The Volume of a certain gas is 29.3 liters at STP. What would be the volume of the same gas at 1.39 atmospheres and -23.0oC? (R= 0.0821 L. atm/mol. k)

V₂ =19.30 L

Explanation:

Given data:

Initial volume of gas = 29.3 L

Initial temperature = standard = 273 K

Initial pressure = standard = 1 atm

Final volume = ?

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

Final pressure = 1.39 atm

Solution:

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 =19.30 L

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .