in the final gas is the same as in the initial gas. Therefore, final moles of gas phase = 0.4125 / 0.8747 = 0.4716
From this it follows that the number of moles of water that condenses to liquid = 1-0.4716 = 0.5284 moles = 9.5 grams. The volume of this liquid water condensate is about 9.5 cc = 0.0095 liters.

The pressure in the initial and final states is 738/760 = 0.9711 atm.

From the ideal gas law, initial volume in the bag is $$V_{85}=\frac{(1)(0.082)(273+85)}{0.9711}=31.2 \ liters$$

From the ideal gs law, the final volume in the bag is $$V_(50)=\frac{(0.4716)0.082)(273+50)}{0.9711}+0.009=12.9\ liters$$

The volume ratio is 12.9/31.2 =0.414
 
Last edited:
  • #11
Chestermiller said:
1 - 0.5975 = 0.4125
I think your calculator needs servicing.
 
  • Like
Likes Nik_2213
  • #12
Tom.G said:
I think your calculator needs servicing.
That 0.5975 was a typo. It should have been 0.5875. I've corrected it.
 
  • Like
Likes Nik_2213
  • #13
Chestermiller said:
I can't really follow what you did, but here is my approach:

$$P_{w,85}=433.6\ torr$$$$P_{w,50}=92.5\ torr$$Assume an initial number of moles of gas = 1 mole.

Initial moles water vapor = 433.6/738=0.5875
Initial moles air in gas = 1 - 0.5875 = 0.4125

At 50 C, mole fraction water in vapor = 92.5/738= 0.1253
mole fraction air in gas phase = 1 - 0.1253 = 0.8747

Since none of the air condenses, the number of moles of air in the final gas is the same as in the initial gas. Therefore, final moles of gas phase = 0.4125 / 0.8747 = 0.4716
From this it follows that the number of moles of water that condenses to liquid = 1-0.4716 = 0.5284 moles = 9.5 grams. The volume of this liquid water condensate is about 9.5 cc = 0.0095 liters.

The pressure in the initial and final states is 738/760 = 0.9711 atm.

From the ideal gas law, initial volume in the bag is $$V_{85}=\frac{(1)(0.082)(273+85)}{0.9711}=31.2 \ liters$$

From the ideal gs law, the final volume in the bag is $$V_(50)=\frac{(0.4716)0.082)(273+50)}{0.9711}+0.009=12.9\ liters$$

The volume ratio is 12.9/31.2 =0.414
Hello,

Thank you very much for your detailed calculation.

$$V_{85}=\frac{(1)(0.082)(273+85)}{0.9711}=31.2 \ liters$$ seems to contain a small error or typo
$$V_{85}=\frac{(1)(0.082)(273+85)}{0.9711}=30.2 \ liters$$ is the correct number.
So the the volume ratio is 12.9/30.2 =0.43 instead of 0.414

In the attachment, please find my updated calculations which give the same result as yours. I chose to also calculate the moist air densities (like Baluncore's approach) and compared them with the literature to check my calculations. I prefer your approach via the moles which is more elegant than my calculations.

In addition, I attached a graph that shows the partial and final volumes of 100 % RH air as temperature changes.

Thank you everyone for your help. This will allow me to calculate the mass flow changes in a dryer scrubber system at work with confidence:) !

graph.jpg

Densities_calulated.jpg

1708312519174.png

1708312547099.png
 
  • Like
Likes Chestermiller
  • #14
Thank you for catching my typo. My eyes are not working as well as they used to because of Parkinson's.
 
  • Like
Likes Nik_2213 and Tom.G
  • #15
Sorry to hear about that, either way, your explanations are most helpful! :)
 
  • Like
Likes Nik_2213

FAQ: Volume change when cooling a saturated volume of air

What happens to the volume of air when it is cooled?

When air is cooled, its volume decreases. This is because the molecules in the air move slower and come closer together at lower temperatures, leading to a reduction in the space they occupy.

Why does the volume of saturated air change when cooled?

The volume of saturated air changes when cooled because the temperature drop causes the water vapor in the air to condense into liquid water. This condensation reduces the amount of water vapor, which in turn reduces the overall volume of the air.

How does cooling affect the relative humidity of saturated air?

Cooling saturated air increases its relative humidity until it reaches 100%, at which point any further cooling will cause condensation. This is because cooler air can hold less water vapor, so as the temperature drops, the air becomes fully saturated more quickly.

What is the dew point in relation to cooling saturated air?

The dew point is the temperature at which air becomes fully saturated and water vapor begins to condense into liquid water. When cooling saturated air, the temperature at which condensation starts is the dew point.

Can the volume of air be reduced indefinitely by cooling?

No, the volume of air cannot be reduced indefinitely by cooling. As the temperature approaches the condensation point, water vapor will condense out of the air, and the remaining dry air will continue to decrease in volume. However, at extremely low temperatures, the air will eventually reach a point where further volume reduction is minimal due to physical limits and the properties of gases at low temperatures.

Similar threads

Replies
1
Views
1K
Replies
29
Views
2K
Replies
2
Views
602
Replies
1
Views
3K
Replies
16
Views
31K
Replies
3
Views
1K
Replies
4
Views
10K
Back
Top