A basketball is inflated to a pressure of 1.50 atm in a 20.0°C garage. What is the pressure of the basketball outside
where the temperature is -5.00°C?

P = 1.37 atm

Explanation: For this problem we will use the Gay Lussac Law which is P1/T1 = P2/T2 then derive for P2 which is the outside pressure. Also remember to convert the units of temperature from °C to Kelvin.

T1 = 20°C + 273 = 293 K

T2 = -5°C + 273 = 268 K

P2 = P1 T2 / T1

= 1.50 atm ( 268 K ) / 293 K

= 1.37 atm

Cancel out the units of K so that the remaining unit will be in atm.

P₂ = 1.4 Atm

Explanation:

Volume is assumed to be constant => Gay-Lussac Law => P ∝ T (Decreasing Temperature => Decreasing Pressure expected).

=> Empirical Relationship => P₁/T₁ = P₂/T₂

P₁ = 1.5Atm                    P₂ = ?

V₁ = V₂                           V₂ = V₁

T₁ = 20.0°c (=293K)       T₂ = -5°C (=268K)

P₁/T₁ = P₂/T₂ => P₂ = P₁(T₂/T₁) = 1.5Atm(268K/293K) = 1.4Atm

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