Eos van der waals

Páginas: 6 (1319 palabras) Publicado: 29 de noviembre de 2011
Van der Waal’s Equation: This gives an explanation of the nonideal behavior of gases. The plot of Z versus P shows that, gases do not conform the ideal-gas equation PV = nRT. CH4 H2 Ideal

2.0 Z 1.0

0.0 p

We saw that, the kinetic molecular theory of gases explains the ideal-gas PVT behavior based on the assumption that gases are composed of randomly moving, infinitely small particles. Vander Waal added two more items to this model and showed that they can account for deviation of real gases from ideal behavior. They are: 1. Volume occupied by the gas molecules. 2. The attractive forces between the molecules.


Volume occupied by the gas molecules, i.e. Excluded Volume.

When n mol of gas is placed in a container of volume V, the volume in which the molecules are free tomove is equal to V only if the volume occupied by the molecules themselves is negligible. The presence of molecules of this nonvanishing size means there is a certain volume, called excluded volume, is not available for the molecules to move in. If this extended volume of 1 mol of a gas sample is represented by b, then We can write a new corrected volume as (V – nb).

Thus, the PV = nRTequation becomes P (V – nb) = nRT The value of b can be treated as constant and is characteristics of each gas. Van der Waal’s Excluded Volume and Molecular Diameter: The molecules are assumed to be spherical and to have diameter d or radius r. A bigger circle shows the volume in which the centers of two molecules cannot move because of each others presence

Excluded volume

Since the diameter ofmolecule is d. The radius of bigger circle is also d. Then Excluded Volume for a pair of molecule = 4 π d3 3 Therefore, Excluded Volume for one molecule = 1 2 4 π d3 3 4 π (d / 2)3 3 4 π r3 3

Excluded Volume for one molecule



Excluded Volume for one molecule



since r = d/2

Thus the excluded volume is 4 times the volume of a molecule.

Since b is the excluded volumeper mole, Then, b = 4 NA 4 π r3 3

where NA = avogadro’s constant.


The attractive forces between the molecules, i.e. Molecular Attractions

The second Van der Waal’s correction term comes from the attractive forces between the molecules. When a pressure is applied to the gas, it condenses to liquid. This can be accounted for pressure as well as the attractive forces between themolecules. Thus the effective pressure will be less than the actual pressure required to confine the gas molecules, since the attractive forces will also help in keeping the gas molecules together. The attraction forces of the molecules attract the nearby molecules. So, if there is n mol of gas in volume V, then the number of molecules that are near any given molecules is proportional to n/ V i.e. numberof moles per unit volume. Thus, for any given molecule which helps in holding the gas together through the molecule-molecule attractions is proportional to the number of near by molecules i.e. n/V At the same time, the other neighboring molecules will also attract its neighbors. Thus, the total pull of gas due to these molecular interactions is proportional (n / V)2. We have, P (V – nb) = nRT nRTP = (V – nb) But if we include the molecular interactions term, the equation becomes nRT P = (V – nb) an2

V2 where a = proportionality factor.

Rearranging the above equation gives Van der Waal’s equation

an2 P + V

(V – nb) = nRT

For 1 mole of gas sample, the equation becomes, a P + V2 (V – b ) = RT

Problem 1 At 2000C a pressure of 42.4 bar of pressure is required to reducethe molar volume of NH3 to 0.85 L. What pressure would have been calculated on the basis of (a)ideal-gas behavior and (b)van der Waal’s equation with a = 4.25 bar L2 mol-1, b = 0.0374 L mol-1, R = 0.0831 L bar K-1 mol-1. T = 273 + 200 = 473 K From ideal-gas law equation: PV = nRT P = nRT / V = (1 mol) (0.0831 L bar K-1 mol-1) (473 K) /(0.851 L) = 46.2 bar From Van der Waal’s equation: an2 P + V
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