CHEM 331

Physical Chemistry

Fall 2014

Gas Models

As we have seen, Real Gases display distinctly non-Ideal behavior:

Z =

= 1 (Ideal Behavior)

Z =

= 1 + B'(T) P + C'(T) P

2 + ….. (Real Behavior)

This means we must resort to Data Tables of experimentally determined P, , T data or an

Equation of State in the form of a Virial expansion to describe the behavior of a given real gas.

In either case, this makes the handing of many thermodynamic problems difficult. Modeling of

the behavior of gases can be important in understanding real gases and in circumventing this

problem. It can also help in building a molecular picture of real gases.

The first "realistic" model of a Real Gas was developed by Johannes Diderik van der Waals in

1873. His model builds in two key features concerning the molecular picture of gases.

1. He allowed molecules to occupy volume, thereby reducing the overall free volume

available for molecular motion. Modifying the Ideal Gas equation of state to

account for this reduced free volume gives:

P ( - b) = RT

where b represents the excluded volume due to molecular volume. In effect, this

builds molecular repulsions into the equation of state. When the gas is compressed

sufficiently, the molecules will be forced together such that they cannot be

compressed any further, at which point they effectively exert a repulsion against

each other.

Note: At T = 0 Kelvin, = b.

and

At P = ∞, = b.

If we impose a molecular picture on the van der

Waals gas, then b can be related to the molecular

volume. Around each molecule is a region in

which other molecules are excluded.

This exclude volume can be calculated as:

Volexcl =

(2r)

3 = 8

r

3

= 8 Volmol

However, as calculated, we have overestimated Volexcl because each molecule in a

pair has a mutual region of excluded volume. So, we have overestimated the

excluded volume by a factor of two.

Volexcl = 4 Volmol

If b represents the molar excluded volume, then:

b = 4 Volmol NA

2. He allowed for intermolecular attractions by building in a term which reduces the

effective pressure of the gas, the gas molecules do not strike the walls of the

container with as much force as they otherwise would:

P =

a represents these intermolecular attractive forces. These forces will fall off as the

molecules become further apart; which will occur proportionally to the inverse of

the molar volume . (These forces will be proportional to the density of the gas

and the density is proportional to 1/ .) Because the forces are due to bimolecular

interactions, the reduction in the pressure will be proportional to (1/ )2.

So a van der Waals gas will follow the following Equation of State:

(P +

) (

Two Notes:

1. As Large ( P ~ Small), we see P = RT / . The gas behaves Ideally.

2. As T Large, we see P = RT / ( . The gas is essentially Ideal.

This equation of state can be rearranged as a polynomial in :

This cubic equation will have three real roots when T is below Tc. This can be visualized by

examining a low temperature isotherm of a van der Waals gas; see below. We observe the van

der Waals equation of state predicts a type of condensation behavior, with part of the isotherm

representing metastable behavior (super saturation) of the gas, liquifaction and metastable

behavior of the liquid (super heating) and another part of the isotherm representing non-physical

behavior.

Above the critical temperature, the equation has a single real root. This is evident in the plot of

the van der Waals isotherms given below. This van der Waals Gas has a critical temperature of

647 K.

The critical isotherm Tc has an inflection point at the critical point (Pc, ). The mathematical

requirements for an inflection point are:

= 0

=

= 0

=

These equations, along with the equation for the isotherm:

=

give us three equations in three unknowns; a, b and R. (Here we are treating R as though its

value were unknown.) Solving for these three unknowns gives us one method for determining

the van der Waals constants a and b from experimental data; Pc, Tc and . (The equation of state

will not be of any use unless the constants can be determined experimentally.)

a =

b =

R =

This method of determining a and b is problematic because experimental results for are

typically poor, infected with considerable uncertainty. (This is due to the large density

fluctuations which exist within the gas as it approaches the critical point.) This can be dealt with

by treating R as a "known" experimental result and as an unknown. Now, our three equations

can be solved for a, b and in terms of Pc, Tc and R:

=

a =

b =

At this point, it should be mentioned, the van der Waals equation of state does not represent

experimental data very well near the critical point, so these values of a and b do not tend to be

good estimates. There are other methods for determining these "constants". Thus, values of the

constants depend on the experimental method used to extract them. Some representative values

for these constants are given below.

Physical Chemistry, 2

nd Ed.

J. Philip Bromberg

Plots of the van der Waals isotherms using the constants derived from data for Ethane gas (see

Table) are provided for 300K and 350K.

Note the agreement with the experimental data. Also note the Ethane gas will condense to a

liquid at 300K if compressed sufficiently.

Next we can examine the Virial expansion of the van der Waals Equation of State.

Z =

=

=

=

This last manipulation is made because we expect (b/ ) to be small. So, we can use an

expansion of the form:

when x is small

So,

Z =

=

This gives us the Virial Coefficients in terms of the van der Waals constants:

B(T) =

C(T) = b2

In terms of our expansion in pressure:

B'(T) =

B(T)

=

C'(T) =

=

So,

Z =

A couple of notes:

1. If T is relatively low, then

dominates B'(T) and this term in Z is negative.

Hence, attractive forces are dominating and Z < 1.

2. If T is relative high, then b dominates B'(T) and this term in Z is positive. Hence,

repulsive forces are dominating and Z > 1.

3. The Boyle Temperature TB occurs when B'(T) = 0. Thus,

0 =

And,

TB =

Again, we turn to our data for Ethane gas and examine how the Virial expansion truncated at 1st

order and truncated at 2nd

order, assuming Ethane is behaving as a van der Waals gas, compares

with experimentally determined values for Z.

Finally, it should be noted that the van der Waals equation of state is not very good at

representing Real Gas behavior. This equation of state is useful because it allows us to model

the effects of attractive and repulsive forces between molecules.

The whole trouble is that the van der Waals equation is not very accurate near the critical state.

This fact, together with the fact that the values of these constants nearly always calculated (one

way or another) from the critical data, means that the van der Waals equation cannot be used for a

precise calculation of the gas properties - although it is an improvement over the ideal gas law.

The great virtue of the van derWaals equation is that the study of its predictions gives an excellent

insight into the behavior of gases and their relation to liquids and the phenomena of liquifaction.

Physical Chemistry, 3rd

Ed.

Gilber W. Castellan

Over the years, other Equations of State have been developed to model the behavior of real

gases. Each of these has its uses and its successes. None is universal. Some of the more commn

equations of state are listed below.

Dieterici Equation of State

P =

Berthelot Equation of State

P =

Beattie – Bridgeman Equation of State

=

+

+ ’P + ’P

2

= Bo -

-

= -Bob +

-

=

’ =

’ =

The Dieterici and Berthelot equations are, like the van der Waals equation, two parameter

equations of state. The Dieterici equation is better than the van der Waals equation near the

critical point. However, it is not often used because it is a transcendental function. The

Berthelot equation is useful for calculations involving . Finally, the Beattie-Bridgeman

equation is used when precision is needed.

It is believed that this equation has to unique degree the following desirable qualities. (a) The

numerical values of the constants can be determined easily and uniquely from the pressure-

volume-temperature data, and with one exception can be obtained as the intercepts or slopes of

straight lines by suitable treatment of the data. (b) The equation is readily amenable to

mathematical manipulation since it is completely algebraic. Hence, most of the usual

thermodynamic relations can be integrated in terms of the elementary functions as, for example,

pdv (and therefore vdp) at constant temperature. (c) The equation fits the data over a wide range

of pressures and temperatures, reproducing the trends as well as the actual pressures. (d) The

volume and temperature functions, A, B and can be expanded further, if necessary, without

alteration of the general form of the equation.

A New Equation of State for Fluids

James A. Beattie and Oxcar C. Bridgemann

Journal of the American Chemical Society

July 1927

A comparison of the results of calculations of for Carbon Dioxide gas at 400K and 100 atm are

given below for the Ideal, van der Waals and Beattie-Bridgemann equations of state. Needed

constants are obtained by consulting appropriate Tables.

Ideal Gas

=

=

= 3.282 x 10

-4 m

3/mole = 0.3282 L/mole

van der Waals Gas

a = 0.364 m6 Pa /mol

2

b = 4.27 x 10-5

m3/mol

-

+

-

= 0

-

+

-

= 0

So, = 3.818 x 10-4

m3/mol = 0.3818 L/mol

Beattie-Bridgeman Gas

Ao = 507.31 x 10-3

Pa m6/mol

2

a = 71.32 x 10-6

m3/mol

Bo = 104.76 x 10-6

m3/mol

b = 72.35 x 10-6

m3/mol

c = 660.0 K3 m

3/mol

=

+

+ ’P + ’P

2

= Bo -

-

= (104.76 x 10-6

) -

-

= - 5.80919 x 10

-5

= -Bob +

-

= - (104.76 x 10-6

)(72.35 x 10-6

) +

-

= 2.21938 x 10-9

=

=

= 7.81624 x 10

-14

’ =

=

= -3.47374 x 10-13

’ =

=

= 6.58778 x10

-21

So, =

- 5.80919 x 10

-5 + (-3.47374 x 10

-13)(10132500) + (6.58775 x 10

-21)

(10132500)2

= 2.673 x 10-4

m3/mol = 0.2673 L/mol

These results cry out for a comparison with the experimental data. Returning to Din's Thermodynamic

Functions of Gases, we have for CO2 gas:

We will have to interpolate the data as 400 K (126.85oC) lies between 120

oC ( = 5.884 L/kg) and

130oC ( = 6.189 L/kg). So,

= 5.884 +

(6.85) = 6.09 L/kg (1 kg / 1000 g) (44.011 g/mol)

= 0.2680 L/mol

Comparison with the Beattie-Bridgeman model is quite favorable. It should also be noted that the Ideal

Gas model is better at describing this gas than is the van der Waals model.

Finally, it is noted that if we define Reduced State Variables as:

PR =

TR =

VR =

that many gases behave similarly when viewed using these reduced variables. This observation is

known as the Law of Corresponding States. A plot of Z vs. PR and TR demonstrates this for a number of

gases:

As an example of the use of the Law of Corresponding States, again consider the calculation of

for Carbon Dioxide at 400K and 100 atm. Data for CO2 is:

Pc = 72.83 atm

Tc = 304.16 K

Thus, PR = (100/72.83) = 1.31 and TR = (400/304.16) = 1.37. From the above chart, we can then

estimate Z as 0.77. Then,

=

=

This result should be compared with those results determined above as well as the

experimentally determined result.

Any equation of state that involves only two constants in addition to R can be written in terms of

the reduced variables only. For this reason equations that involved more than two constants were,

at one time, frowned upon as contradicting the law of corresponding states. At the same time,

hopes were high that an accurate two-constant equation could be devised to represent the

experimental data. These hopes have been abandoned; it is now recognized that the experimental

data do not support the law of corresponding states as a law of great accuracy over all ranges of

pressure and temperature. Although the law is not exact, it has a good deal of importance in

engineering practice; in the range of industrial pressures and temperatures, the law often holds

with accuracy sufficient for engineering calculations.

Physical Chemistry, 3rd

Ed.

Gilbert W. Castellan