# Question 8b3ee

Jun 10, 2017

$\text{3 L}$

#### Explanation:

You know that the temperature and the number of moles of gas are being kept constant, so you can calculate the volume by using Boyle's Law.

According to Boyle's Law, the pressure and the volume of a gas have an inverse relationship when the temperature and the number of moles of gas, i.e. the amount of gas, are kept constant.

In other words, when volume goes down, pressure goes up and when volume goes up, pressure goes down.

In your case, the pressure is increasing

$\text{123 kPa " < " 207 kPa}$

so you should expect the volume to decrease

${\text{5 L " > " V}}_{2}$

You can find the new volume of the gas by using this equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{P}_{1} {V}_{1} = {P}_{2} {V}_{2}}}}$

Here

• ${P}_{1}$ and ${V}_{1}$ represent the pressure and volume of the gas at an initial state
• ${P}_{2}$ and ${V}_{2}$ represent the pressure and volume of the gas at a final state

Rearrange to solve for ${V}_{2}$

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

Plug in your values to get

V_2 = (123 color(red)(cancel(color(black)("kPa"))))/(207color(red)(cancel(color(black)("kPa")))) * "5 L" = "2.97 L"#

Rounded to one significant figure, the number of sig figs you have for the initial volume of the gas, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{V}_{2} = \text{3 L}}}}$

As predicted, we got

$\text{5 L " > " V"_2 = "3 L}$