A small hot-air balloon is filled with 1.02*106 l of air (d = 1.20 g/l). as the air in the balloon is heated, it expands to 1.09*106 l. what is the density of the heated air in the balloon?

The first thing you should know for this case is that density is defined as the quotient between mass and volume.

d = m / v

We have two states:

State 1:

d1 = 1.20 g / l

v1 = 1.02 * 106 l

State 2:

v2 = 1.09 * 106 l

Since the mass remains constant, then:

m = d1 * v1

Then, the density in state two will be:

d2 = m / v2

Substituting the value of the mass we have:

d2 = (d1 * v1) / v2

Substituting the values:

d2 = ((1.20) * (1.02 * 10 ^ 6)) / (1.09 * 10 ^ 6) = 1.12 g / l

answer:

The density of the heated air in the balloon is 1.12 g / l

1.12 g/L

The total mass of the air will remain constant, but since the volume changes and density is defined as mass per volume, we can simply calculate the new density of the heated air.

variables

d0, d1 = density cold, density hot

m = mass of air

v0, v1 = volume cold, volume hot

d0 = m/v0 = 1.20 g/L

d1 = m/v1

m/v0 = 1.20 g/L

m = v0 * 1.20 g/L

m/v1 = v0 * 1.20 g/L / v1

d1 = v0 * 1.20 g/L / v1

d1 = 1.02x10^6 * 1.20 g/L / 1.09x10^6

d1 = 1.02x10^6 * 1.20 g/L / 1.09x10^6

d1 = 1.12 g/L

So the density of the heated air is 1.12 g/L