Question

Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to double its volume? Assume no change in temperature or pressure. g

Answer 1

Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

`\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=(6.24g)/(4g/mole)=1.56moles`

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

`V\propto n`

or,

`(V_1)/(V_2)=(n_1)/(n_2)`

where,

`V_1` = initial volume of gas = V

`V_2` = final volume of gas = 2V

`n_1` = initial moles of gas = 1.56 mole

`n_2` = final moles of gas = ?

Now we put all the given values in this formula, we get

`(V)/(2V)=(1.56mole)/(n_2)`

`n_2=3.12mole`

Now we have to calculate the mass of helium gas at doubled volume.

`\text{Mass of }He=\text{Moles of }He* \text{Molar mass of }He`

`\text{Mass of }He=3.12mole* 4g/mole=12.48g`

Therefore, the mass of helium gas added must be 12.48 grams.