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Gas Laws -- why calculate the mean square speed at 273K?

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  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    Why is the formula ##p = \frac{1}{3}\rho<c^2>## used to calculate the mean square speed at 273K?
    Why 273K?
     
  2. jcsd
  3. Mar 21, 2017 #2

    BvU

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    It works at any temperature. Could you give a bit more context ? Is this a section in your textbook you have a question about or is it in an exercise ?
     
  4. Mar 21, 2017 #3
    The density of a gas at a temperature of ##273~K## and a pressure of ##1.02*10^5~ Pa## is ##0.9kgm^{-3}##. It may be asssumed to be an ideal gas.
    Calculate rms speed at 273K and 546K.

    For 273, we were supposed to use the given equation and for second we were supposed to used the ratio of mean square speed and temperature.
    My thinking is the equation only applies for those temperature for which we are using the pressure values.
     
  5. Mar 21, 2017 #4

    Borek

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    You are given pressure and density at 273 K and asked to calculate rms at exactly this temperature, I don't see where is the problem?
     
  6. Mar 21, 2017 #5
    The confusion I had was why is that formula only applicable for the 273K. That's to say why I can't just put that value down for 546K? Later I realized, the pressure inserted in those equations are the indicator of what temperature should the RMS value be for.
     
  7. Mar 22, 2017 #6

    BvU

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    We're still missing context. Please be a bit more complete in formulating problem statements. This way we spend way too much time dragging out what is troubling you.
    Missing:
    2. Relevant equations

    3. The attempt at a solution
    Accidentally ? Or did you erase them ? That's a no-no in PF !

    And anyway, 1. should probably have been
    Any word or sentence missing from this full problem statement ? If no, then a list of all variables is good to have at hand. For you too.
    What given equation ?

    Did you do anything at all with the information that the gas can be considered an ideal gas ?

    Where did the
    come from ? From the text book or from the problem statement ? (or perhaps from the solution manual :rolleyes: ?)
     
  8. Mar 22, 2017 #7
    1. No words or sentences are missing.
    2. The given equation is ##p = \frac{1}{3}\rho<c^2>##
    3. By telling us that it could be considered an ideal gas, permission was given to use the ideal gas law and the provided equation.
    4. The formula was supposed to be derived by using the equation ##pV = \frac{1}{3}Nm<c^2>## and ##\rho = \frac{Nm}{V}##

    Btw my confusion has been sorted out. Thanks for the help though
     
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