A 6.60 L weather balloon at 1.00 atm and 289 K is released. If it shrinks to a size of 6.20 L at 0.93 atm, what is the temperature?

Answer 1

Based on the information given, you need to use the combined gas law P₁V₁ ⁄ T₁ = P₂V₂ ⁄ T₂

This law can also be written as P₁V₁T₂ = P₂V₂T₁

You are given information before (label 1) and information after. What is not clear is the unit for temperature, do they want in Kelvin or ^o C (Celsius) ?

Plugging in the information P₁= 1.00 atm; V₁= 6.60 L; T₁ = 298 K; P₂ = 0.93 atm; V₂ = 6.20 L,

You can calculate T₂... The answer will be in Kelvin... if needed in ^o C will need to do conversion...

Answer : T₂ = (0.93 atm x 6.20L x 298 K ) ⁄ (6.60 L x1.00 atm) = 260 K