# Question e1c04

Nov 20, 2015

Here's how you can do that.

#### Explanation:

As you know, Boyle's Law states that pressure and volume have an inverse relationship when temperature and number of moles are kept constant.

Mathematically, Boyle's Law can be written like this

$\textcolor{b l u e}{{P}_{1} {V}_{1} = {P}_{2} {V}_{2}} \text{ }$, where

${P}_{1}$, ${V}_{1}$ - the pressure and volume at an initial state
${P}_{2}$, ${V}_{2}$ - the pressure and volume at a final state

Now, regardless of what the problem asks you to determine, this will always be your starting equation when it comes to Boyle's Law problems.

To set up the equation for any of the four terms, you can use classic algebraic manipulation. For example, to solve for ${P}_{2}$, the new pressure of the gas, you would need to isolate this temr on one side of the equation.

To do that, divide both sides by ${V}_{2}$

$\frac{{P}_{1} \cdot {V}_{1}}{V} _ 2 = {P}_{2} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{2}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{2}}}}}$

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

Now, if is very important to keep track of your units. In this case, both volumes are given in mililiters, so you don't need to do any conversions here.

The pressure will come out to have the same units as the initial pressure, which in this case is measured in mmHg.

So, plug in your values and solve for ${P}_{2}$

P_2 = (55.0color(red)(cancel(color(black)("mL"))))/(30.0color(red)(cancel(color(black)("mL")))) * "750 mmHg" = "1375 mmHg"#

You would then round this value off to two sig figs, the number of sig figs you have for the initial pressure of the gas

${P}_{2} = \textcolor{g r e e n}{\text{1400 mmHg}}$

SIDE NOTE The exact same approach applies when solving for volume. Let's say that you know the two pressure and the final volume.

To find ${V}_{1}$, you would rearrange the equation like this

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{1}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{1}}}}} {V}_{1} = \frac{{P}_{2} {V}_{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{1}}}}}$

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