Question

5. A balloon used for atmospheric research has a volume of 1.0 x 106 L. Assume that the balloon is filled with helium gas at STP (standard temperature and pressure) and then allowed to ascend to an altitude of 10 km, where the pressure of the atmosphere is 243 mm Hg and the temperature is -33°C. What will the volume of the balloon be under these atmospheric conditions? (Hint: Three variables have been given so what equation will you use?)

Answer 1

The new volume will be 2,749,512.5 L

Step-by-step explanation:

Step 1: Data given

Volume of the balloon = 1.0 * 10^6 L

At STP = 1 atm, 273 K

Pressure = 243 mmHg = 0.319737 atm

Temperature = -33 °C = 240 K

Step 2: Calculate the volume

p1*V1 / T1 = p2*V2 /T2

⇒with p1 = the initial pressure = 1 atm

⇒with V1 = the initial volume 1* 10^6 L

⇒with T1 = the initial temperature = 273 K

⇒with p2 = the new pressure = 0.319737 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the new temperature = 240 K

1 * 10^6 / 273 = 0.319737 * V2 / 240

V2 = 2749512.5 L

The new volume will be 2,749,512.5 L

Answer 2

The volume of the balloon under these atmospheric conditions will be 3x10⁶ L

Step-by-step explanation:

Given data:

Initial pressure of a balloon P₁ = 760 mmHg

Initial volume of a balloon V₁ = 1x10⁶ L

Initial temperature of a balloon T₁ = 0°C + 273 = 273 K

Final pressure of a balloon, P₂ = 243 mm Hg

Final temperature of a balloon T₂ = -33°C + 273 = 240 K

Therefore,

Initial conditions: P₁ = 760 mm Hg, V₁ = 1x10⁶ L, T₁ = 273 K

Final conditions: P₂ = 243 mm Hg, V₂ = ?, T₂ = 240 K

according to combine gas law: P₁V₁ / T₁ = P₂V₂ / T₂

V₂ = P₁V₁T₂ / T₁P₂

Substituting and solving for the parameters

V₂ = 3x10⁶ L