The volume of a certain gas is 29.3 L at stp what would be the volume of a gas at 1.39 atm and -23.0oc?

V₂ = 19.3 L

Explanation:

Given data:

Initial volume = 29.3 L

Initial pressure = standard = 1 atm

Initial temperature = standard = 273 K

Final temperature = -23°C (-23+273 = 250K)

Final volume = ?

Final pressure = 1.39 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1 atm × 29.3 L × 250 K / 273 K × 1.39 atm

V₂ = 7325 atm .L. K / 379.47  K.atm

V₂ = 19.3 L

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the ratio of the conjugate base to the acid required to attain the desired ph.