Question

You are trying to determine the volume of the balloon needed to match the density of the air in the lab. You know that if you can get the balloon's density below this value, it will float. You measure both the temperature in the room as well as the mass of the balloon you will use and find they are 23.5°C and 0.587 grams, respectively. What volume of balloon will generate a density equal to that of the air around it? (Hint: Use the relationship you calculated above to determine the air density at the temperature given)

Answer 1

The equation to be used are:

PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles

The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.

PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa

n = 0.587 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.02024 mol
(101,183.9 Pa)V = (0.02024 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.931×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.931×10⁻⁴ m³ * 1000
V = 0.493 L