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Ideal Gas Law/Estimate of Cooling Power

  1. Mar 2, 2012 #1
    Oh you who are wise in the ways of Physics, I beg a moment of your time.

    I've poked around wikipedia and found my way to PV=nRT and Thermal Mass and some other basics, and Googled "compressed air" "ice" "expansion" and similar, but can't quite find anything that helps me answer my question, so I have come back to my colleagues here at PhysicsForum.

    We "all know" that if you take a can of compressed air (say, a keyboard de-duster) and spray it out slowly through a straw into a cup of water, ice will form.

    As the air leaves the can, the pressure decreases to ambient (1atm), the volume of the bubbles increase as they pass upward through the water, and the air itself absorbs heat from the water to cover the difference. This thermal transfer takes place at some level off real-world efficiency, and the air will be slightly cooler than ambient air when the bubble bursts at the water's surface. Once enough heat has been removed from the water, ice begins to form. Also, as anyone who's done the experiment will attest, the can gets colder as well.

    Suppose
    1. I have 1000L of air at a pressure of 1.09 atm, which is at 20°C.
    2. I allow this to bubble out of the tank (into the STP environment) through an expansion valve that is immersed in 1 liter of water which starts at 20°C.
    3. The thermal mass of air is approximately 1/4 that of water.
    How do I figure out (even roughly, within an order of magnitude) how much air I will have to bubble in order to freeze (say) half of the water? Assumptions like "perfect heat transfer" are fine, as long as we state them clearly. Neither the volume nor the pressure in item 1 are sacrosanct. If it's easier to solve the problem with more/less volume or pressure, great. (Solving for 0L or 1atm doesn't count)

    -Jeff Evarts
     
  2. jcsd
  3. Mar 3, 2012 #2

    russ_watters

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    Staff: Mentor

    You are misunderstanding what is going on. Most (all?) "compressed air" cans do not contain compressed air, they contain a pressurized liquid hydrocarbon - with some gas on top - similar to propane. When you open the valve on the can, the little bit of gaseous hydrocarbon comes out and the pressure drops in the can, causing the remaining liquid to boil. As the remaining liquid boils, its temperature drops until it reaches the boiling temperature of the hydrocarbon at the new, lower pressure.

    So the water in your cup freezes not due to adiabatic expansion of air through a valve, but rather through the lost latent heat of vaporization of a boiling hydrocarbon.
     
  4. Mar 3, 2012 #3
    Russ: Thank you for your reply.

    OK, so the quick example I gave is bogus.

    I think the question is still valid, though, right? If I keep bubbling a compressed gas through a volume of water, the temperature SHOULD drop until the water freezes, yes? If so, then I'm still interested in quantifying that process
     
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