Chemistry Question

Question

A small weather balloon launched at sea level starts out at a volume of 1,000 liters at 25 degrees C and 1.00 atm of pressure. What is its volume when it reaches an altitude of 40,000 feet where the pressure is only 125 mmHg and the temperature is -56 degrees C?

Answer 1

Using the Ideal Gas Law, in a closed system (like a balloon, the number of moles of gas remains the same from time 1 (at sea level) to time 2 (at 40,000 ft).

So, for time 1: P₁V₁ = nRT₁ or P₁V₁/T₁ = nR

And for time 2: P₂V₂ = nRT₂ or P₂V₂/T₂ = nR

Since "nR" is the same for both times, then:

P₁V₁/T₁ = P₂V₂/T₂

We are looking for P₂, so:

V₂ = T₂P₁V₁/(P₂T₁)

Before plugging in the numbers, lets adjust the units. Make the temperatures absolute:

T₁ = 25 + 273.15 = 298.15 K

T₂ = -56 + 273.15 = 217.15 K

P₂ = (125 mmHg) x (1 atm/760 mmHg) = 0.16447 atm

So V₂ = (217.15•1.00•1000)/(0.16447•298.15) = 4428.2 liters I'll let you round to the appropriate number of sig figs.