Statistical Physics

1.3 Kinetic Theory of Gas Molecules 19walls are perfectly flat and are so hard that a collision is a perfect elasticreflection.Let us assume that the gas is contained in a rectangular box whose threesides have lengths L x , L y ,andL z as shown in Fig. 1.5. We calculate thecontribution of the ith molecule to the pressure on the wall that is perpendicularto the x-axis. This molecule has a velocity whose x-component is v ix .Ina collision, this molecule gives an impulse 2mv ix to the wall. Such a collisionoccurs once in a time interval ∆t =2L x /v ix , so the impulse per unit time,that is, the time average of the force on the wall, is¯f i =2mv ix= mv2 ix. (1.30)2L x /v ix L xSummation of this force over all of the molecules gives the force on the wall,and when divided by the area S = L y L z it gives the pressure P :P =N∑i=1= 13V¯f N iS = ∑i=1mv 2 ixVN∑m(vix 2 + viy 2 + viz) 2 = 2 13 Vi=1N∑i=112 mv2 i= 2 13 V E = N Vβ . (1.31)In the third equality in (1.31), we have used the fact that the motion of the gasmust be isotropic, so that ∑ i v2 ix = ∑ i v2 iy = ∑ i v2 iz . In the final equality, wehave noticed the fact that the sum of the kinetic energies is the total energy E,and is equal to 3N/2β according to (1.28).An ideal gas is known to satisfy the Boyle–Charles equation of state PV =Nk B T . Combining this with the present result gives β =1/k B T , as mentionedbefore. However, you should remember that we have not defined the absolutetemperature yet.Fig. 1.5. Collisions of molecules with the walls are the origin of the pressure ofa gas. In a collision with the right wall, a molecule gives an impulse 2mv x to thewall