We have Ideal gas constant for ideal gases. But ideal gas is very hard to make or to be found. So we have real gases in nature. Now there is 'real gas equation' (also called 'Van Der Waal's equation') for real gases which is given by:

$$(p + a/V^2)(V-b) = nRT$$

Here a and b are constant parameters that are determined empirically for each gas. Also 'a' tells you about attraction of molecules and 'b' tells you about volume covered by the molecules of gas.
Now couldn't we just change the value of gas constant to meet the ideal gas equation. Why there is need for two constants if we could just change the value of the constant which was already there.

  • 3
    $\begingroup$ ...because changing just $R$ doesn't give the correct behaviour? Just hold the r.h.s., i.e. the temperature, constant. Surely you see that the van der Waals equation gives a different relation between $p$ and $V$ than just the ideal gas law with another $R$. $\endgroup$
    – ACuriousMind
    Jan 7, 2016 at 19:03

1 Answer 1


You mentioned the answer yourself, in a way. $a$ and $b$ are corrections to two different terms and have an entirely different origin and function. Suppose you want a gas to have pointlike particles but having some degree of attraction between the particles. So only $a$ is required. Now if you want them to have finite size but no attraction, then only $b$ is required. In case you want to have both these features, you need to tae into account both of these parameters.

Now in your question, you want to change the value of the ideal gas constant in order to account for the real behaviour. The answer is you can't. Holding the temperature to be constant, you can compress the ideal gas to zero volume, and similarly your "modified $R$ real gas" to the same volume too. But we know that is not the answer, the real gas can be compressed only upto a finite volume. So we need $b$ and a shift in $R$ won't do the job. Hence we can't model real gas behaviour just by changing $R$ A similar argument can be made for $a$ too.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.