Download Condensed Phases: Liquids and Solids

CHEM 331
Physical Chemistry
Fall 2014
Condensed Phases: Liquids and Solids
A distinguishing feature of liquids and solids is that they are typically much more dense than
gases under ordinary conditions. At 0oC and 1 atm pressure:
Gases
= 22.4 L/mol
Liquid Water
= 0.018 L/mol
Thus, the molar volume of a gas is about ~1000 times that of a liquid or solid. This means the
molecules in a liquid or solid are much closer together, by a factor of ten, than they are in a gas.
If the ratio of the gas volume to liquid volume is 1000, then the ratio of the distance between the
molecules in the gas to that between those in the liquid is the cube root of the volume ratio, that is,
ten. The molecules of the gas are ten times farther apart on the average than are those of the
liquid. The distance between molecules in the liquid is roughly equal to the molecular diameter;
hence, in the gas the molecules are separated by a distance which is, on the average, ten times their
diameter. This large spacing in the gas compared with that in the liquid results in the
characteristic properties of the gas and the contrast of these properties with those of the liquid.
This comes about simply because of the short-range nature of the intermolecular forces, the van
der Waals forces. The effect of these forces decreases very sharply with increase in distance
between the molecules and falls to an almost negligible value at distances of four to five times the
molecular diameter. If we measure the forces by the magnitude of the term a/ 2 in the van der
Waals equation, then an increase in volume by a factor by a factor of 1000 in going from liquid to
gas decreases the term by a factor of 106. Conversely, in the liquid the effect of the van der Waal;s
forces is a million times larger than it is in the gas.
Physical Chemistry, 3rd Ed.
Gilbert W. Castellan
The upshot is, it is not reasonable to find a generalized Equation of State for a liquid or solid like
it is for a gas. We must instead rely on the primary data for a relationship between and T or P.
This data is usually summarized in the form of Thermal Expansion Coefficients () and
Isothermal Compression Coefficients (). These are defined according to:
 =
 =
For Water, this data is as follows:
(Table cont'd below.)
Generally,
 > 0
(This is a requirement of the 2nd Law of Thermodynamics)
 > 0
(This is an experimental observation; note, from the data for Water above,
this is not universally true. Water contracts as it is heated when below
4oC.)
We can describe Volume changes with Temperature and Pressure, based on this data, as below.
Volume Changes with Pressure
At constant Pressure:
=  dT
Integrating from a reference temperature To at which the volume is Vo, we have:
Or, if we assume  is fairly constant:
Solving for V gives us:
V =
Taking our reference temperature to be To = 0oC, this gives:
V =
Since  is relatively small, we can use the following expansion:
ex = 1 + x +
+
+ …..
Truncating after the linear term gives us our final result:
V = Vo (1 + T)
Let's see how this works for Water. From the data table above:
Vo = 1.000160 g/cm3

= 257.05 x 106 oC-1
(I've taken  to be the value at 25oC. At 0oC, , for Water,
has the wrong trend.)
So,
V
= (1.000160) (1 + 257.05 x 106 T[oC])
A plot of the predicted linear behavior, compared with the experimental results, is:
We can certainly see some significant departures from the data. But, remember, Water is unusual around 4 oC and the
fit is getting better by 50oC.
If we wish to do better in representing the data, we could:
1. Recognize that  depends on T. This means we cannot assume  is constant in the above
derivation. Instead, it is typical to assume  has a quadratic dependence on T:
 = o + 'T +
T2
2. In the above derivation, truncate the expansion of the exponential at higher powers of T:
V = Vo (1 + aT + bT2 + cT3 + …..)
Volume Changes with Pressure
At constant Temperature:
=  dP
Integrating, assuming  is constant and taking our reference point to be Po = 1 bar with V = Vo,
we have:
V = Vo (1 - (P -1))
(It should be noted, in older data tables Po = 1 atm. The difference is slight but noticeable.)
General T and P Dependence
Putting this all together, we have:
V =
(1 + (P - 1))
In terms of derivatives, a direct measurement of:
is generally not possible for liquids and solids. Heating the system to determine the change in
pressure with temperature will usually result in a slight deformation of the container holding the
liquid/solid. This means V is not constant. (This is not a problem for gaseous systems. Why?)
We generally use  and  data to obtain this derivative. By the Cyclic Rule of calculus:
= -1
Rearranging:
= -
= -
=
Example
For Mercury, we have:

= 2 x 10-4 K-1

= 3.9 x 10-11 Pa
How much would the pressure of a sample of Mercury increase if we heat it from 50 to 51 oC?
Assume:
=
Then,
P =
T =
(1 K) = 5.1 x106 Pa = 51 atm