Physical Chemistry math not working out

AI Thread Summary
The discussion revolves around a user struggling to derive equation b from equation a using equation c in a physical chemistry assignment. The user attempted to integrate the probability function P(v) but encountered difficulties in simplifying the constants to match the desired form. Other participants pointed out potential mistakes in the user's algebra and suggested corrections to the constants used in the derivation. The user acknowledged these suggestions and expressed a commitment to learning LaTeX for clearer mathematical communication. Overall, the conversation highlights the challenges of mathematical derivation in physical chemistry and the collaborative effort to resolve them.
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
 
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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.
 
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