Ideal gas: total kinetic energy of molecules striking a vessel's wall

by mSSM
Tags: energy, ideal, kinetic, molecules, striking, vessel, wall
mSSM is offline
Apr28-12, 05:21 PM
P: 24
Molecules in an ideal gas contained in a vessel are striking the vessels wall. I am trying to find the total kinetic energy of gas molecules striking a unit area of that well per unit time.

The number of collisions per unit area per unit time is derived from the normalized Maxwellian distribution of molecules per unit volume:
\mathrm{d}N_v = \frac{N}{V} \frac{m^{3/2}}{(2\pi T)^{3/2}} \exp\left[ -m(v_x^2 + v_y^2 + v_z^2)/2T \right] \mathrm{d}v_x \mathrm{d}v_y \mathrm{d}v_z

The number of collisions per unit area per unit time is then just obtained by multiplying [itex]dN_v[/itex] by the volume of a cylinder of unit base area and height [itex]v_z[/itex] (we imagine that an element of surface area of the vessel wall is perpendicular to some coordinate system's z-axis):
\mathrm{d}\nu_{\vec{v}} = \frac{N}{V} \left(\frac{m}{2\pi T}\right)^{3/2} \exp\left[ -m(v_x^2 + v_y^2 + v_z^2)/2T \right] v_z \mathrm{d}v_x \mathrm{d}v_y \mathrm{d}v_z[/tex]

The total number of impacts of gas molecules per unit area per unit time on the vessel is then just obtained by integrating over all velocities; the z-component of velocity is integrated only from [itex]0[/itex] to [itex]\infty[/itex], because negative velocities would mean that a molecule is going away from the wall:
\nu = \frac{N}{V} \sqrt{\frac{T}{2\pi m}}

Now, from here I actually want to calculate the total kinetic energy of the gas molecules striking unit area of the wall per unit time. I thought this would just be:
[tex]E = \nu \frac{1}{2} m \overline{v^2}[/tex].

However, I am not sure how to properly calculate the average velocity, since I have to take care of the integration of the z-component.

I know the result is:
[tex] E = \frac{N}{V} \sqrt{\frac{2T^3}{m\pi}}[/tex],
which, if my above idea is correct, would just mean that my mean-square velocity would have to be:
[tex]\overline{v^2} = 4 \frac{T}{m}[/tex]

However, I have no idea how I am supposed to obtain that.

The x and y components of velocity would both give [itex]\frac{T}{m}[/itex], of course assuming Maxwellian distribution. That leaves me with [itex]2\frac{T}{m}[/itex], which I have no idea where to take from.
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mSSM is offline
Apr29-12, 11:24 AM
P: 24
I actually found something similar to my problem here:

I think that would give me a velocity distribution close to what I need; I just don't understand how to get there... Any ideas?

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