A gas‑filled weather balloon with a volume of 59.0 L is held at ground level, where the atmospheric pressure is 752 mmHg and the temperature is 24.3 ∘ C . The balloon is released and rises to an altitude where the pressure is 0.0708 bar and the temperature is − 5.41 ∘ C . Calculate the weather balloon's volume at the higher altitude.

At the higher altitude, the new volume is 750.2L

Explanation:

If we decompose the Ideal Gases Law for the two situations (initial and final), we can determine this relaion:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

Number of moles does not change, and R stays the same.

Let's make some conversions, before:

752 mmHg . 1 atm / 760mmHg = 0.989 atm

24.3°C + 273 = 297.3K

0.0708 bar . 0.986 atm / 1bar = 0.070 atm

-5.41°C + 273 = 267.59 K

We replace: (0.989 atm . 59L) / 297.3K = (0.070 atm . V₂) / 267.59K

[(0.989 atm . 59L) / 297.3K] . 267.59K = 0.070 atm . V₂

52.5 atm.L = 0.070 atm . V₂

V₂ = 52.5 atm.L / 0.070atm = 750.2L