Question

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?

Answer 1

I got `25.2 \ "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 \ "g"` of `"He"` into moles.

`"He"` has a molar mass of `4 \ "g/mol"`.

So, `0.86 \ "g of He"` will be `(0.86cancel"g")/(4cancel"g""/mol")=0.215 \ "mol of He"`

Now, if we remove `0.205 \ "g of He" = 0.05125 \ "mol of He"`,

the new amount becomes `0.215 - 0.05125=0.16375 \ "mol of He"`

Plugging in the values, we get

`(19.2 \ "L")/(0.215 \ "mol")=(V_2)/(0.16375 \ "mol")`

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

Answer 2

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