Volume change when cooling a saturated volume of air

Gstoettenmayr · Feb 15, 2024

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.

Baluncore · Feb 15, 2024

Welcome to PF.

Are you aware that some water will condense, to liquid in the bag, when the bag is cooled?

Gstoettenmayr · Feb 15, 2024

Thank you for your reply. Yes, I am aware of this. What shrinkage did you calculate for the question I posted?

Chestermiller · Feb 16, 2024

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.

Baluncore · Feb 16, 2024

@Gstoettenmayr
Why do you ask such a specific question?
What accuracy does the answer require?
What is your application?

Gstoettenmayr · Feb 17, 2024

Here are my calculations, what are your thoughts?

Gstoettenmayr · Feb 17, 2024

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.

Baluncore · Feb 17, 2024

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.

Gstoettenmayr · Feb 17, 2024

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?)

Chestermiller · Feb 17, 2024

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

Tom.G · Feb 17, 2024

Chestermiller said:

1 - 0.5975 = 0.4125

I think your calculator needs servicing.

Chestermiller · Feb 18, 2024

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.

Gstoettenmayr · Feb 18, 2024

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:) !