Boltzmann Distribution: Solving 1D Ideal Gas Homework

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

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