Modern Thermodynamics
- Chapter 1
32
in any other direction. Let us assume the average speed of the gas molecules is vavg. We denote its x, y
and z components of the by vxavg, vyavg, and vzavg. Thus
vavg
2
= vxavg
2
+ vyavg
2
+ vzavg
2
(1.6.1)
Because all directions are equivalent, we must have,
vxavg
2
= vyavg
2
= vzavg
2
= (vavg
2
/ 3)
(1.6.2)
The following quantities are necessary for obtaining the expression for pressure:
N = amount of gas in moles
V = gas volume
M = Molar mass of the gas
m = mass of a single molecule = M/NA
n = number of molecules per unit volume = NNA/V
NA= Avogadro number (1.6.3)
Now we can calculate the pressure by considering the molecular collisions with the wall. In doing so we
will approximate the random motion of molecules with molecules moving with an average speed vavg. (A
rigorous derivation gives the same result.) Consider a layer of a gas, of thickness ∆x, close to the wall of
the container (see Fig. 1.8). When a molecule collides with the wall, which we assume is perpendicular to
the x-axis, the change in momentum of the molecule in the x-direction equals 2mvxavg. In the layer of
thickness ∆x and area A, because of the randomness of molecular motion, about half the molecules will
be moving towards the wall, the rest will be moving away from the wall. Hence, in a time ∆t=(∆x/vx)
about half the molecules in the layer will collide with the wall.
The number of molecules in the layer = (∆xA) n
The number of molecules colliding with the walls = (∆xA) n/2
Since each collision imparts a momentum 2mvxavg, in a time ∆t, the total momentum imparted to the wall
= 2mvxavg (∆xA) n/2.
Thus the average force F on the wall of area A is:
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