# If the pressure exerted on a gas in a weather balloon decreases from 1.01 atm to 0.562 atm as it rises, by what factor will the volume of the gas in the balloon increase as it rises?

Jun 16, 2016

This is a simple manifestation of Boyle's Law The volume of the balloon should almost DOUBLE.

#### Explanation:

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$ at constant temperature.

Thus ${V}_{2}$ $=$ $\frac{{P}_{1} {V}_{1}}{P} _ 2$

$=$ $\frac{1.01 \cdot a t m \times {V}_{1}}{0.562 \cdot a t m}$

$\cong$ $2 {V}_{1}$ as required.

Note that given Boyle's Law, the proportionality of $P$ and $V$ at constant $T$, I am quite justified in using non-standard units such as atmospheres, and litres. I could even use $\text{pounds per square inch}$ and $\text{pints}$ and $\text{gallons}$, and I presume old Boyle did use these units. When these problems are applied to other situations, the utility of using standard SI units becomes more obvious.