Physical Chemistry week 9 Wednesday March 6, 2013 page 1
1⁄
2 1 2
e2bvx
-b
g(vx )= ( )
2π
⟨εtr ⟩=
3
kT
2
What is b=?
k is Boltzmann constant , molecular scale
⟨εtr ⟩=
1
m⟨v 2 ⟩
2
v is speed
⟨v 2 ⟩=3⟨vx2 ⟩
3
3
kT= m⟨vx2 ⟩
2
2
kT
m
⟨vx2 ⟩=
X
⟨f(x)⟩= ∑ f(w)g(w)dw
g(w)dw represents probability
w
That sum is an infinite sum over infinitesimal quantities.
wmax
⟨f(w)⟩= ∫
f(w) g(w)dw
wmin
⟨f1 (w)+f2 (w)⟩= ⟨f1 (w)⟩ + ⟨f2 (w)⟩
⟨f1 (w)f2 (w)⟩≠ ⟨f1 (w)⟩⟨f2 (w)⟩
⟨cf1 (w)⟩=c⟨f1 (w)⟩
⟨vx2 ⟩
vx =w
f(w)=vx2
w is the continuous variable
∞
⟨vx2 ⟩= ∫ vx2 g(vx )dvx
-∞
Substitute in g(vx) and use a table of integrals to evaluate.
∞
2n -ax 2
∫ x e
dx =
0
(2n)!π
1⁄
2
1
22n+1 n!an+2
Set n=1 to match x2.
∞
2 -ax 2
∫ x e
0
dx =
1⁄
2
2!π
23 a
3⁄
2
wmin =-∞
wmax =∞
⟨vx2 ⟩=2 (
1⁄
2
-b
)
2π
⟨vx2 ⟩=
∞
2
∫ x e
bx2
2
1⁄
2
-b
2 (2π)
dx =
3⁄
2
and
⟨vx2 ⟩=
kT
m
so
1⁄
2
-b
2(2)
=
-b
23 ( )
2
0
-1
b
1
2!π ⁄2
3⁄
2
1⁄ -b
2( )
8π
-1 kT
=
b
m
1⁄
2
2π
and
=
-1
b
2
b=
-m
kT
m 1⁄2 -mvx2 1 dNvx
g(vx )= (
) e 2kT =
2πkT
N dvx
m 1⁄2
At constant temperature, the (
) term is constant.
2πkT
k = 1.38 * 10-23 J/k so kT has units of energy.
kT is the characteristic energy of the system.
G(v)dv ∝ g(vx )g(vy )g(vz )dvx dvy dvz
∝means proportional to, not equal to
dvxdvydvz is the volume element in velocity space.
Imagine a sphere within a sphere. The inner sphere has radius v. The slightly larger sphere has radius
v+dv. We want the volume inside the outer sphere but outside the inner sphere. The volume of the
outer sphere is (4/3)π(v+dv)3. The volume of the inner sphere is (4/3)πv3. The volume we want is the
difference of those two: (4/3)π(v+dv)3 - (4/3)πv3, which equals (4/3)π(v3+3v2dv+3vdv2+dv3)-(4/3)πv3,
which equals (4/3)π(3v2dv)+(4/3)π(3vdv2)+(4/3)πdv3. The (4/3)π(3vdv2) and (4/3)πdv3 terms are
negligible because of the exponents on the dv terms, so that simplifies to: 4πv2dv, which is the formula
for the volume element.
-mv 2
m 3⁄2 -m(vx +vy +vz )
m
2kT
G(v)dv= ∑ (
) e
dvx dvy dvz = (
) e 2kT 4πv 2 dv
2πkT
2πkT
2
2
2
shell
g(vx) is distribution of velocity. G(v) is distribution of speed.
2
g(vx )=Ae-avx
gaussian function, symmetrical about the y axis
Consider a diagram with vx on the horizontal axis and g(vx)/(s/cm) on the vertical axis, where s is seconds
and cm is centimeters. On curve has a peak at 0cm/s and curves down to the right and also curves down
to the left. This is the curve for T=300K(Kelvins). A second curve also has a peak 0cm/s but is not as tall
and is wider. This is the curve for T=1000K. Both curves have the same area underneath. Since each
curve has a peak at vx=0cm/s, for each temperature the most probable speed is 0, but that doesn’t mean
that nothing is moving.
Consider a diagram with v in cm/s on the horizontal axis and G(v)/(s/cm) on the vertical axis. One curve
starts at the origin and goes up and right to a peak at roughly 50,000cm/s then comes back down. This
is the curve for T=300K. A second curve also starts at the origin and goes up and right but not as steeply,
so it peaks farther to the right but not as high, but then curves down, but not as quickly. This is the
curve for T=1000K. Just like with the previous diagram, the top of the curve reveals the most probable
speed.
m 3⁄2 -mv2
G(v)dv= (
) e 2kT 4πv 2 dv
2πkT
The 4πv2dv term dominates at low speeds.
Consider a diagram with v on the horizontal axis and G(v)/(s/cm) on the vertical axis. One curve starts at
the origin and goes up and right to a peak, then curves down and right. This is the curve for argon,
which has a molar mass of 40. A second curve starts at the origin and goes up and right, but not as
steep and peaks to the right of the first curve’s peak and not as tall as the first curves peak, then curves
back down. This is the curve for neon, which has a molar mass for 20. A third curve starts at the origin
and goes up even less steep than the other two and peaks farther to the right than the other two and
not as high as the other two, but then curves down and right. This is the curve for neon. What these
curves tell us is that a gas with heavy molecules will have slower molecules than a gas with light
molecules.
⟨v 2 ⟩=
3kT 3RT
=
m
M