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

"2.43 L"

Explanation:

The first thing to do here is to convert the mass of oxygen as, "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

"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 * V_1

Plug in your values to find

V_2 = (34.375 color(red)(cancel(color(black)("moles"))))/(34color(red)(cancel(color(black)("moles")))) * "2.4 L" = color(green)(bar(ul(|color(white)(a/a)color(black)("2.43 L")color(white)(a/a)|)))

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.