How can I derive the Van der Waals equation?

1 Answer
Feb 28, 2016

First off, the van der Waals equation looks like this:

#\mathbf(P = (RT)/(barV - b) - a/(barV^2))#

where:

  • #P# is the pressure in #"bar"#s.
  • #R# is the universal gas constant (#0.083145 "L"*"bar/mol"cdot"K"#).
  • #T# is the temperature in #"K"#.
  • #barV# is the molar volume of the gas in #"L/mol"#.
  • #b# is the volume excluded by the real gas in #"L/mol"#.
  • #a# is the average attraction between gas particles in #("L"^2"bar")/("mol"^2)#.

#\mathbf(a)# and #\mathbf(b)# depend on the gas itself .

For example, #a = 13.888# and #b = 0.11641# for butane (Physical Chemistry: A Molecular Approach, McQuarrie).

We can derive it by starting from the ideal gas law:

#PbarV = RT#

#P = (RT)/(barV)#

But the actual derivation is fairly involved, so we won't do it the complete way, but in more of a conceptual way.

By assuming the gas is a hard sphere, we say that it takes up the space it has available to move, i.e. it decreases #barV# by #b#. Again, this depends on the exact gas.

Therefore, the first half becomes:

#P = (RT)/(barV - b) pm ?#

Now we examine #a#; #a# essentially accounts for the different extent of the attractive intermolecular forces present in each gas. Basically, when accounting for that we subtract #a/(barV^2)#.

#color(blue)(P = (RT)/(barV - b) - a/(barV^2))#

If we note that the compressibility factor #Z = (PbarV)/(RT)# is #1# for an ideal gas, it is less for a real gas if it is easily compressible and greater for a real gas if it is hardly compressible. As such, we have these two relationships:

  • When #Z > 1#, #barV# is greater than ideal, which means the gas is harder to compress. It corresponds to a higher pressure required to compress the gas.
  • When #Z < 1#, #barV# is smaller than ideal, which means the gas is easier to compress. It corresponds to a lower pressure required to compress the gas.

From this we should see that :

  • For a gas that is easier to compress than the ideal form, with a smaller #barV#, the magnitude of #a/(barV^2)# increases more than the magnitude of #(RT)/(barV - b)# increases, thereby decreasing the pressure #P# acquired from the van der Waals equation than from the ideal gas law.
  • For a gas that is harder to compress than the ideal form, with a larger #barV#, the magnitude of #a/(barV^2)# decreases more than the magnitude of #(RT)/(barV - b)# decreases, thereby increasing the pressure #P# acquired from the van der Waals equation than from the ideal gas law.