User:
Momentum delivered to the wall
Equation
So the average moment\\n\\n
$2,m,v_x$
\\n\\nwhich is transferred to the wall in
$\displaystyle\frac{1}{2}c_n,S,v_xdt$
\\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
| $\langle dp_x\rangle= m c_nS\langle v^2\rangle dt$ |
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)
Number of particles that reach the wall
Equation
In a time
| $dN_x=\displaystyle\frac{1}{2}c_n\langle v_x\rangle dt S$ |
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)
Pressure
Equation
On the other hand the pressure is the force
$p=\displaystyle\frac{\langle F_x\rangle}{S}$
\\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
| $p=\displaystyle\frac{2}{3}c_n\langle\epsilon\rangle$ |
ID:(3937, 0)
Average speed of the particles
Equation
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
| $\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle$ |
ID:(824, 0)