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## New answers tagged kinetic-theory

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Perhaps a better way is to redefine your energy equation to: $$m \ c^{2} \left( \gamma_{rms} - 1 \right) = \frac{3}{2} k_{B} \ T$$ where $\gamma_{rms}$ is the Lorentz factor of the rms speed of the relativistic gas. This can be shown to be: $$\left( V_{rms} \right)^{2} = c^2 \left[ 1 - \left( \frac{3 \ k_{B} \ T}{2 \ m \ c^{2}} + 1 \right)^{-2} \right] ... 1 Introduction Let us define the density of particles of species s in a volume element, d\mathbf{x} \ d\mathbf{v}, at a fixed time, t, centered at (x, v) as the quantity f_{s}(\mathbf{x},\mathbf{v},t). I assume this function is non-negative, contains a finite amount of matter, and it exists in the space of positive times and \mathbb{R}^{3} and ... 0 Kinetics are focused on the rate and mechanism of chemical processes, so you are definitely right to say that you gain a lot of insight about mechanism from kinetics. Many kinetic theory make extensive use of statistical thermodynamics methods, and that's why you perceive a resemblance. However, keep in mind that in kinetics the system is not in ... 1 Benard convection involves fluid flow on a horizontal plane heated from below. - Scholarpedia (peer-reviewed open access online encyclopedia) Ever since antiquity, geometers have known that only three regular polygons can tile a plane, without overlap or extra space. These are the equilateral triangle, the square, and the hexagon. Interference patterns on ... 2 Assuming the oxygen and lead vapour can be treated as ideal gases then yes the partial pressures of the two will be the same. This happens because the pressure is proportional to the momentum of the atoms multipled by their velocity (shout if you want me to explain why this is):$$\begin{align} P &\propto pv \\ &\propto mv \times v \\ ...

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The second theorem of König states that the kinetic energy in a reference frame $E_c$ is related to the kinetic energy relatively to the center of mass $E^*_c$ in the following way: $$E_c=E_c^*+\frac12Mv_G^2,$$ where $v_G$ is the velocity of the center of mass in the reference frame and $M$ is the total mass. To answer your question, remark that the energy ...

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