Before it was launched, a helium-filled balloon had a pressure of 201 kpa at a temperature of 27°c. at an altitude of 15,000 m, the pressure had decreased to 2.5 kpa and the temperature had dropped to -14 °c. the volume of the balloon increased to 59.3 m3. what is the original volume of the balloon?

0.99 m³  

Step-by-step explanation:

This looks like a case where we can use the Combined Gas Law to calculate the volume

p₁V₁/T₁ = p₂V₂/T₂               Multiply both sides by T₁

   p₁V₁ = p₂V₂T₁/T₂             Divide each side by p₁

      V₁ = V₂ × p₂/p₁ × T₁/T₂  

Data:

p₁ = 201    kPa; V₁ = ?;             T₁ =  27 °C = 300.15 K

p₂ =    2.5 kPa; V₂ = 59.3 m³; T₂ = -14 °C = 223.15 K

Calculation:

V₁ = 59.3 × 2.5/201 × 300.15/223.15  

V₁ = 20 × 0.0124 × 1.345  

V₁ = 0.99 m³  

a:

by the combination of an unlimited number of elements

the dependant variable is spilled water

the independent is the heat