To a good approximation the atmosphere behaves as an ideal gas, with each mole of gas obeying the law:


We can obtain the corresponding law for unit mass of air using the density $\rho$.

I was wondering why, in this case, the parameter density is so relevant ?

Maybe 'cause here the purpose is to study gases with volumes that are difficult to quantify, so we can make use of calculations employing the physical property of density, defined as the mass of the gas per unit volume, to derive a form of the ideal gas equation that has broader applications.


2 Answers 2


You are right: if the gas you are studying is not in a container, it is difficult to attribute a volume to it.

The key here is to realize that on the scale of the atmosphere, temperature, pressure and density change - by a lot. So you can't think of "all of the atmosphere" as a single body of air with uniform properties - the properties change locally. As such, you want to express the behavior in terms of "local" properties only. This makes volume not a suitable candidate - but temperature, pressure and density are all locally defined, so ideal for describing such a system. As @t.c. points out n another answer, the formal name for such "locally defined" properties is "intensive", to contrast with "extensive" properties which apply to a system. Note that while the wiki entry says

An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system.

that doesn't mean it cannot change with location in the system - and indeed in the atmosphere they change a lot, which is why they are useful for describing the system.

Per David Hammen's suggestion, going from the ideal gas law formulation:

$$pV = nRT$$

to the formulation in terms of density, you replace $n$ with $\frac{m}{M}$, then divide both sides by $V$:

$$p = \frac{m}{V} \frac{R}{M} T$$

We recognize $\frac{m}{V}$ as the density $\rho$, and $\frac{R}{M}$ as the specific gas constant, sometimes written as $R^\ast$. This leads to

$$p = \rho R^\ast T$$

Now all parameters ($p, \rho, T$) are intensive quantities.

  • 1
    $\begingroup$ +1. A suggestion for improvement: Show how dividing both sides of the ideal gas law $pV=nRT$ by volume $V$ results in an alternative expression of the gas law $p=\rho R^{\ast}T$ in which all of the parameters are now intensive. $\endgroup$ Commented Sep 27, 2014 at 20:05
  • $\begingroup$ @DavidHammen - good suggestion. Done. $\endgroup$
    – Floris
    Commented Sep 28, 2014 at 0:12

Because Pressure and Temperature are intensive properties. Intensive properties do not depend on mass or number of moles.

Whereas Volume is a extensive property, so it is divided by n to make it an intensive property (often to facilitate calculations).



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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