Chemistry Reference
In-Depth Information
v
l
Figure 1.2
Single molecules oscillating between two walls.
By considering the collision between the molecule and one of the walls, the
momentum lost by the molecule and the wall is
p
=
p
initial
−
p
final
=
m
v
x
−
(
−
m
v
x
)
=
2
m
v
x
.
(1.9)
The time between successive collisions on this particular wall will be
2
l
v
x
t
=
(1.10)
Force is the rate of change of momentum, so the force on the wall from the single
molecule is
x
=
p
2
m
v
x
m
v
F
t
=
/v
x
=
.
(1.11)
2
l
l
For a large number (
j
) of molecules and collisions with the wall, this becomes
m
j
v
2
jx
F
=
.
(1.12)
l
Now, by adding in collisions with walls in all six directions this gives
l
j
jz
2
m
2
jx
2
jy
2
F
=
v
+
v
+
v
.
(1.13)
For equilibrium conditions and a sufficiently high collision rate with the walls,
the force on all six walls can be assumed to be the same. Therefore the force on a
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