Question 29cee

Jan 7, 2017

A $P V$ vs. $P$ graph simply shows that the relationship ${P}_{1} {V}_{1} = {P}_{2} {V}_{2} \approx \text{const}$ holds. If it were really the case that $\left(0 , 0\right)$ is a point on the graph, then it would not matter what the volume is, as long as it is finite.

But Boyle's law states that volume must affect pressure if temperature and mols of gas are held constant.

In other words, you would then be ignoring the dependence of $V$ on $P$; if it were true that $P = 0$ implies $P V = 0$, then changes in volume would not affect the pressure under the conditions of Boyle's law (constant $T$ and $n$), which directly contradicts Boyle's law, or $0 \cdot \infty = 0$, which is mathematically incorrect.

I also graphed the data above myself and replicated the graph, so that at truly constant $n$ and $T$, we have the true values of $V$ for $n = \text{0.0025 mols}$ and $T = \text{298.15 K}$:

Really, how you should view $P V$ vs. $P$ is that as $P \uparrow$, $V \downarrow$ and vice versa. No matter how $P$ changes, $P V$ should not.

Thus, as $P \to 0$, $V \to \infty$ such that $P V = \text{const}$, i.e. ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$.

So, if we had:

${P}_{1} = \text{2 atm}$
${V}_{1} = \text{5 L}$
${P}_{2} = \text{0 atm}$

Then

${V}_{2} = {\lim}_{{P}_{2} \to 0} \left({P}_{1} / {P}_{2}\right) {V}_{1}$

but since ${P}_{2} = \frac{n R T}{{V}_{2}}$, we have that ${V}_{2} \to \infty$ when $n$ and $T$ are constant (as usual in Boyle's law). On the other hand, given a nonzero ${P}_{2}$, say, ${P}_{2} = \text{0.05 atm}$, we at least have a finite volume, which clearly increases as ${P}_{2}$ decreases:

${V}_{2} = \left(\text{2 atm")/("0.05 atm")("5 L}\right)$

$=$ $\text{200 L}$

If you repeat with smaller and smaller ${P}_{2}$, you can see the pattern:

For ${P}_{2} = \text{0.005 atm}$, V_2 = (("2 atm")/("0.005 atm"))("5 L")

$=$ $\text{2000 L}$

For ${P}_{2} = \text{0.0005 atm}$, V_2 = (("2 atm")/("0.0005 atm"))("5 L")#

$=$ $\text{20000 L}$

etc.