# Acontainer originally contains 34 moles of oxygen gas at a volume of 2.4 L. If the container has 12 more grams of oxygen gas pumped into it, what will be in the new volume of the container? Assume constant temperature and pressure.

##### 1 Answer
Sep 12, 2016

$\text{2.43 L}$

#### Explanation:

The first thing to do here is to convert the mass of oxygen as, ${\text{O}}_{2}$, to moles by using oxygen's molar mass

12 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.375 moles O"_2

This will bring the number of moles of oxygen gas present in the container to a total fo

${\text{34 moles " + " 0.375 moles" = " 34.375 moles O}}_{2}$

Now, the volume of a gas and the number of moles of gas it contains have a direct relationship as given by Avogadro's Law.

This means that when temperature and pressure are kept constant, increasing the number of moles of gas present in a container will cause its volume to increase as well.

If you take ${V}_{1}$ and ${n}_{1}$ to be the volume and number of mole of gas present in the container at an initial state and ${V}_{2}$ and ${n}_{2}$ their corresponding values at a final state, you can say that

${V}_{1} / {n}_{1} = {V}_{2} / {n}_{2}$

Rearrange to solve for ${V}_{2}$, the volume of the gas after the mass of oxygen is added to the container

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

Plug in your values to find

${V}_{2} = \left(34.375 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(34color(red)(cancel(color(black)("moles")))) * "2.4 L" = color(green)(bar(ul(|color(white)(a/a)color(black)("2.43 L}} \textcolor{w h i t e}{\frac{a}{a}} |}}\right)$

I'll leave the answer rounded to three sig figs, but keep in mind that your values only justify using two sig figs for the answer.