Question

A small hot-air balloon is filled with 1.00×106 L of air (d = 1.20 g/L). As the air in the balloon is heated, it expands to 1.12×106 L. What is the density of the heated air in the balloon?

Answer 1

The volume of air in the balloon is
V = 1 x 10⁶ L.
The density is
d = 1.20 g/L = 1.2 x 10⁻³ kg/L.

The mass of air is
m = V*d = (1 x 10⁶ L)*(1.2 x 10⁻³ kg/L) = 1.2 x 10³ kg

When the balloon s heated and the volume of air expands to 1.12 x 10⁶ L, the mass of air remains the same.
The new density, D, is given by
(1.2 x 10³ kg) = (D kg/L)*(1.12 x 10⁶ L)
D = (1.2 x 10³)/(1.12 x 10⁶) = 1.07 x 10⁻³ kg/L

Answer:  1.07 x 10⁻³ kg/L (or 1.07 g/L)