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Helium Gas: Pressure, root-mean-square velocity, and more

  1. Dec 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A rigid, thermally insulated container with a volume of 22.4 liters is filled with one mole of helium gas (4 grams per mole( at a temperature of 0 Cesius (273K). The container is sitting in a room, surrounded by air at standard temperature and pressure (STP:1atm, 0 Celsius).
    a) Calculate the pressure inside the container in N/m^2
    b) Calculate the root-mean-square speed of the helium atoms.
    c) Now open a tiny square hole in the container, with area 10^-8 m^2. After 5 seconds, how many helium atoms will have left the container?
    d) During the same 5 sec. some air molecules from the room will enter the container _ through the same hole. How many air molecules will enter the container?
    e) Does the pressure inside the container increase or decrease during this 5 second _ period?

    2. Relevant equations

    a) P=nkT where n=N/V, N is # of molecules (using avagadro's number) and V is in m^3. k is Boltzman's constant, T is temp. in Kelvin.
    b) Vrms = sqrt((3kT)/m) where k and T are as above and m is mass.
    c) .25nAv[avg] = number of molecules crossing area A per second. n is as above and v[avg] is average velocity.


    3. The attempt at a solution

    For part one I calculated 102,716 N/m^2, and for part two I calculated 1.69 * 10^-9 m/s. For part c) I got 5.68*10^8 He atoms escape, and part e, that pressure inside decreases. I am not sure if these answers are right so far, especially the Vrms, as it is so small. The value I got for the pressure 102,716 N/m^2 came directly from the pressure equation above, which I thought should have resulted in atm rather than Pa (N/m^2), but seeing that 102,716 was of the order of magnitude of Pa in this case, I assumed that it was, and left it at that. I wanted to check and make sure I haven't made some grave error. I am also not quite sure how to approach d). Could anyone help? Thanks in advance.
     
    Last edited: Dec 11, 2009
  2. jcsd
  3. Dec 11, 2009 #2
    You need to enter the mass in kilograms in the formula for the rms speed.
     
  4. Dec 11, 2009 #3
    Which I did... .004Kg...
     
  5. Dec 11, 2009 #4
    You need to mass of an helium atom, which is approximately 4 u, where u is the atomic mass unit:

    u = 1.660539*10^(-27) kg
     
  6. Dec 11, 2009 #5
    Oh, so you're saying the mass in the equation is of one Helium atom, not the total mass in the container. OH; I used the mass in the container, and seeing that there was one mole of gas, the mass would have been 4g. But I see now. Thank you!
     
  7. Dec 12, 2009 #6

    ideasrule

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    Pressure is indeed approximately 102.7 N/m^2. It isn't it atmospheres because if you do dimensional analysis on PV=nkT with the units that you used, you'll get P in N/m^2.

    For part b, you used v=sqrt(3kT/m), where m is the mass of one helium atom. So you need to find the mass of a single helium atom.

    For d), can you calculate the average (not rms) speed of the air molecules? If so, you can apply the equation N=0.25nAv again. Remember that in the ideal gas approximation, gas molecules don't interact, so the gas molecules exiting the hole has no effect on the gas molecules entering it.
     
  8. Dec 12, 2009 #7

    ehild

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    Gold Member

    This is definitely wrong.

    ehild
     
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