Question

A balloon now occupies 9.7L at 25.0°C and 1.00 atm. What temperature was it initially, if it occupied 9.0 L and was in a freezer with a pressure of 0.92 atm?

Answer 1

-19°C (2 sig.figs.)

Step-by-step explanation:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = 1.00atm

V₁ = 9.7L

T₁ = 25°C = 298K

P₂ = 0.92atm

V₂ = 9L

T₂ = ?

T₂ = T₁P₂V₂/P₁V₁

= 298K(0.92atm/1.00atm)(9.0L/9.7L) = 254K = (254 - 273)°C = -19°C (2 sig.figs.)