5.00 L of a gas is initially collected at 298 K and 2.3 atm. what will be the new volume if conditions are then changed to STP conditions?

Answer 1

P₁V₁ = n₁R₁T₁

P₂V₂ = n₂R₂T₂

n₁R₁ = n₂R₂

(P₂V₂)/(P₁V₁) = T₂/T₁

solve for V₂ = (T₂/T₁) (P₁V₁/P₂) This is the equation we need.

STP is 1 ATM at 273.15 K

so, T₂= 273.15 K, T₁ = 298 K, P₁ = 2.3 atm, V₁ = 5.00 L, P₂ = 1 atm

V₂ = (T₂/T₁) (P₁V₁/P₂) = (273.15/298)(2.3*5.00/1) L = 10.5 L