# On a PV diagram, suppose we place volume on the y axis; is the slope negative or positive?

Jul 21, 2017

Negative... and nonconstant.

Well, you can start from Boyle's law:

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

at constant temperature and mols of gas. Hence,

${V}_{2} = {P}_{1} / {P}_{2} {V}_{1}$

If ${P}_{2} > {P}_{1}$, then we see that the volume decreases from left to right, as it should. That clearly eliminates $c$ and $d$.

The slope would be given by:

$\text{slope} = \frac{\Delta V}{\Delta P} \equiv \frac{{V}_{2} - {V}_{1}}{{P}_{2} - {P}_{1}}$

Now we just need to show that this slope is NOT a constant. Try subtracting ${P}_{1} {V}_{2}$ from Boyle's law:

${P}_{1} {V}_{1} - {P}_{1} {V}_{2} = {P}_{2} {V}_{2} - {P}_{1} {V}_{2}$

$\implies {P}_{1} \left({V}_{1} - {V}_{2}\right) = {V}_{2} \left({P}_{2} - {P}_{1}\right)$

$\implies - {P}_{1} \Delta V = {V}_{2} \Delta P$

$\implies \textcolor{b l u e}{\overline{\underline{| \stackrel{\text{ ")(" "(DeltaV)/(DeltaP) = -V_2/P_1" }}{|}}}}$

The slope is negative, which makes sense. If pressure increases at constant temperature and mols of gas, the volume should compress.

Since ${P}_{1}$ does not ever change, but ${V}_{2}$ will decrease as ${P}_{2}$ increases, we can see that the slope becomes less and less negative, and is surely NOT constant. In fact, it starts at a high negative and becomes a small negative, not unlike a $1 / x$ curve in quadrant I.

Clearly, it means the slope decreases over time and is negative.