Moles of Helium in a Balloon (Ideal Gas Law)

Given

• Balloon volume: V = 2.00 L

• Pressure: P = 1.00 atm

• Temperature: T = 298 K (25 degC + 273.15)

• R = 0.08206 L atm / (mol K)

• Molar mass of helium: M(He) = 4.0026 g/mol

Find

The amount of helium in moles, and its mass in grams.

Principle

The ideal gas law PV = nRT relates the measurable properties of a gas — pressure, volume, and temperature — to the moles present, using the gas constant R as the bridge. It assumes gas particles are point masses with no attraction, which is a fine approximation for helium at ordinary conditions. Solving for n turns three measurements into a mole count.

Formula

Solution

1. Write the ideal gas law and solve for n: n = (P x V) / (R x T).

2. Substitute: n = (1.00 atm x 2.00 L) / (0.08206 L atm mol^-1 K^-1 x 298 K).

3. Compute the denominator: 0.08206 x 298 = 24.454 L atm mol^-1.

4. n = 2.00 / 24.454 mol = 0.0818 mol of He.

5. Convert to mass: m = 0.0818 mol x 4.0026 g/mol = 0.327 g of He.

Unit check

(atm x L) / ((L atm mol^-1 K^-1) x K) simplifies to mol — every unit except moles cancels. Multiplying mol by g/mol leaves grams for the final mass. The unit chain is exact.

What this means physically

A whole party balloon holds less than a tenth of a mole of helium — about a third of a gram, lighter than a paperclip. Gas moles pack loosely: at these conditions each mole fills roughly 24.5 L, which is why balloons are mostly empty space.