atmospheric pressure

Question

A balloon is partially inflated with 5.00 L of helium at sea level where the atmospheric pressure is 1.00 atm. The balloon ascends to an altitude of 1600 m where the pressure is 0.83 atm. (a) What is the volume of the balloon at the higher altitude if the temperature of the helium does not change during the ascent? (b) What is the percent increase in volume?

Answer 1

Temperature is constant and the amount of helium is constant, so this is Boyle's law:

P1 V1 = P2 V2

(a) Volume at altitude

V2 = P1 V1 / P2 = (1.00 atm)(5.00 L) / (0.83 atm) = 6.0 L

The pressure units cancel, so there is no need to convert to pascals or torr - just keep both pressures in the same unit.

V2 = 6.0 L (the unrounded value is 6.024 L; 0.83 atm carries only two significant figures, so two is all you are entitled to report)

(b) Percent increase

percent increase = (V2 - V1) / V1 x 100 = (6.024 - 5.00) / 5.00 x 100

= 1.024 / 5.00 x 100 = 20% (20.5% before rounding)

Carry the unrounded 6.024 into this step rather than the rounded 6.0, and round only at the end. Rounding early here would have given exactly 20.0%, which looks more precise than it is.

A shortcut worth seeing

Part (b) never actually needed part (a). Since V2/V1 = P1/P2, the fractional change depends only on the pressure ratio:

percent increase = (P1/P2 - 1) x 100 = (1.00/0.83 - 1) x 100 = 20%

The 5.00 L cancels out entirely. That is a useful check - if your percent answer changes when you change the starting volume, something went wrong.

Does the direction make sense?

Yes. The balloon rises into thinner air, the outside pressure drops, and the helium pushes outward until the balloon expands enough to match. Lower pressure means larger volume, so V2 must exceed 5.00 L - and it does. If you had divided the other way you would have gotten 4.15 L, a shrinking balloon, which is physically backwards. Checking the direction before checking the arithmetic catches that instantly.

One assumption the problem is quietly making: the temperature really does stay constant. In reality air cools about 6.5 degrees Celsius per kilometer of altitude, so a real balloon at 1600 m would be roughly 10 degrees colder, and that cooling would partly offset the expansion. Problems say "if the temperature does not change" precisely so you can use Boyle's law alone instead of the combined gas law.