Boltzmann Distribution: Solving 1D Ideal Gas Homework

[Tikkelsen]

Homework Statement

I have to find the Boltzmann ditribution of a 1 dimensional ideal gas.
The answer is given as:

$\frac{dn}{n}=\sqrt{\frac{m}{2piKT}}e^{(\frac{-mc^2}{2KT})}$

For the second part I have to find the mean kinetic energy.

2. Homework Equations / Attempt

For part 1:
I know how to work out the Boltzmann distribution for a 3D and 2D gas. However, for a 1D gas, I can't figure out what the constant has to be. I know the form to solve it is:

$\int_0^\infty C e^{(\frac{-mc^2}{2KT})} dv = 1$ (1)

Where $C$ is a constant.
However, when I do this and solve for $C$, I get a factor of 2 in front of my equation. Is there something wrong in my logic here? Am I meant to use a factor infront of my $C$

For the second part, I know that I need to get a $v^2$ infront of the exponential, but I cannot figure out how to do this for the 1D case and even for the 3D case.

Any help would be much appreciated and please tell me if I need to clarify anything.

 

[Tikkelsen]

New progress:

Is it right to say that equation (1) should actually be equal to a half since the probability of the particle being on the positive side of the line is actually a hal and not 1?

 

[Orodruin]

1. What do you have against negative velocities? Half the phase space consists of negative velocities...

2. How are thermal averages defined? What is the expression for kinetic energy?

 

[Tikkelsen]

1. It's the speed distribution, so I am looking for the absolute value of the velocity. This crashes my idea of it being equal to a half I realize.

2. $Average KE = \frac{1}{2}m\bar{v^2}$
I now realize that the equation for $\bar{v^2}$ would be:

$\bar{v^2}=\int_0^\infty v^2\frac{dn}{n} dv$

Then equate to 1 and solve.

 

[Tikkelsen]

Looking at my OP, I can see that what I said can be a bit confusing, so here is the question.