# Question 25589

Apr 14, 2014

Boyle's law developed by Robert Boyle in 1662, states that if we keep the temperature of a gas constant in a sealed container. Its pressure (P) varies inversely with its volume (V). In other words at any given temperature. If we pressurize a gas its volume will be reduced proportionately to the pressure change. If we increase the volume of a gas its pressure will increase.

I am not sure what you mean by the steps, but I can help you identify a Boyle's Law problem, and tell you how to solve it.

Boyle's Law is that Pressure, P is inversely related to volume, V at constant temperature and number of moles. The mathematical relationship is
$P = k \cdot \frac{1}{V}$
which rearranges to

$P \cdot V = k$,

where $k$ is a constant.

Since $P V = k$, you get ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$.

You can tell a problem will be a Boyle's law problem, if there is a change in pressure or volume and it asks you about the other.

For example, you could be asked the following:
If a gas at atmospheric pressure, has its container expand to three times its volume, what is the new pressure.

Solution
${P}_{1} = 760 m m H g$
${V}_{1} = {V}_{1}$

P_2 = ?#
${V}_{2} = 3 {V}_{1}$ because the new volume is 3 times bigger.

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$ Boyle's Law

${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2$ Divide both sides by ${V}_{2}$

${P}_{2} = \frac{\left(760 m m H g\right) {V}_{1}}{3 {V}_{1}}$ Substitute givens.

${P}_{2} = \frac{760 m m H g}{3} = 253.3 m m H g$