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Homework Help: Calculating the root mean square speed from pressure and density.

  1. Jan 7, 2014 #1
    1. The problem statement, all variables and given/known data

    A tyre contains gas at a pressure of 150 kPa. If the gas has a density of 2.0 kg m-3, find the root mean square speed of the molecules.

    2. Relevant equations

    These are the equations I believe to be relevant:

    [tex]c_{rms} = \frac{\sqrt{<c^2>}}{N}[/tex]

    [tex]pV = \frac{1}{3}Nm<c^2>[/tex]

    [tex]p = \frac{1}{3}ρ<c^2>[/tex]

    3. The attempt at a solution

    [tex]\frac{3p}{ρ} = <c^2>[/tex]
    [tex]\frac{3 * 150000}{2} = <c^2> = 225000ms^{-1}[/tex]

    But I'm not sure how to work out N as I don't any volume or temperature. I'm not quite sure how they get their answer in the back of the book which is 474 ms-1 (which is the square root of 225000) meaning that N = 1? How can that be?

    Thank you for reading!
    Last edited: Jan 7, 2014
  2. jcsd
  3. Jan 7, 2014 #2


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    If c^2 = 225000, then the units can't be m/s
  4. Jan 8, 2014 #3
    Sorry yes <c^2> should be in m^2s^-2 shouldn't it?
  5. Jan 8, 2014 #4


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    Are you sure there should be an ##N## here?
  6. Jan 8, 2014 #5
    I'm sure, atleast that's how they have quoted it in my text book:

  7. Jan 8, 2014 #6


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    How does your book define <c^2>?

  8. Jan 8, 2014 #7
    It defines it as:

    [tex]<c^2> = \frac{{c_1}^2 + {c_2}^2 + {c_3}^2 + ....{c_N}^2}{N} = \frac{{c_i}^2}{N} [/tex]

    I'm not exactly sure what the right-hand most part means with the subscript i.
  9. Jan 8, 2014 #8


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    It should be

    [tex]<c^2> = \frac{{c_1}^2 + {c_2}^2 + {c_3}^2 + ....{c_N}^2}{N} =\frac{\sum{{c_i}^2}}{N} [/tex]

    i is the summation index, and √<c2> itself is the rms speed. No need to divide it by N. [tex]\frac{3p}{ρ} = <c^2>[/tex]

  10. Jan 8, 2014 #9


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    yeah, I think what ehild said is right. The equation
    [tex]c_{rms} = \frac{\sqrt{<c^2>}}{N}[/tex]
    is most likely a mistake in the book. It doesn't make sense, if you think about it. If you had a bunch of atoms all moving at the same speed, then this equation would give an rms value that is smaller for a larger population of these identical atoms. Which doesn't make sense at all.
  11. Jan 8, 2014 #10

    Okay, I've got it now, thanks for the explanation.
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