# Question #9f08e

Jan 30, 2016

Boyle's Law states that pressure is indirectly proportional to volume when the temperature and moles of the gas are constant.

This relationship can be modelled by the equation:

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

where:
${P}_{1} =$initial pressure
${V}_{1} =$initial volume
${P}_{2} =$final pressure
${V}_{2} =$final volume

When pressure is indirectly proportional to volume, this means that
say, for example, if the pressure was increased by a factor, the volume would decrease by the $\textcolor{red}{\text{same}}$ factor.

Similarly, if the volume was increased by a factor, the pressure would decrease by the $\textcolor{red}{\text{same}}$ factor.

For example, if the initial pressure was $\textcolor{b l u e}{2}$ $\textcolor{b l u e}{a t m}$, the initial volume was $\textcolor{b l u e}{4}$ $\textcolor{b l u e}{L}$, and the pressure decreased by a factor of $\textcolor{b l u e}{\frac{1}{2}}$, the volume would increase by a factor of $\textcolor{b l u e}{2}$.

Algebraically, we can solve for the final volume using Boyle's Law formula, assuming that the temperature and moles of the gas are constant:

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

$\left(2 a t m\right) \left(4 L\right) = \left(1 a t m\right) \left({V}_{2}\right)$

${V}_{2} = \frac{\left(2 \textcolor{red}{\cancel{\textcolor{b l a c k}{a t m}}}\right) \left(4 L\right)}{\left(1 \textcolor{red}{\cancel{\textcolor{b l a c k}{a t m}}}\right)}$

${V}_{2} = 8$ $L$

Thus, the volume has increased from $4$ $L$ to $8$ $L$.

Graphically, the relationship between pressure and volume can be represented as: