# Question c793c

Mar 6, 2017

$\text{3.3 moles}$

#### Explanation:

The thing to remember about a gas that is kept under constant temperature and pressure is that the volume it occupies is directly proportional to the number of moles present in the sample.

This is known as Avogadro's Law and can be written as

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

Here

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

In your case, both gases are kept under the same conditions for pressure and temperature, which means that the balloon that contains more moles of gas will also have the higher volume.

So, right from the start, you can say that the second balloon will contain fewer moles of gas, since

${V}_{2} < {V}_{1} \implies {n}_{2} < {n}_{1}$

To find the exact number of moles present in the second balloon, rearrange the equation for Avogadro's Law and solve for ${n}_{2}$

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

Plug in your values to find

n_2 = (14.9 color(red)(cancel(color(black)("L"))))/(23.5color(red)(cancel(color(black)("L")))) * "5.2 moles" = color(darkgreen)(ul(color(black)("3.3 moles")))#

The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of gas present in the first balloon.

As predicted, the number of moles of gas present in the second balloon is smaller than the number of moles of gas present in the first balloon.