Maxwell-Boltzmann Spherical Cap

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1. Apr 26, 2015

J_M_R

1. The problem statement, all variables and given/known data

An ideal gas satisfying the Maxwell-Boltzmann distribution is leaking from a container of the volume V through a circular hole of area A'. The gas is kept in the container under pressure P and temperature T. The initial number density (concentration) is given by n0=N/V.

Find the number of particles leaking from the container per unit of time and reaching a spherical cap with the radius R and the height h, h<R. The centre of the sphere is positioned exactly at the centre of the hole and the base of the cap is in the plane parallel to one of the container's wall.

2. Relevant equations

Maxwell-Boltzmann Distribution: f(vi) = √(m/(2πkT))*(e^[(mvi^2)/(2kT)])

where vi stands for either vx, vy, or vz.

k - Boltzmann constant.

3. The attempt at a solution

When looking at the hole of the box, if the hole has length vxdt, then the portion of particles with velocity vx getting out of the hole (of area A') in time dt is: n0A'vxdt = dN(vx).
Because the volume covered by particles in a moment of time, dt = A'vxdt. Where no=N/V

I have completed the work with a disk instead of a spherical cap at a distance from the box, removing the y and z velocity components and replacing them with polar coordinates for velocity and think that to solve this problem spherical coordinates are required to replace the x, y and z velocity components associated with the cap.

I have got to this point below but do not know how to derive the limits of the triple integral:

N = no * A' * dt * (m/(2*pi*k*T))^(3/2) ∫ (v * d^3v * e^-(m*v^2/(2*k*T)))

The upper limit of integration is v2 and the lower limit is v1 which are 3-dimensional velocities.

where v^2 = vx^2 + vy^2 + vz^2

and d^3v = dvx dvy dvz = v^2 sinӨdӨdφ

I understand that I need to use a triple integral but I am unsure how to obtain the limits using spherical coordinates, if anybody is able to give me some hints that would be much appreciated.

Last edited: Apr 26, 2015
2. Apr 29, 2015

Staff: Mentor

There is a dv missing in the last equation.

If you plug in your expression for d3 into the integral, you get two integrals that are easy to solve and one where you have to integrate a function of the type of $v^3 e^{-v^2}$, which can be integrated with standard methods.

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