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Thermo: Lab vacuum

  1. Sep 10, 2014 #1
    The best laboratory vacuum has a pressure of about 1.00 x 10-18 atm, or 1.01 x 10-13 Pa. How
    many gas molecules are there per cubic centimeter in such a vacuum at 293K?

    PV = nkT
    P/kT = n/V = N (# of molecules per cm^3)

    Since V is typically in units of m^3 or liters, should I make my volume .01V to account for the cm^3?
  2. jcsd
  3. Sep 10, 2014 #2


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    When making unit conversions, use the trick of multiplying by "1" to help you do the unit conversion.

    So to convert from cm^3 to m^3 multiply by "1" like this:

    [tex]1cm^3 * \frac{(1m)^3}{(100cm)^3}[/tex]

    The cm^3 unit terms in the numerator and denominator cancel (just like numbers cancel if they are identical in the numerator and denominator of a fraction), and you are left with what in units of m^3? Hint -- It's not 0.01 ... :smile:
  4. Sep 10, 2014 #3


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    Staff: Mentor

  5. Sep 10, 2014 #4


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    You are using the right equation. People are conditioned to see n as number of moles and N as number of molecules, so better use N.

    Now try to work in a consistent set of units. SI units is what every reasonable person would use.
    p in Pascal, Pa with 1Pa = 1 N/m[sup2[/sup]. Conversion: 1 atm = 101325 N/m[sup2[/sup].
    k Boltzmann constant, J/K 1.3806488 10-23 J/K
    T 293 K

    The equation gives you n/V in molecules/m3.
    Since the exercise asks for molecules/cm3, all you have to do is multiply by

    ( molecules/cm3 ) / (molecules/m3 ) = cm3 / m3 = (cm / m3) = (10-2)3

    [I see berkeman beat me to it, well, good for you!

    But I don't agree with him (/her?): pV = NkT is just fine. It's the same as pV = nRT since n = N/NA and R = kB*NA ]
  6. Sep 10, 2014 #5
    Ok, sweet. Thank you!
  7. Sep 10, 2014 #6


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    Staff: Mentor

    Great post BvU. Thank you.
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