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Thermal Dynamics- heating gas in a flask.

  1. May 10, 2007 #1
    1. The problem statement, all variables and given/known data
    A flask with a volume of 2.70 L, provided with a stopcock, contains ethane gas (C_2 H_6) at a temperature of 297 K and atmospheric pressure 1.01×10^5 Pa. The molar mass of ethane is 30.1 g/mol. The system is warmed to a temperature of 396 K, with the stopcock open to the atmosphere. The stopcock is then closed, and the flask cooled to its original temperature.

    1-What is the final pressure of the ethane in the flask?

    2-Find the mass of ethane remaining in the flask.
    Use 8.31 J/(mol * K) for the ideal gas constant.

    2. Relevant equations
    M=n * mm

    3. The attempt at a solution
    I used pV=nRT to get the final pressure to be 7.6 * 10^4 Pa for part one, but when I use this to work out the answer to part two I keep getting it wrong.

    I used n = (pV)/(RT) and then M = n * mm so M = (30.1pV)/(RT)

    The advice that was given was that You have to watch the units, do I have to convert Pa to atm or something, what am I missing?
  2. jcsd
  3. May 10, 2007 #2


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    30.1 gmol: maybe you need to change that to kg to be consistent with SI units? Just a quick guess.

    And L to m^3. You didn't show very much of your calculation so it's hard to tell what you did. But if you are using that value for the gas constant, you need to make sure everything else fits with it.
    Last edited: May 10, 2007
  4. May 10, 2007 #3
    I tried n= (pV)/(RT) = [(7.6*10^4)*(2.7)] / [(8.31)*(297)] = 83.14 moles, that means that there is 83.14*30.1 = 2502.57g of gas left in the flask, that just doesn't seem right.
  5. May 10, 2007 #4


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    Convert 2.7 L to m^3. They are not the same and m^3 is consistent with the 8.31 J/(mol K).
  6. May 10, 2007 #5

    Andrew Mason

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    You have to figure out the initial number of moles. You then work out the number of moles in the flask after heating. Then work out the pressure of that quantity at the original temperature.

    n = PV/RT = 1.01e5*2.7e-3/8.31*297 = .11 moles. (you know this is about right since one mole at STP occupies 22.4 L).

    After heating:

    n = PV/RT = 1.01e5*2.7e-3/8.31*397 = 8.3e-2 moles

    So what pressure does this amount of ethane in a volume of 2.7 L produce at 297 K? What is the mass of the ethane?

  7. May 11, 2007 #6
    Thanks everyone, that was really helpful, one quick question though, should Liters always be converted to meters cubed when Pascals are used for the pressure, and Liters used when the pressure is in atm?
  8. May 11, 2007 #7


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    Yes. It helps to check the units of the value of the gas constant you are using, to be sure you get everything consistent.
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