Physical Chemistry math not working out

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Homework Help Overview

The discussion revolves around deriving an equation in physical chemistry, specifically relating to the velocity distribution of particles. The original poster is attempting to derive equation b from equation a using equation c, which involves integrating a probability function.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster outlines their approach to the derivation, including their expressions for constants and the integral involved. They express confusion over the algebraic manipulation required to reach the desired form of the equation.
  • Some participants question the correctness of the constants used in the original poster's equations and suggest potential errors in the integral evaluation.
  • Others raise points about the algebraic steps taken, particularly regarding the treatment of constants and the integration process.

Discussion Status

Participants are actively engaging with the original poster's work, offering corrections and suggestions for re-evaluating certain expressions. There is recognition of typographical errors and algebraic challenges, but no consensus has been reached on the correct path forward.

Contextual Notes

The original poster is working under time constraints and has expressed a desire for guidance within a limited timeframe. There is an acknowledgment of the need to clarify mathematical expressions and constants used in the derivation.

Roo2
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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

Homework Statement


Derive equation b from equation a using equation c

Homework 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

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?
 
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Roo2 said:
c1 = {[m/(2*pi*Kb*T)]^(3/2)}/(4*pi)

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


Roo2 said:
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"

I'm looking at it right now, and it says that it's equal to 1/(2a2).
 
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.

Roo2 said:
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

Homework Statement


Derive equation b from equation a using equation c

Homework Equations



\textrm{(a)}~~~P(v) = \left[\frac{m}{2\pi k_B T}\right]^{3/2} \exp\left(-\frac{mv^2}{2k_B T}\right) 4 \pi v^2\textrm{(b)}~~~\langle v \rangle = \sqrt{\frac{8 k_B T}{\pi m}}

\textrm{(c)}~~~\langle v \rangle \equiv \int_0^\infty v P(v)\, dv

The Attempt at a Solution



P(v) = (C_1 v^2) e^{-C_2 v^2}

C_1 \equiv 4\pi \left[\frac{m}{2\pi k_B T}\right]^{3/2}

C_2 \equiv \frac{m}{2k_B T}

Therefore,

\int_0^\infty vP(v)\,dv = C_1 \int_0^\infty v^3 e^{-C_2 v^2}\,dv
 
Last edited:
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.
 
Roo2 said:
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.

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:

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