# If a balloon containing 3000 L of gas at 39 C and 99 kPa rises to an altitude where the pressure is 45.5 kPa and the temperature is 16 C, the volume ofthe balloon under these new conditions would be calculated using what conversion factor ratios?

Feb 11, 2017

By the combined gas equation: $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$.

#### Explanation:

This assumes constant amount of gas. And we solve for ${V}_{2}$ to get:

${V}_{2} = \frac{{P}_{1} \times {V}_{1} \times {T}_{2}}{{T}_{1} \times {P}_{2}}$, which equation, clearly, has units of volume.

Do you agree?

So V_2=(99*kPaxx3000*Lxx289*K)/(312*Kxx45.5*kPa)=??L

So the volume has effectively doubled. Of course, this is probably a CLOSED weather balloon. I have never been in a hot-air balloon, but I suspect (and I may be wrong), that doubling the volume of a hot-air balloon might dangerously stretch the fabric. If there are any hot-air balloonists on these boards, their insight would be welcome.