Ideal Gas Law — Moles in a Party Balloon (Worked Example)

Medium

Problem

A party balloon holds 10.0 L of helium at 300 K and 1.00 atm (101325 Pa). How many moles of helium are inside?

The helium inside the balloon: P, V and T linked by the ideal gas law.

Given

V = 10.0 L = 0.0100 m³
T = 300 K
P = 101325 Pa
R = 8.314 J/(mol·K)

Find

Amount of gas n in moles.

Principle

The ideal gas law links pressure, volume, temperature and amount of gas in a single equation of state.

Formula

PV = nRT \quad\Rightarrow\quad n = \frac{PV}{RT}

PV = nRT → n = PV/RT

Solution

Convert volume: $10.0\ \mathrm{L} = 0.0100\ \mathrm{m^3}$
Rearrange: $n = PV/RT = (101325 \times 0.0100)/(8.314 \times 300)$
Numerator: $1013.25$ ; denominator: $2494.2$
Divide: $n = 1013.25/2494.2 = 0.406\ \mathrm{mol}$

≈ 0.406 mol

Unit check

(Pa·m³)/(J/(mol·K)·K) = mol. Correct — since 1 J = 1 Pa·m³.

What this means physically

Less than half a mole — about 1.6 g of helium (0.406 × 4.00 g/mol). A balloon feels light because it holds so little mass, yet that gas still pushes outward with a full atmosphere of pressure.

Common mistake

Leaving volume in liters while R is in SI units. Convert L → m³ (÷1000) first, or the answer will be off by a factor of 1000.