Question

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?

Answer 1

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_1V_1 = P_2V_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.