How would the van der Waals equation of state look like when molar volume of gas is high? Does high molar volume correspond to low pressure?

My textbook says that

A real gas behaves similar to an ideal gas in the limit of large molar volumes.

What is the reason behind that?

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    $\begingroup$ What does the van Der Waals equation look like in the limit of $P<<\frac{a}{V^2}$ and $V>>b$? $\endgroup$ Feb 14, 2019 at 14:28
  • 1
    $\begingroup$ Sorry. I meant $P>>\frac{a}{V^2}$. $\endgroup$ Feb 14, 2019 at 18:34

1 Answer 1


The ideal gas equation of state assumes all of the particles are not interacting. This is a bad assumption when particles are close together, such as in a liquid. However as particles get further apart this becomes a more realistic model since particles collide less and less often.

In general low pressure and high temperature leads to conditions where the ideal gas equation will work. We see this in combustion reactions actually. You can model a gasifier fairly well with the ideal gas equation of state because of how high the temperatures are.

At high temperatures, molecules move with higher energy, so they bounce of walls harder. Thus to attain a given pressure, you need fewer molecules if they are hitting the walls harder. Thus, at a higher temperature, there are fewer molecules in your vessel, making them further apart, which makes them more ideal.

In terms of molar volumes (I prefer pressure but won't begrudge your textbook), yes as the molar volume gets bigger that means the same number of particles take up more space, meaning they are farther apart from each other. In the extreme limit they will become infinitely far apart and be exactly like an ideal gas. In practice the molar volume does not have to be extreme for the ideal gas to start modeling the system well.


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