# Flexible plastic container with 0.86 g of 19.2 L helium gas. If you remove 0.205 grams of helium gas in continuous pressure and temperature, what is the new size?

Feb 17, 2018

I got $25.2 \setminus \text{mol}$ of helium gas.

#### Explanation:

This is a case of Avogadro's law, that is

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

We first need to convert $0.86 \setminus \text{g}$ of $\text{He}$ into moles.

$\text{He}$ has a molar mass of $4 \setminus \text{g/mol}$.

So, $0.86 \setminus \text{g of He}$ will be (0.86cancel"g")/(4cancel"g""/mol")=0.215 \ "mol of He"

Now, if we remove $0.205 \setminus \text{g of He" = 0.05125 \ "mol of He}$,

the new amount becomes $0.215 - 0.05125 = 0.16375 \setminus \text{mol of He}$

Plugging in the values, we get

$\left(19.2 \setminus \text{L")/(0.215 \ "mol")=(V_2)/(0.16375 \ "mol}\right)$

:.V_2=(19.2 \ "L" * 0.16375cancel "mol")/(0.125cancel "mol")=25.152 \ "mol"~~25.2 \ "mol"

Feb 17, 2018

14.62 liters

#### Explanation:

The requested answer was the new size reached by the flexible container (presumably a balloon)
Removing 0.205 grams of helium brings the mass to 0.655 grams.
By a simple proportion:
0.86 g : 0.655 g = 19.2 L : x L
x= 14.62
The new size of the container is 14.62 L
No need to calculate moles number.
The above calculation holds true for every gas