How many liters of pure methanol and pure water must be
mixed to form four liters of antifreeze at 25 degrees Centigrade? Partial molar volumes of methanol and water
in a 30 mole percent methanol solution are given as:
Methanol (1): partial molar volume, 38.633 cubic centimeter
per mol, molar volume 40.277 cubic centimeter per mole, mole fraction 0.3
Water (2): partial molar volume, 17.765 cubic centimeter per mol, molar volume 18.068 cubic centimeter per
mole, mole fraction 0.7
Solution:
Partial
Molar Properties:
Each of
the derivatives in brackets is a partial derivative of the form [partial (nM) over partial ni]
holding T, P, and nj constant. Such derivatives are accorded special
significance in solution thermodynamics, and are called the partial molar
properties of component i in solution, designated by
symbol M overbari. Thus, by definition,
M overbari is equal to [partial (nM) over partial ni]
holding T, P, and nj constant.
Where M
represents any molar thermodynamic property of a solution in which i is a component.
The overbar on M overbari
identifies it as a partial molar property.
The name implies that a mole of component i in
a particular solution at specified temperature, pressure, and composition has
associated with a set of properties (such as H overbari,
S overbari, etc) that is partially
responsible for the properties (H, S, etc) exhibited by the solution of which i is a component.
The
properties to which above equation is applicable are known to be extensive;
that is, if a solution made up of n1 moles of component 1, n2
moles component 2, ni moles of component i, etc. has an initial total property nM,
where n is equal to sum of ni, then
increasing each ni by the factor alpha at
constant and P produces a solution with a total property alpha (nM).
nM is equal
to sum of (ni M overbari).
Division
by n yields the alternative expression.
M is equal
to sum of (xi M overbari).
where xi is the mole fraction of component i in the solution.
For a
binary mixture, M overbar1 = M minus x2 dM over dx2
Since x2
is equal to 1 minus x1 and dx2 is equal to minus dx1,
this becomes
M overbar1
is equal to M plus (1 minus x1)dM
over dx1
Similarly
M overbar2 is equal to M minus x1 dM
over dx1
Figure shows a representative plot of M versus x1
for a binary system. Values of the
derivative dM over dx1 are given by the
slopes of lines drawn tangent to the curve M versus x1. One such
line drawn tangent at a particular value of x1 is shown in
Figure. Its intercepts of the figure at
x1 equal to 1 and x1 equal to zero are labeled as I1
and I2 respectively.
It is seen
from the figure that two equivalent expressions can be written for the slop of
this line:
dM over dx1
is equal to (M minus I2) over x1
and dM over dx1 is
equal to (I1 minus I2) over (1 minus 0) or is equal to I1
minus I2.
Solving
the first equation for I2 and the second for I1 (with
elimination of I2) gives
I1
is equal to M plus (1 minus x1) dM over dx1
and
I2
is equal to M minus x1 dM over dx1
Comparison
of these expressions with the definition of M over bar1 and M
overbar2 shows that
I1 is equal to M overbar1
And I2 is equal to M overbar2.
Thus the
tangent intercepts directly give the values of two partial properties. These intercepts of course shift as the point
of tagency moves along the curve, and the limiting
values are indicated by the constructions shown in Figure. The tangent drawn at x1 equal to zero (pure
component 2) gives M overbar2 equal to M2, consistent
with our earlier conclusion regarding the partial molar property of a pure
material.
The opposite intercept gives M overbar1 equals to
M overbarinfinity1, the partial property of component 1
when it is present at infinite dilution (x1 equals to zero). Similar comments apply to the tangent drawn at
x1 equal to 1 (pure component 1).
In this case, M overbar1 is equal to M1 and M
overbar2 is equal to M overbarinfinity2, since
it is component 2 that is present at infinite dilution (x1 is equal
to 1, x2 is equal to zero).
Total volume, VT is 4 liters.
Specific volume, V is equal to x1V overbar1
plus x2V overbar2.
Plugging the values, we get V equals to 24.025 cubic centimeters per
mol.
Total number of volumes is equal to VT over V or
166.49 moles
Number of moles of methanol, n1, is equal to x1
N or 0.3 time 166.49 or 49.95 moles.
Number of moles of water, n2, is equal to 16.49
minus 49.95 or 116.54 moles.
Volume of methanol, V1T or n1
V1 or 49.95 time 40.277 or 2.03 liters
Volume of water, V2T or n2
V2 or 116.54 time 18.068 or 2.11 liters