# Question a13a7

Aug 29, 2017

${\text{1.10 g L}}^{- 1}$

#### Explanation:

The trick here is to assume that the mass of air remains unchanged.

You know that you start with a volume of "1.03 * 10^6 $\text{L}$ of air and that the density of air before you heat it is equal ${\text{1.20 g L}}^{- 1}$.

You can use the density and the volume of the balloon to figure out how many grams of air are trapped inside.

1.03 * 10^6 color(red)(cancel(color(black)("L"))) * "1.20 g"/(1color(red)(cancel(color(black)("L")))) = 1.236 * 10^6color(white)(.)"g"

Now, when the air inside the balloon is heated, it expands to a volume of $1.12 \cdot {10}^{6}$ $\text{L}$. Right from the start, the fact that the volume of the balloon increases as the mass of air remains unchanged should tell you that the density of the gas will decrease as you heat it.

To find the density of the heated air, calculate the number of grams of air present in $\text{1 L}$ by using the fact that the same mass of air occupies $1.12 \cdot {10}^{6}$ $\text{L}$

1 color(red)(cancel(color(black)("L"))) * (1.236 * color(blue)(cancel(color(black)(10^6)))color(white)(.)"g")/(1.12 * color(blue)(cancel(color(black)(10^6))) color(red)(cancel(color(black)("L")))) = "1.1036 g"#

You can thus say that the density of the heated air is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{density heated air = 1.10 g L}}^{- 1}}}}$

The answer is rounded to three sig figs.

As you can see, the density of the air inside the balloon decreased as a result of the fact that the volume increased while the mass of air remained unchanged.