Question

Need some help finishing a problem, and also check if right.
http://assets.openstudy.com/updates/attachments/51a4eb5ce4b0aa1ad887f665-frostbite-1369764604391-unavngivet.png

Answer 1

Problem is found in the attackment

Answer 2

@amistre64

Answer 3

And I like to do it by hand and not using a solving tool.

Answer 4

@abb0t

Answer 5

Might fast need to say that:
\[c ^{*}=v=\sqrt{\frac{ 2RT }{ M }}\]

Answer 6

@chmvijay

Answer 7

what u want in this u want to derive this

Answer 8

The last expression is the derived Maxwell distribution, now that need to equal 0 in order to find the maximum (remember the graph for the distribution). Just doing so that gets me nowhere.

Answer 9

I am bad at maths dude i am sorry :(

Answer 10

Darn, no problem :/

Answer 11

What I want to show:

Answer 12

@satellite73

Answer 13

@Hunus

Answer 14

Got any idea how to solve that dang thing?

Answer 15

And we ignore any negative solutions as we can't talk about a negative velocity :)

Answer 16

Get rid of the e^((-Mv^2)/2RT) by dividing both sides by it

Answer 17

Why did I not come to think of that, good idea.

Answer 18

And then you will have
\[\Large \frac{2Mv \sqrt{\frac{2M}{RT}}}{\sqrt{\pi}RT}=\frac{M^2v^3 \sqrt{\frac{2M}{RT}}}{\sqrt{\pi}R^2T^2}\]

Answer 19

It will all cancel you and you will come to your solution

Answer 20

2 sec, need to follow with pen just so I don't miss anything.

Answer 21

Awesome, it sure do. Thanks just saved my butt :)

Answer 22

Yup :)

Answer 23

How are you in PDE's then?

Answer 24

I'm alright with PDE's

Answer 25

familiar with the diffusion equation then? similar to the heat equation

Answer 26

Yup

Answer 27

Perfect think you can help me a bit with that too. I haven't leaned how to solve PDEs except over a set.

Answer 28

I'll try :)

Answer 29

I'll upload that one in the physics section :)