Calculating Pressure in a 2D Gas: Find Expression for <v>

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SUMMARY

The discussion focuses on deriving an expression for the pressure of a two-dimensional gas in terms of velocity . The velocity distribution is defined as f_{(v^2)} = (\frac{a}{\pi}) e^{-av^2}, where a = \frac{m}{2kT}. The speed distribution is obtained by integrating over all angles, resulting in w_{(v)} = 2a v e^{-av^2}. The pressure is calculated using the change in momentum of molecules, leading to the expression dP = \frac{n \frac{a}{\pi}ve^{-av^2} dv dθ * (L v cosθ dt) * (2mv cosθ)}{L dt}, which is then integrated over the specified limits.

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  • Basic principles of momentum and pressure in physics
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Homework Statement



Find an expression for the pressure of a 2-D gas in terms of <v>.

Homework Equations


The Attempt at a Solution



In 2D, velocity distribution is:

f_{(v^2)} = (\frac{a}{\pi}) e^{-av^2} , a = \frac{m}{2kT}

Integrate all possible angles to get speed distribution and normalize to get speed distribution:

w_{(v)} = 2a v e^{-av^2}
Number of molecules traveling between speed v and v+dv, at angles between θ and θ + dθ per unit area = n \frac {dθ}{2\pi}w_{(v)} dv
= n \frac{a}{\pi}ve^{-av^2} dv dθ

To find pressure, take the above expression * change in momentum of one molecule / (L dt)

dP = \frac{n \frac{a}{\pi}ve^{-av^2} dv dθ * (L v cosθ dt) * (2mv cosθ) }{L dt}

Then we integrate v from 0 to ∞, θ from 0 to ∏/2.

Is that right?
 
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That looks right to me.
 

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