# What is the new volume of the gas in a "33.0-L" balloon that rises from an altitude with a pressure of "100.4 kPa" into the stratosphere where the pressure is "21.8 kPa"?

May 13, 2017

The Hard-Working Chemist is getting a little tired of seeing all these gas law questions, but presses on.

Seeing the volume $V$ and pressure $P$ in one sentence, and the absurd assumption that the temperature remains constant at such a high altitude, he assumes that the ideal gas law is, as usual, the go-to equation.

$P V = n R T$

He recognizes at once that the mols of gas are assumed constant, and thus that he can write two states of the gas:

${P}_{1} {V}_{1} = n R T$
${P}_{2} {V}_{2} = n R T$

With a steady eye, he constructs the relation now known as Boyle's law

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$,

and determines the final volume of the gas to be

$\textcolor{b l u e}{{V}_{2}} = \left({P}_{1} / {P}_{2}\right) {V}_{1}$

$= \text{100.4 kPa"/"21.8 kPa" xx "33.0 L}$

$=$ $\textcolor{b l u e}{\text{152 L}}$

As a physical chemist, he considers for a moment whether he made any sense at all, or if the calculation was complete hogwash.

He concludes that since the pressure has decreased, it is apparent the volume must have increased, so that the forces of the gas particles on the balloon's inside surface are less frequent and less powerful.

He, however, doubts that the balloon would have survived the climb, unless the temperature decrease were to be accounted for.