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Physical Chemistry math not working out

  1. Sep 14, 2010 #1
    I put off an assignment until the last minute and on the very last question it came back to bite me in the butt. I'm supposed to derive an equation from another equation, and the math is not working out for me. If there's anyone still up and reading this and that can point me in the right direction within the next 9 hours, I'd appreciate it :D

    1. The problem statement, all variables and given/known data
    Derive equation b from equation a using equation c

    2. Relevant equations

    equation a: P(v) = [m/(2*pi*Kb*T)]^(3/2) * e^-[(mv^2)/(2*Kb*T)] * 4*pi*v^2

    equation b: <v> = sqrt[(8*Kb*T)/(pi*m)]

    equation c: <v> = integral (from 0 to infiniti) v*P(v) dv

    3. The attempt at a solution

    P(v) = (c1*v^2) * e^-(c2*v^2)

    c1 = {[m/(2*pi*Kb*T)]^(3/2)}/(4*pi)

    c2 = m/(2*Kb*T)

    Therefore, int(v*P(v)dv) = c1 * int[(v^3)*e^(-c2*v^2)]

    According to wiki, http://en.wikipedia.org/wiki/Lists_of_integrals (which our prof said to use):

    integral (x^3) e^-ax^2 = 1/a^2
    The integral is the 5th one down in the section "Definite integrals lacking closed-form antiderivatives"

    Therefore, my integral evaluates to c1/(c2^2)

    However, when recompile my constants, they are not in a form that is very easily rearrangeable to the desired form. Furthermore, I tried plugging in values to both formulas and obtained different results. Could someone please tell me where I made a mistake?
     
  2. jcsd
  3. Sep 15, 2010 #2

    cepheid

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    Welcome to PF!

    Shouldn't the first factor be multiplied by 4pi and not divided by it?


    I'm looking at it right now, and it says that it's equal to 1/(2a2).
     
  4. Sep 15, 2010 #3

    cepheid

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    I took the liberty of reformatting your post. Here at Physics Forums, we use the LaTeX typesetting system to create mathematical equations, which is very powerful. It might be useful for you to learn. There's a thread that goes over some of the basics, but for now, you can click on the equation graphics that are generated to see the LaTeX code that was used to generate them.

     
    Last edited: Sep 15, 2010
  5. Sep 15, 2010 #4
    Thank you for the notice! I tried to do two steps in one and spliced it. I still can't get the 8 to come through, nor can I isolate a square root. Also, you are right about the integral; it was a typo on my part and I used the correct integral in my work. In the interest of saving time, here's is my simplified work so far (sorry, no latex but it simplified down a lot):

    Z = m/kBT

    <v> = c1/2*c2 = sqrt(Z/2pi) / Z

    The 8 stubbornly refuses to show up. Am I forgetting some basic algebra?

    P.S. I'll learn LaTeX for future posts; I'm still trying to crank out this problem though.
     
  6. Sep 15, 2010 #5

    cepheid

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    Shouldn't that be: c1 / 2c22
    (meaning that your c2 is supposed to be squared)?

    EDIT: also, from your definition of Z, shouldn't it be that:

    c1 = 4pi * (Z/2pi)3/2

    EDIT: and,

    c2 = Z/2

    EDIT: Yes, I got the algebra to work out for me using these expressions.
     
    Last edited: Sep 15, 2010
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