A small hot-air balloon is filled with 1.03×106 L of air (d = 1.20 g/L).

Question

As the air in the balloon is heated, it expands to 1.15×10⁶ L. What is the density of the heated air in the balloon?

Answer 1

The relationship you need is: density = ^mass / _volume or D = ^m/_V

If you have values for two of the three variables in the equation, you can solve for the third:

You are given: V = 1.03 x 10⁶ L and D = 1.20 g/L

You can solve for mass by rearranging the eq'n: m = D x V = 1.20 ^g/_L x 1.03x10⁶ L = 1.24 x 10⁶ g

Given a new volume: V = 1.15 x 10⁶ L and the mass just calculated, you can solve for the new density:

D = ^m /_V = ^{1.24 x 106 g} /_{1.15x10^6 L} = 1.08 ^g/_L