Question

A weather balloon is inflated to a volume of 27.4 L at a pressure of 739 mmHg and a temperature of 30.2 ∘C . The balloon rises in the atmosphere to an altitude where the pressure is 370. mmHg and the temperature is -13.4 ∘C .

Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude (in L).

Answer 1

Use the ideal gas law: P₁V₁/T₁ = P₂V₂/T₂

P₁ = 739 mmHg = (739 mmHg)*(1 atm/760 mmHg) = 0.97237 atm (1 atm = 760 mmHg)

V₁ = 27.4 L.

T₁ = 30.2⁰C = (30.2 + 273) K = 303.2 K

P₂ = 370 mmHg = (370 mmHg)*(1 atm/760 mmHg) = 0.48684 atm

T₂ = -13.4⁰C = (-13.4 + 273) K = 259.6 K.

Plug in values to get V₂.

(0.97237 atm)*(27.4 L)/(303.2 K) = (0.48684 atm)*V₂/(259.6 K)

===> V₂ = (0.97237 atm)*(27.4 L)*(259.6 K)/(0.48684 atm).(303.2 K) = 46.85666 L ≈ 46.9 L.

The balloon can expand to a volume of 46.9 L (ans).