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Homework Help: Standard deviation of V_x of He gas

  1. Dec 26, 2017 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    ## \frac 1 2 m<v_x^2> =\frac 1 2 k_BT ,

    \\ \sqrt{ <v_x^2>} = 556~ m/s ## So, I guess that the standard deviation should be less than rms speed.

    So, the option is (a)

    ## \left< ax + b \right> = a\left<x\right> + b ##

    So, ## \left< v_x^2 - <v_x^2> \right> = \left< v_x^2\right> - \left< v_x^2\right> = 0 ##

    Is this correct?
     
    Last edited: Dec 26, 2017
  2. jcsd
  3. Dec 26, 2017 #2

    Charles Link

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    I think it your last statement, you are attempting to write ## \sigma^2=E(X^2)-(EX)^2 ##. If that is the case, and you are saying ## EX=0 ## , yes, it is correct. Please put in the mass and temperature so that we can check your arithmetic. For the final answer, the standard deviation, ## \sigma_{v_x}=\sqrt{E(v_x)^2}=v_{x \, rms} ##. Also, the statement of the problem is incomplete. The available answers did not show up.
     
    Last edited: Dec 26, 2017
  4. Dec 26, 2017 #3
    The question is not showing in the OP. So, posting it again.
    upload_2017-12-26_22-4-13.png
    I didn't understand what you mean by the above line. Please explain the symbol EX and so on.
     
  5. Dec 26, 2017 #4

    Charles Link

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    In probability theory, the ## E ## means expectancy. ## EX ## is the same thing as ## <X> ##.
     
  6. Dec 26, 2017 #5
    In the above statement, as ## < v_x ^2> ## is constant, I am taking it out of ## \left< v_x^2 - <v_x^2> \right>## and hence I got ## \left< v_x^2 - <v_x^2> \right> = \left< v_x^2\right> - \left< v_x^2\right> = 0 ##.
    But, this way I will always get 0 standard deviation. So, something is wrong with it. Right?
     
  7. Dec 26, 2017 #6

    Charles Link

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    The equation correctly reads ## \sigma^2_{v_x}=<v_x^2>-<v_x>^2 ##. ## \\ ## Editing: This comes from ## \sigma^2_{v_x}=<(v_x-<v_x>)^2> ##. And I think your arithmetic in the OP is incorrect.
     
    Last edited: Dec 26, 2017
  8. Dec 26, 2017 #7
    ## \sqrt{ <v_x^2>} = \sqrt{ \frac { k_BT = 4.14 * 10^{-21}}{m=4* m_p = 6.69 * 10^{-27}}}= 786~ m/s ##

    Is this correct?
     
  9. Dec 26, 2017 #8

    Charles Link

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    Yes. Very good. Round that off and (d) becomes the correct answer. See also my edited additions to post 6.
     
  10. Dec 26, 2017 #9
    I reloaded the page three times. Still, no edition to post #6 is shown.
    What I get in post # 6 is:
     
  11. Dec 26, 2017 #10

    Charles Link

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    I didn't use the word "edit". Let me put it in there for clarity. After I wrote the original post, I came back a couple of minutes later and added the last two sentences.
     
  12. Dec 26, 2017 #11
    Thanks.
     
  13. Dec 26, 2017 #12
    Now I understood.

    ## \sigma_{v_x} = \sqrt{ \left< (v_x - \left <v_x \right >)^2 \right> = \left < v_x^2 \right > – 2 \left < v_x \right >^2 + \left < v_x \right >^2 = \left < v_x^2 \right > – \left < v_x \right >^2 }##

    Since , ## \left < v_x \right > = 0 ##

    ## \sigma_{v_x} = v_{rms} ##
     
  14. Dec 26, 2017 #13

    Charles Link

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    Very good. :) (You need a parenthesis { } around v_x in the subscript of ## \sigma_{v_x} ##. Then your Latex will work.)
     
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