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 dа ta:

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