# How can I derive the Van der Waals equation?

##### 1 Answer

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)# .

**and** **depend on the gas itself** .

For example, *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*

Therefore, the first half becomes:

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

Now we examine **different extent of the attractive intermolecular forces** present in each gas. Basically, when accounting for that we subtract

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

If we note that the **compressibility factor** **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.