Real Gases

Deviations from Ideal Behavior
Real Gases fail to obey the Ideal Gas equation of state exactly. Why?

For exactly one mole of an ideal gas:

Plotting the experimentally determined value of (pV/RT) for exactly one mole of various REAL GASes as a function of pressure, p, shows a deviation from ideality (The quantity (pV)/(nRT) = Z is called the COMPRESSIBILITY FACTOR and should be unity for an Ideal Gas):


The deviation from ideal behavior is large at high pressure and low temperatue At lower pressures and high temperatures, the deviation from ideal behavior is typically small, and the ideal gas law can be used to predict behavior with little error.

Deviation from ideal behavior can also be shown for a given gas (Nitrogen, in this example) as a function of temperature:


As temperature is decreased below a critical value, the deviation from ideal gas behavior becomes severe, because the gas CONDENSES to become a LIQUID.


The van der Waals Equation (A fix to the Ideal Gas Equation)

One of the most useful equations to predict the behavior of real gases was developed by Johannes van der Waals (1837-1923) It takes into account Molecular Stickiness and Molecular Size

The Ideal Gas Equation:

Has been presented as an Empirical relation, but it doesn't work perfectly for all gases under all conditions because it is based on imperfect assumptions.

Remember the conditions assumed for an Ideal Gas?:

  1. Molecules are perfectly elastic (no STICKINESS)
  2. Molecules are point masses (no SIZE)
  3. Molecules move at random

The first two of these assumptions is clearly wrong for all gases, because at low temperatures all gases CONDENSE, or form a liquid phase. This must happen because the molecules stick to one another, at least a little. We can correct the Ideal Gas equation for stickiness. Moreover, the liquid has measurable molar volume, and this volume is simply the size of the close-packed molecules of the liquid.

At high pressures, and thus high densities, the intermolecular distances can become quite short, and attractive forces between molecules becomes significant. Neighboring molecules exert a relatively long-ranged attractive force on one another, which will reduce the momentum in which they tranfer to the container walls. The observed pressure exerted by the gas under these conditions will be less than that for an Ideal Gas

To put the observed (real) pressure into the Ideal Gas expression, we must correct for the decrease in pressure due to molecular stickiness. For Stickiness to be a factor, the two gas molecules must have a collision. The probability of a collision is the probability of two molecules being in the same place at the same time. The probability of the first molecule being at the place of the collision is proportional to the number density (n/V). The probability of the second one being in the same place is the same, (n/V). Thus the reduction in pressure due to stickiness should be proportional to (n/V)2. If the proportionality constant is called a, then the ideal pressure is

Thus, correcting for molecular stickiness alone, the Ideal Gas equation would become:

As pressures and density increase, the volume of the molecules themselves becomes significant relative to the size of the container.

To correct for the effect of finite molecular volume, we must recognize that in the ideal gas equation the volume used is the "free volume" that the molecules find themselves in. The free volume is just the real (container) volume minus the volume that is taken up by the molecules of the gas itself.

where b is a constant representing the volume of a mole of gas molecules at rest. Thus, the ideal gas equation, if corrected for molecular size and stickiness, looks like:

We can rearrange this expression slightly to give the familiar form of the van der Waals equation.


Unlike the universal gas constant, R, The van der Waals constants a and b are different for different gases.

Substance
a (L2 atm/mol2)
b (L/mol)
He
0.0341
0.0237
H2
0.244
0.0266
O2
1.36
0.0318
H2O
5.46
0.0305
CCl4
20.4
0.1383

Numerical Example:

Use the van der Waals equation to calculate the pressure of a sample of 1.0000 mol of oxygen gas in a 22.415 L vessel at 0.0000°C (Note: For an ideal gas, this temperature and volume would lead to conditions of STP, i.e. a pressure of exactly 1.0000 atm)

V = 22.4 L
T = (0.000 + 273.15) = 273.15K

a (O2) = 1.36 L2 atm/mol2 (From Table)
b (O2) = 0.0318 L /mol (From Table)

(nRT)/(V-nb) = 1.0014 atm
- a (n/V)2 = -0.0027 atm
p = (nRT)/(V-nb) - a (n/V)2 = 0.9987 atm


Here is a calculator for the numerical evaluation of ideal and real gas equations of state.
Syllabus || Staff || Operations || TOP
PJ Brucat || University of Florida