Ideal vs. Real Gases

All gases obey the ideal gas equation of state provided they are sufficiently dilute. The ratio Z = PV/nRT = 1 at all pressures for an ideal gas. For a real gas Z is deviates from one as the pressure increases. First Z < 1 at intermediate pressures of tens to several hundreds of bar due to the attractive forces between molecules. At higher pressure the repulsive forces begin to dominate resulting Z > 1.

There are non-ideal equations of state designed to account for the deviations of gases from ideal behavior. The best known of these is the van der Waals equation

Where V bar designates the molar volume. The constants a and b are called van der Waals constants, and they depend on the gas being described. The parameter a depends on the strength of intermolecular interactions. The parameter b depends on the size of the molecules.

The van der Waals equation can be rewritten a number of ways:

(a)
in terms of pressure

(b)
in terms of the compressibility factor Z

(c) in terms of the molar volume

The last form shows clearly that the van der Waals equation of state is a cubic polynomial. This is important since it implies that below a critical temperature there will be a phase transition. In other words this simple equation also serves as an equation of state for the liquid!

Equations of State: Virial, van der Waal's and Redlich-Kwong

The ideal gas law (PV = nRT) assumes that atoms and molecules have no extent (no finite size) and there are no interactions between particles.

The virial equation of state shows a natural connection between the microscopic and macroscopic view.

The coefficients B(T) are called virial coefficients.

Z is called the compressibility. The second virial coefficient is negative at low temperature.

A negative B_2V represents the dominance of intermolecular attractions. B_2V can be related to the intermolecular potential through

N_A is Avagadro’s number. k_B is the Boltzmann constant.

k_B = R/N_A. k_B is a microscopic constant that corresponds to the macroscopic universal gas constant, R.

If u(r) is the Lennard-Jones potential the equation is not analytic, but has been solved numerically. This is an important equation since it establishes a connection between a microscopic potential energy function and macroscopic term in the compressibility expansion.

The second virial coefficient can be expressed in terms of molecular volume analytically if we use a hard sphere potential. The hard sphere potential is

u(r) = infinity for r < s and u(r) = 0 for r > s.

which is equal to four times the volume of N_A hard spheres.

The van der Waals equation has a natural interpretation in terms of finite size since b is the volume reduction due to the size of the gas molecules and a is a pressure reduction due to interactions between molecules.

The interpretation the van der Waals parameters can be obtained by writing a virial expansion

Microscopic Interpretation of the van der Waals Equation

The compressibility can be expressed in terms of the van der Waal’s equation expanded in a power series in the molar volume. By comparing the appropriate derivatives we obtain:

The second virial coefficient is obtained by comparison

To obtain a connection with molecular properties we can use a hard sphere L-J potential. In this potential

u(r) = infinity for r < s

u(r) = -c₆/r^-6 for r > s.

The Redlich-Kwong Equation

The Redlich-Kwong equation is more accurate than the van der Waals equation. The Redlich-Kwong equation is also a cubic polynomial in V bar. It turns out that this type of equation can describe both the gaseous and liquid states.

Isotherms

Isotherms are constant temperature curves in pressure-volume space (PV). Using the ideal gas law we can plot a number of isotherms. To evaluate simple expression and plot functions while at the computer we will use the program Maple. To plot an ideal gas isotherm (P = RT/V) we go to the Maple prompt and type in the command

Ø with(plots):plot(0.082*300/v,v=0.1..1);

where we have used R = 0.082 L-atm/mol-K and T = 300 K. To see three isotherms on a plot type

Ø plot({0.082*300/v,0.082*200/v,0.082*100/v},v=0.05..0.3);

Your plot should look like this

where the red, yellow, green curves are isotherms.

The {} bracket allows more than one function to be

plotted on the same graph as {function1, function2,..}.

We can make the same plots for a van der Waals gas. Let’s look at ammonia (a = 4.25 L²atm/mol² and b = 0.037 L/mol). At 298 K we have

Ø with(plots):plot(0.082*298/(v-0.037)-4.248/(v*v),v=0.05..1);

You will note that this isotherm has a strange shape. The reason is that the cubic polynomial is giving rise to inflection. At high enough temperature this will no longer be important (and the van der Waals equation will begin to approach the ideal gas law).

Ø plot({0.082*298/(v-0.037)-4.248/(v*v),0.082*400/(v-0.037)-4.248/(v*v),0.082*500/(v-0.037)-4.248/(v*v)},v=0.05..0.3);

Note that we have changed the range on the abcissa (the molar volume axis) to better see how the shape of the curves is changing. This behavior is shown in Figures 16.7 – 16.9 of McQuarrie and Simon as well. You can compare the ideal gas law to the van der Waals equation at 300 K

Ø

with(plots):plot({0.082*300/v,0.082*300/(v-0.037)-4.248/(v*v)},v=0.05..0.3);

and 500 K

Ø with(plots):plot({0.082*500/v,0.082*500/(v-0.037)-4.248/(v*v)},v=0.05..0.3);

The plot at 500 K shows more similarity than the plot at 300 K. Above a critical temperature there is no inflection point and the curves for real and ideal equations of state begin to have the same shape. This behavior corresponds to the known behavior of substances. They all have a critical temperature (and pressure) beyond which there is no distinction between liquid and gas. Below the critical temperature there will be a phase transition for a given temperature and pressure. Since the molar volume changes (think about liquid water turning to vapor) there should be an abrupt change in the molar volume as shown in Figure 16.7 in your text for isotherms below the critical point.

The critical point is defined mathematically as an inflection point. At the critical temperature the curve will no longer turn down. For ammonia the critical point must be between 300 K where the isotherm shows a change in the sign of curvature and 500 K where it does not. At the inflection point the slope of the curve will be zero and the curvature will be zero. In pressure-volume space this means:

McQuarrie and Simon have an elegant and simple way of solving this. They note that at the critical point there is only one root. Therefore, the cubic equation can be written

The coefficient V_c bar can be equated with the coefficients above for the van der Waals equation of state expressed as a cubic polynomial. A single point in P-V space is solved for, thus, there is a critical temperature, T_c, pressure, P_c, and volume, V_c. These are:

in terms of the van der Waals parameters. For example, we can now calculate the critical temperature of ammonia, T_c = 8(4.25)/(27(0.039)(0.082)) = 393.7 K.

We did not explicity write out the units. I used R = 0.082 L-atm/mol-K in this case since we are working in the L-atm units of P-V space.

In practice, experimental critical data are used to obtain the parameters a and b (or A and B for Redlich-Kwong). We will see further that the virial equation of state can be related to the van der Waals equation of state and to parameters that describe molecular interactions.

The Law of Corresponding States

We can define reduced quantities

P_R = P/P_c, V_R = V/V_c, T_R = T/T_c

By substitution into the van der Waals equation we find

This equation is a universal equation for all gases. Although the actual pressures and volumes may differ, two gases are said to be in corresponding states if their reduced pressure, volume, and temperature are the same.

The compressibility factor Z can also be cast into the form of corresponding states showing that Z also can be expressed as a universal function of V_R and T_R or any
other two reduced quantities.