A 3.2 liter sample of a gas is at 40 oC and 1.0 atmosphere of pressure. If the temperature decreases to 20 oC and the pressure decreases to 0.60 atmospheres, what is the new volume of the gas in liters?

V₂ = 5 L

Explanation

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given dа ta:

Initial volume = 3.2 L

Initial pressure = 1 atm

Initial temperature = 40 °C (40 + 273 = 313 K)

Final temperature = 20°C (20+273 = 293 K)

Final pressure = 0.60 atm

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂ / T₁ P₂

V₂ = 1 atm * 3.2 L * 293 K / 313 K * 0.60 atm

V₂ = 937.6 atm. L. K / 187.8 K. atm

V₂ = 5 L