Three gases (8.00 g of methane, ch4, 18.0 g of ethane, c2h6, and an unknown amount of propane, c3h8) were added to the same 10.0-l

Question

Three gases—8.00 g of methane (CH₄), 18.0 g of ethane (C₂H₆), and an unknown amount of propane (C₃H₈)—were introduced into a single 10.0-L container. At a temperature of 23.0 °C, the total pressure inside the container is measured to be 5.50 atm. Calculate the partial pressure of each gas within the container.

Anonymous · Accepted Answer

Assume ideal gas behavior, then solve for the total number of moles:PV = nRT(5.50 atm)(10 L) = n(0.0821 L-atm/mol-K)(23+273 K)n = 2.263 molMoles methane: 8 g ÷ 16.04 g/mol = 0.499 molMoles ethane: 18 g ÷ 30.07 g/mol = 0.599 molMoles propane: 2.263 - (0.499+0.599) = 1.165 molApplying Raoult's Law: Partial pressure = Mole fraction * Total PressurePartial Pressure of Methane = (0.499/2.263)(5.5 atm) = 1.21 atmPartial Pressure of Ethane = (0.599/2.263)(5.5 atm) = 1.46 atmPartial Pressure of Propane = (1.165/2.263)(5.5 atm) = 2.83 atm

Lonzo Jast · Answer

Partial pressure of methane: 1.18 atmPartial pressure of ethane: 1.45 atmPartial pressure of propane: 2.35 atmExplanation: Let the total moles of gases in a container be n.Total pressure of the gases in a container =P = 5.0 atmTemperature of the gases in a container =T = 23°C = 296.15 KVolume of the container = V = 10.0 L (Ideal gas equation)Moles of methane gas =Moles of ethane gas =Moles of propane gas =Partial pressure of all the gases can be calculated by using Raoult's law: = partial pressure of 'i' component. = mole fraction of 'i' component in mixtureP = total pressure of the mixturePartial pressure of methane:Partial pressure of ethane:Partial pressure of propane:

Breanne Hegmann III

Three gases (8.00 g of methane, ch4, 18.0 g of ethane, c2h6, and an unknown amount of propane, c3h8) were added to the same 10.0-l

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