An open container of volume 50 cc holds air (mean molecular weight = 28.8 g ) and is exposed to the atmosphere where the pressure is 100 kPa . The container is initially placed in a bath containing boiling water ( 100^ C ) until thermal equilibrium is reached. The container is then closed and transferred to a bath containing melting ice ( 0^ C ). Determine the pressure of the air inside the container when thermal equ

Step-by-step solution

Initially, the open container is in thermal equilibrium with boiling water. Since it is open to the atmosphere, the pressure inside is equal to the atmospheric pressure. P₁ = 100 kPa T₁ = 100^ C = 373.15 K The container is then closed, trapping a fixed mass of air, and placed in a melting ice bath. The volume of the container remains constant. T₂ = 0^ C = 273.15 K Using Gay-Lussac's law for a fixed mass of an ideal gas at constant volume: P₁ T₁ = P₂ T₂ P₂ = P₁ T₂ T₁ P₂ = 100 273.15 373.15 P₂ 73.2 kPa