Physical Chemistry week 8 Monday February 25, 2013 page 1
Ternary liquid-liquid equilibrium
Consider an equilateral triangle with acetone at the top point, water at the bottom left point, and ether
at the bottom right point. Just to the right of the left side on the bottom is a curve that going up and
right until roughly the middle then down and right to the bottom just left of the right side. Outside of
this curve is the one phase region since water and acetone are completely soluble and acetone and
ether are completely soluble. Inside this curve is the two phase region since water and ether are
partially soluble. Inside the two phase region are phases alpha and beta. The tie lines in the two phase
region in this example are diagonal going up and left. For this two phase region: c=3 and p=2 so f=32+2=3 but T and P are constant on this diagram so f=1 on this diagram for the two phase region. Going
up and left in the two phase region the tie lines get shorter. Where the tie line is zero length is point k.
Point k is the isothermal critical point. Assume point G is in this two phase region somewhere and
͞ nα=G͞Hnβ. As shown in this example,
points F and H are the end points of the tie line for point G then FG
tie lines can be diagonal, not just horizontal.
Begin chapter 15: statistical mechanics: kinetic molecular theory of gases
Statistical mechanics is the bridge connecting the microscopic viewpoint to the macroscopic viewpoint.
Microscope viewpoint → statistical mechanics → macroscopic viewpoint
Pressure of an ideal gas (derivation)
Consider a graph in 3 dimensions: x, y, and z. v⃑ is velocity. 𝓋 is speed. Point C is an arbitrary point.
The velocity vector v⃑ is the one from the origin to C. Point B is just point C projected down onto the xy
plane. Speed can be zero or positive but not negative. Velocity can be positive, negative, or zero. Point
O is the origin. The length of segment O͞C2 = 𝓋2. Using the Pythagorean Theorem:
𝓋2 = 𝓋z2 + O͞B2 = 𝓋z2 + 𝓋x2 + 𝓋y2.
𝓋x, 𝓋y, 𝓋z can be zero, positive, or negative, but 𝓋 is always positive or zero, never negative.
ε is lower case epsilon = kinetic energy of motion = translational energy
εtr is translational energy
εtr =
1
1
1
mvx2 + mvy2 + mvz2
2
2
2
Let F(t) = some time dependent function or property (force in this case)
<F> is average value of F
n
1
<F> = ∑ Fi Fi are the observed values of F, n is the number of observations
n
i=1
As n→∞, ∆t→0 and the sum becomes an integral
t2
1
mathematically: <f> =
∫ F(t)dt
t 2 -t 1 t 1
equation I
Consider a box(cube for mathematically simplicity). l (lower case l) is the length of a side. i is the atom.
Atom i is moving to the left(y component positive, other components 0). The area of the wall on the left
side of the box is lxlz. a is acceleration.
Fy,i =may,i =m
d
d
vy,i = (mvy,i )
dt
dt
p (lower case p) is momentum.
py is linear momentum along the y axis.
define: py = m𝓋y
dpy,i = Fy,idt then integrate:
∫ dpy,i = ∫ Fy,i dt
t” – t’ is collision time
t"
py,i (t")- py,i (t')= ∫ Fy,i dt
t'
t"
-mvy,i - mvy,i = ∫ Fy,i dt
t'
t"
-2mvy,i = ∫ Fy,i dt
t'
t"
∫ Fy,i dt is impulse, units:
t'
impulse is vector quantity, essentially average force
Newton’s 3rd law: Fy,i = -Fw,i
Fy,i is y component of force of particle i
-Fw,i is perpendicular force on wall due to particle i
t2
2mvy,i = ∫ Fw,i dt
t1
t1 to t2 includes the t’ to t”
t’ to t” is when particle i is colliding with the wall
kg m s
kg m
=
2
s
s
t1 to t2 includes the travel time and the collision time
2m𝓋y,i = <Fw,i>(t2-t1)
y= vy t ∴ t 2 -t 1 =
2ly
vy,i
2ly since it makes full round trip
2
2mvy,i 2mvy,i mvy,i
<Fw,i > =
=
=
2ly
t 2 -t 1
ly
⁄v
y,i
average force for one molecule
N
<Fw > = ∑ <Fw,i >
N is the number of molecules
i=1
N
N
i=1
i=1
mvy,i m
2
<Fw > = ∑
= ∑ vy,i
ly
ly
N
2
<vy,i
>=
∑
i=1
<Fw > =
2
vy,i
N
mN 2
<vy >
ly
V = lylxlz = volume of box
P=
Fave <Fw > mN<vy2 > mN<vy2 >
=
=
=
A
lx lz
ly lx lz
V