# Which of these graphs would be a pure isotherm? A V vs. n graph or a P vs. V graph?

## Why is it not the $V$ vs. $n$ graph? It was made at constant temperature, wasn't it?

Jul 26, 2017

It can be, but it is also an isobar, so it is not just at constant temperature, but also constant pressure. It is also a mechanically-open system, i.e. it allows transfer of gas in and out so that $n$ varies, changing the size of the container (e.g. a balloon).

On the other hand, the $P - V$ graph has constant mols of ideal gas and constant temperature, i.e. it follows Boyle's law:

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, constant $n$ and $T$

We don't really think about the constant mols of gas in this case, as that can easily be accomplished with a closed system containing a single gas, making this a pure isotherm.

But for the $V - n$ graph, according to Avogadro's principle:

${V}_{1} / {n}_{1} = {V}_{2} / {n}_{2}$, constant $T$ and $P$

So, the $V - n$ graph is not a pure isotherm, but it does involve constant temperature (and pressure). In this case, mols of gas can vary, and thus the system is not mechanically-closed.