# If given the following, what is the change in volume of the balloon as it ascends from sea level to 6000 ft?

## A certain flexible weather balloon contains helium gas at a volume of 855 L. Initially, the balloon is at sea level where the temperature is 25 C and the barometric pressure 730 torr. The balloon then ises to an a titude of 6000 ft, where the pressure is 605 torr and the temperature is 15 C.

Oct 11, 2017

The new volume is $980 L$
The change from the original volume is $125 L$

#### Explanation:

The altitude is irrelevant, except as needed to define the pressure. In this example all three parameters of the Ideal Gas Laws are changing. We set up the equation to solve for the desired final volume. Start with the Ideal Gas Law: $P \cdot V = \left(n \cdot R \cdot T\right)$ where the number of moles of gas and the gas constant do not change.
${\left(\frac{P V}{T}\right)}_{1} = {\left(\frac{P V}{T}\right)}_{2}$
Rearrange for the desired ${V}_{2}$ , Temperatures must be in ‘K: 298 & 283.
$\left(\frac{{P}_{1} {V}_{1}}{T} _ 1\right) \times {T}_{2} / \left({P}_{2}\right) = {V}_{2}$

${V}_{2} = \frac{{P}_{1} \times {V}_{1} \times {T}_{2}}{{T}_{1} \times {P}_{2}}$

${V}_{2} = \frac{730 \times 855 \times 283}{298 \times 605} = 980 L$

“Change” in volume is 980 – 855 = 125L