Question

A balloon has a volume of 1.34 liters with 2.00 moles of helium gas. How many moles of helium gas must be added to get the volume to 7.00 liters.

Answer 1

Avogadro's law
$\frac{ V1}{V2} = \frac{n1}{n2} \\ \frac{1.34}{7} = \frac{2}{x} \\ x = \frac{7 \times 2}{1.34} \\ x = 10 \: mol$

Answer 2

Answer:

8.4 mol.

Explanation:

The pressure and temperature are constant. The only variables are the volume and number of moles, so we can use Avogadro’s Law:

The volume of a gas is directly proportional to the number of moles.

V = kn

or

V₂/V₁ = n₂/n₁     Multiply both sides by n₁ and transpose

n₂ = n₁ × V₂/V₁

n₁ = 2.00 mol; V₁ = 1.34 L

n₂ = ? mol;       V₂ = 7.00 L     Solve for V₂

n₂ = 2.00 × 7.00/1.34

n₂ = 2.00 × 5.223

n₂ = 10.4 mol

===============

The balloon originally held 2.00 mol of helium. It now holds 10.4 mol.

Moles added = 10.4 – 2.00

Moles added = 8.4 mol