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

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SUMMARY

The discussion centers on calculating the total kinetic energy of gas molecules striking a vessel's wall in an ideal gas scenario. The number of collisions per unit area per unit time is derived using the normalized Maxwellian distribution, leading to the formula for impacts: ν = (N/V) √(T/(2πm)). The total kinetic energy is expressed as E = (N/V) √(2T³/(mπ)), with the average velocity needing clarification. The mean-square velocity is proposed to be 4T/m, but the derivation of this value remains unclear to the participants.

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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:
<br /> \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<br />

The number of collisions per unit area per unit time is then just obtained by multiplying dN_v by the volume of a cylinder of unit base area and height v_z (we imagine that an element of surface area of the vessel wall is perpendicular to some coordinate system's z-axis):
<br /> \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

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 0 to \infty, because negative velocities would mean that a molecule is going away from the wall:
<br /> \nu = \frac{N}{V} \sqrt{\frac{T}{2\pi m}}<br />

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:
E = \nu \frac{1}{2} m \overline{v^2}.

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

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

The x and y components of velocity would both give \frac{T}{m}, of course assuming Maxwellian distribution. That leaves me with 2\frac{T}{m}, which I have no idea where to take from.
 
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