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Thermal Physics (GAS)

  1. Dec 14, 2006 #1
    1. The problem statement, all variables and given/known data

    The density of helium gas at T = 0°C and atmospheric pressure is 0 = 0.179 kg/m3. The temperature is then raised to T = 105°C, but the pressure is kept constant. Assuming that helium is an ideal gas, calculate the new density f of the gas. kg/m3

    2. Relevant equations

    PV = nRT
    m/v = weight in grams?

    3. The attempt at a solution

    T1 = 0*C T2 = 105*C
    Po = .179 Pf = ?
    Pa1 = 1 atm? Pa2 = 10 atm?

    I think that is how im suppose to set it up but from there I am at a complete lost on what to do. But i am not even sure where to start on this problem.
     
  2. jcsd
  3. Dec 15, 2006 #2

    Andrew Mason

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    The key here is to find the volume change per unit mass.

    They give you the density of He at STP. But let's calculate it. Assume you have one mole of gas. Calculate the volume of one mole at 273K (0 C) and 1 atm (101325 Pa) using the Ideal gas law:

    [tex]V = nRT/P = 1*8.314*273/101325 = .0224 m^3 = 22.4 L[/tex]

    Since one mole of He has mass of 4g, the density is .004/.0224 = .179 kg/m^3.

    Do the same for He at a pressure of 378K (105C) and 10 atm. That is all the question is asking.

    AM
     
  4. Dec 15, 2006 #3
    I have a question on how did you get R to equal to 8.314?

    I thought R would be .0831 or am i looking at a different R value?
     
  5. Dec 15, 2006 #4

    Andrew Mason

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    In MKS:

    [tex]R = N_A k_B = 8.31451 m^2kg/s^2 K mol[/tex]

    The .0831 must be in units of [itex]m^2kg/s^2 K mol \times 10^3[/itex] or kilomoles

    AM
     
    Last edited: Dec 15, 2006
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