Question

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?

Answer 1

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

Explanation:

`P_1V_1=P_2V_2` at constant temperature.

Thus `V_2` `=` `(P_1V_1)/P_2`

`=` `(1.01*atmxxV_1)/(0.562*atm)`

`~=` `2V_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 `"pounds per square inch"` and `"pints"` and `"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.