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Thermal energy in a room.

  1. Apr 22, 2007 #1
    1. The problem statement, all variables and given/known data

    A 6.40m by 7.80m by 3.10m room contains air at 20C .

    What is the room's thermal energy?

    2. Relevant equations

    3. The attempt at a solution

    I started looking at this problem and to me it is missing information. I know that because it's a homework question it probably isn't. I tried to find how many mols are in the room but that dosen't make sense since I don't know what the air is made up of inside the room. Could someone give me a push in the right direction.

  2. jcsd
  3. Apr 22, 2007 #2

    Chi Meson

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    Step 1: consider air to be an ideal gas (it isn't, but if you wanted to be very precise, you'd have to be given the concentrations of of nitrogen, oxygen, CO2, H2O, etc).

    Step 2:Ideal gas law formula

    Step 3: energy of an ideal gas formula
  4. Apr 22, 2007 #3


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    Even if you consider it as an ideal gas you still need a molecular weight for it to get the number of moles. Air is roughly 2/3 N2 (28) and 1/3 O2 (32) and other stuff. So it's average molecular weight ought to be somewhere between O2 and N2 but closer to N2. I found a more detailled inventory on the web that put the mean at 28.97. For dry air.
  5. Apr 22, 2007 #4
    Thank you very much that helped a ton. You don't need the mols because I have the following

    p=1 atm
    v= 154.752m^3

    If I use the ideal gas formula I can solve for n which gives me the mols in the room. Then I use the equation

    (5/2) nRT

    I get the thermal energy.

  6. Apr 22, 2007 #5


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    Duh. Right. Good work. If you were given the mass of air you would need the weight. Don't know what I was thinking.
  7. Apr 22, 2007 #6

    Chi Meson

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    That's the spirit! It's good that you used the "5/2" bit instead of "3/2" since air is mostly diatomic. NOtice that you never needed to know the # of moles since if

    so just by calculating (5/2)PV, you are all done.
  8. Apr 23, 2007 #7

    That makes sense. I tend to take to many steps in my math anyways. I find that it helps me to understand exactly what the equation means and then I can use it any way I want to. This section is the first time I have seen these equations so I am still in the mode of over using them.
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