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ID:(1613, 0)

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So the average moment\\n\\n

$2,m,v_x$

\\n\\nwhich is transferred to the wall in dt time by\\n\\n

$\displaystyle\frac{1}{2}c_n,S,v_xdt$

\\n\\nparticles gives a moment equal to:\\n\\n

$\langle dp_x\rangle=\langle 2,m,v_x\displaystyle\frac{1}{2}c_n,S,v_x dt\rangle$

Since the mass, the section and the time are independent of the average, it has to be

The factor 1/2 is due to the fact that half of the particles move towards the wall while the other half move away.

ID:(3934, 0)

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In a time dt all the particles that are at a distance equal to or less than \ langle v_x \ rangle dt will hit the wall. If the surface is S and the concentration of particles c_n , we will have that the number of particles is

The factor 1/2 is due to the fact that half of the particles move towards the wall while the other half move away.

ID:(3935, 0)

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Particles with f degrees of freedom

ID:(1959, 0)

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Particles with five degrees of freedom

ID:(1958, 0)

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Partículas con tres grados de libertad

ID:(1957, 0)

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On the other hand the pressure is the force F_x per area S \\n\\n

$p=\displaystyle\frac{\langle F_x\rangle}{S}$

\\n\\nand the force is the variation of the moment in time\\n\\n

$\langle F_x\rangle=\displaystyle\frac{\langle dp_x\rangle}{dt}$

\\n\\nwe have with\\n\\n

$\langle dp_x\rangle=c_nmS\langle v^2\rangle dt$

\\n\\nand\\n\\n

$\langle\epsilon\rangle=\displaystyle\frac{1}{2}m\langle v^2\rangle$

than

ID:(3937, 0)

Equation

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Como la suma de los cuadrados de la velocidad en cada componente es igual al cuadrado de la magnitud\\n\\n

$v^2=v_x^2+v_y^2+v_z^2$

y como por simetría todas las componentes tiene que ser iguales

ID:(824, 0)