Van der Waals Equation of State 

Back to previous section:  Ideal Gas Mixtures
Continue to next section:  Van der Waals Roots

We mentioned before that the Pv product of gases is different for each compound at higher pressures, but that in the limit of low pressures (low density), the product becomes the same for all compounds, as shown in the plot below. This is because at low densities, the intermolecular interactions are negligible because of the separation distances between molecules.

.                                               . 

The reason that the ideal gas equation is valid at very low densities is that at those conditions (1) the molecules occupy an insignificant portion of the gas volume and (2) molecular interactions are negligible. Van der Waals (1873) tried to relax these two assumptions by postulating that the real pressure of the gas would be the ideal gas pressure minus the contracting forces per unit area due to the intermolecular attractions

P = P^IG - a/v²

v = v^IG + b

Various forms of van der Waals equation of state

Cubic Equations of State
Note that the van der Waals (vdw) equation of state is cubic in volume. The first equation above can be multiplied out to give:

Pv³ - (Pb + RT)v² + av - ab = 0

To see what all this means, look at the figure below.  The red curves are isotherms, constant temperature lines, that are obtained from the vdw equation of state (eos).  Each point on the line is generated by plugging in the molar volume and calculating the pressure from the vdw eos.  Notice that for temperatures below the top isotherm labeled with a Tc, that the three different volumes give the same pressure; i.e., there are three real roots for volume when solving the vdw eos for a given P and T.  This is the cubic eos way of modeling the coexistence of liquid and vapor in the two-phase region.

The eos tries to model the discontinuity that occurs at a phase boundary with a continuous line that goes down, back up, and then down again on the PV plot. (We will see an isotherm for ammonia in the two-phase region as calculated by several different equations of state in a few more pages.)

The real behavior of the fluid however is discontinuous. If we focus attention on the lowest isotherm shown and gradually decrease the pressure from a very high value, the liquid begins to expand as we follow the isotherm down.  Liquids are relatively incompressible, so the isotherm is very steep; i.e., it takes the release of quite a bit of pressure to get the liquid to expand very much.  (In the figure below, we are following the purple line.) However, when we reach the blue line, called the binodal curve, the first bubbles of vapor appears as the liquid boils.  For this reason, the left branch of the binodal curve is called the bubble line.  Instead of following the curved line of the eos, the real fluid now follows the horizontal line shown in the figure below. What happens is that as we try to relieve additional pressure, the pressure doesn't change even though the molar volume increases. The liquid molar volume doesn't change - it is still at the value given by the intersection of the binodal (blue) curve with the purple isotherm - but the mixture molar volume continues to increase because we are forming more and more vapor which has a much higher molar volume.  In fact, the vapor molar volume is given by the intersection of the purple isotherm with the blue coexistence curve on the right.  The right side of the coexistence curve is called the dew line because it represents the case where we have all vapor and the first drops of liquid just begin to form if we try to increase the pressure.  Thus, as we try to relieve pressure on a two-phase fluid, we simply evaporate more and more of the liquid instead of decreasing the pressure, and the molar volume of the system increases because we are changing some liquid (with a small molar volume) into vapor (with a large molar volume).  The binodal curve encloses the region (colored) where two phases (liquid and vapor) will coexist. As long as we are in this two-phase region, the pressure and temperature will be constant, and we simply evaporate more liquid or condense more vapor when we try to expand or compress, respectively, the system. When we have vaporized all of the liquid and the molar volume is now that of the saturated vapor (on the dew line), then when we try to lower the pressure again, we find that the pressure will again go down and the volume will increase as the gas expands.

So, how does the eos of isotherm try and mimic the purple line above that represents the real fluid behavior through the two phase region?  The red line on the figure below represents unstable behavior in the region where the slope is positive.  This is an instability with respect to staying in one phase.  Therefore the fluid will separate into two phases in this region.  It can be shown that the way in which the fluid separates must be such that the tie line, the horizontal line through the two-phase region that connects the bubble point of the fluid to the dew point, divides the isotherm into equal areas.  These equal areas are shown in the diagram below. Thus, the eos predicts that the binodal or coexistence curve will go through all of the isotherms at the points where a horizontal line will bisect the area created by the up and down behavior of the eos isotherm.  The blue coexistence curve is actually the locus of all such points where the tie line bisects the eos isotherms.

More importantly for our discussion here, the binodal curve represented by the ends of the purple tie lines are the molar volumes of the liquid and vapor.  Thus, the vapor pressures for isotherms are represented by the horizontal purple tie lines.  The ends of these tie lines represent the liquid and vapor molar volumes at this temperature and pressure.

This means that to find the molar volume of the liquid from the vdw eos, we plug in the vapor pressure at the given temperature and solve for the smallest root.  On the other hand, if we want the vapor molar volume, we plug into the vdw eos the vapor pressure at the given temperature and solve for the largest root.  What about the middle root?  It is an unstable molar volume and does not represent real behavior.

Notice that the phase dome narrows with increasing temperature.  As we go up in temperature, the molar volume of the saturated liquid (saturated means that it is on the phase dome at the limit between single- and two-phase behavior) gets larger and the molar volume of the saturated vapor gets smaller.  The smallest and largest roots begin to converge.  At the top of the phase dome, all three roots converge into a single root.  Above this temperature, we have only one real root when solving the vdw eos for volume; the other two roots are imaginary.  The isotherm that goes through the point where the three roots converge to one real root has an inflection point at this locus.  This point is called the critical point.  The isotherm that goes through it is at the critical temperature, Tc, the maximum pressure at which we can have three real roots is called the critical pressure, Pc, and the molar volume of the fluid at this point is the critical volume, vc.

At the critical point v_L = v_V. The molar volumes, hence densities, of the liquid and vapor are identical. Thus, there is no distinction between a liquid and vapor.  You can not tell them apart because they are identical in every way at that point.  Moreover, the fact that the critical point is an inflection point for the critical isotherm, allows us to evaluate the two constants in the vdw eos in terms of the critical temperature and pressure. The mathematical criteria for inflection points are (dP/dv)_T = 0 = (d²P/dv²)_T. This gives us two equations with which to evaluate the two unknowns, a and b, in the vdw eos.  We could differentiate the vdw eos with respect to volume and set the first and second derivatives to zero.  These two equations could then be rearranged to obtain expressions for a and b in terms of T_c and P_c. When this is done, we get:
 

Continue to next section:  Van der Waals Roots