Expansion of liquefied air to what volume when heated to 20 degrees C

  • #1
I have a physics thought experiment I can't seem to figure out. I need to know what .0417 grams of liquefied air will expand to when it is introduced to room temperature, or approximately 20 degrees C (293.15 K).

When I say liquefied air, I mean regular earth atmospheric air that has been super cooled and condensed to a liquid. I know that the earths atmosphere is 78% Nitrogen, 21% Oxygen, and for this experiment I am to assume 1% water vapor (H2O). The consistencies do not change from the gas to liquefied form or liquefied form to gas in my thought experiment. The percentages always stay the same and do not change.

Upon cooling Oxygen would liquefy before Nitrogen at -297 degrees F, or -183 degrees C (90.2 K).
Then upon more cooling Nitrogen would then liquefy at -320 degrees F, or -196 degrees C (77.36 K).

I am to assume that the temperature of my liquefied air is starting at 77.36 Kelvin so that the entire substance is in liquid form at the start of the thought experiment.
I was not given any pressure it is under, I only know that it is liquefied air I am starting with but it is at sea level, so I am assuming the liquefied air is starting and ending at the same pressure, which is 1 bar. I was not given a volume that it is currently residing in. I only know that the mass of this liquefied air is .0417 grams at the start before it expands.
I have researched and found out that liquefied air has a density of 870 kg/m3 (0.87 g/cm3), but I am not sure what temperature this information was cited at. This could be where my equations went wrong. Is liquefied air really this dense or am I incorrect?

I have also researched and found out that regular air has a density of 1.204 kg/m3 (.001204 g/cm3) at 1atm (sea level) or also known as 1 bar and 20 degrees Celsius. Therefore, liquefied air would seem to be 722.59 times more dense than regular atmospheric air at 20 degrees Celsius and at 1 bar according to this information if it is correct. I just need to make sure that I have the density of liquefied air correct. I may have it wrong.

I have found out that Volume = Mass/Density. I did the calculation and found that liquefied air with a density of 870 kg/m3 (0.87 g/cm3), and a mass of .0417 grams, would have a volume of 47.931 Cubic Millimeters or 0.047931 Cubic Centimeters (cc).

Regular air at 1 bar and 20 degrees Celsius with a density of 1.204 kg/m3 (.001204 g/cm3) and also a mass of .0417 grams would have a volume of 34,635 Cubic Millimeters or 34.635 Cubic Centimeters (cc). Therefore, liquefied air would seem to expand 722.601 times when heated to 20 degrees Celsius. I just need to make sure that I have the density of liquefied air correct.

Since the liquefied air is starting out at -196 degrees Celsius, and is introduced to room temperature, which is to be 20 degrees Celsius for this experiment, the temperature change is immediately a 216 degee change.

Are my calculatiosn correct? Would liquefied air expand over 722 times if you raised the temperature by 216 degrees instantly? Or are my calculations wrong somewhere? I really need someone to find out the density of liquefied air at -196 degrees Celsius (77.36). If I have the density wrong my calculations would be all wrong.

The consistency does not change, the percentages of the mixtures in the air stays the same, only the temperature changes.

If someone could help it would be much appreciated because I have been attempting to figure this out for quite some time.
 

Answers and Replies

  • #2
work out how many moles of 'air' you have in the liquid (hint average O2 and N2 to get molecular mass of 'air = 28.9')
Then from PV=nRT you know the volume of this many moles of 'ideal gas' at room temp (about 24l/mol)
 
  • #3
I did read a little about the Ideal Gas Law but can't seem to see how this would answer my question. I have read everywhere that the "molecular mass" or moles of air is approximately 28.966. I have also read everywhere that the Sea Level pressure on earth is about 14.6959488 psi (pounds per square inch). I have even used these numbers to plug into an Ideal Gas Law calculator and found that at 20 degrees Celsius the volume of the air would be 42,520 ci (cubic inches) but what does this mean? Surely the earths atmosphere is not limited to 42,520 cubic inches right? Isn't the entire earth the same psi of pressure all over the place and approximately 20 degrees celsius over most regions?

Also, does the "molecular mass" of air always stay the same in moles? Whether it is in liquid form or air/gaseous form, isn't the number always 28.966 moles or am I wrong?

You stated that the volume of my gas would be 24l/mol. What is this measurement? I am not familiar with this. What would that number be in cubic inches?

Thanks for your help.
 
  • #4
PV = nRT
Pressure (in Pa) * volume (m^3) = n (number of moles) * R (8.31 J/mol/K) * temperature (kelvin)

You can use different units it will just give you a different value of the constant.
 
  • #5
0.0417g of air is 0.0417/28.966 = 0.00143mols

So at 1 atm (100KPa) and 20C (293.15 K)

V = nRT/P = 0.00143 * 8.31 * 293.15/100E3 m^3
You can convert this into any units you like.

You stated that the volume of my gas would be 24l/mol. What is this measurement?
If you put 1mol and 1 atm into the equation.
V = nRT/P = 1 * 8.31 * 293.15/100E3 = 0.024 m^3 = 24 litres (At 0C it's 22.4 litres/mol - worth remembering)
 

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