Page 21 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:35 9in x 6in b985-ch01 Elementary Physical Chemistry

6 Elementary Physical Chemistry

theoppositewall is L, the molecule is reflected in the opposite direction,
having a velocity of −u x . and a change of velocity of 2u x . The molecule will
make u x/2L collisions per unit time with the shaded wall. Accordingly, the
change in momentum per molecule per unit time at the shaded wall will be
2
(2mu x) × (u x/2L)= mu /L.For N molecules, the change in momentum
x
2
per unit time will be Nm u /L where stands for average.
x
In classical mechanics, the momentum change on an area represents
the force exerted on that area. Denoting the force as f we can write
2
f = Nm u /L as the force exerted on the shaded wall. Pressure is
x
2
force per unit area, P = f/A,and so P = Nm u /V ,where V is the
x
volume V = A × L. This oversimplified analysis shows how a macroscopic
(thermodynamic) property, i.e. pressure, can be related to the microscopic
(mechanical) property, i.e. molecular velocity.
Thus,
2
2
P = f/A = Nm u /L 3 (1.11)
x
2
2
2
2
If c denotes the speed in 3-dimensions, c = u + u + u ,we can write
y
x
z
1
2
2
u = c , yielding
x 3
1 2 3 1 2
P = Nm c /L = Nm c /V (1.12a)
3 3
If N A is Avogadro’s number, then Nm = nN Am = nM,where M is the
molar mass. Thus,
1 2
PV = nM c (1.12b)
3
Equating this to the ideal gas law gives, for n = 1, the root-mean-square
velocity:

2
c rms = c = c = (3RT/M) (1.12c)
Conclusion: The root-mean square speed of a molecule in an ideal gas is
proportional to the square-root of the temperature and inversely proportional
to its mass.
Example 1.2. What is the mean square speed of a N 2 molecule (treated
5
◦
as an ideal gas) at a temperature of 25 C and a pressure of 1 bar (10 Pa)?
3
Using R =8.314 Pa m K −1 mol −1 and observing that 1 Pa =
1kgm −1 −2 and that the molar mass of N 2 is M =28.0g/mol or
s