A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased to 50.0 °C?

Respuesta :

Answer:

3.31 atm.

Explanation:

  • Gay-Lussac's law states that for a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

∵ P α T.

∴ P₁T₂ = P₂T₁.

P₁ = 3.00 atm, T₁ = 20.0 °C + 273.15 = 293.15 K.

P₂ = ??? atm, T₂ = 50.0 °C + 273.15 = 323.15 K.

∴ P₂ = (P₁T₂)/T₁ = (3.00 atm)( 323.15 K)/(293.15 K) = 3.307 atm ≅ 3.31 atm.