# 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

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 container. at 23.0 ∘c, the total pressure in the container is 5.50 atm . calculate the partial pressure of each gas in the container.

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 mol

Moles methane: 8 g ÷ 16.04 g/mol = 0.499 mol
Moles ethane: 18 g ÷ 30.07 g/mol = 0.599 mol
Moles propane: 2.263 - (0.499+0.599) = 1.165 mol

Applying Raoult's Law:

Partial pressure = Mole fraction * Total Pressure

Partial Pressure of Methane = (0.499/2.263)(5.5 atm) = 1.21 atm
Partial Pressure of Ethane = (0.599/2.263)(5.5 atm) = 1.46 atm
Partial Pressure of Propane = (1.165/2.263)(5.5 atm) = 2.83 atm

Partial pressure of methane: 1.18 atm

Partial pressure of ethane: 1.45 atm

Partial pressure of propane: 2.35 atm

Explanation:

Let the total moles of gases in a container be n.

Total pressure of the gases in a container =P = 5.0 atm

Temperature of the gases in a container =T = 23°C = 296.15 K

Volume 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 mixture

P = total pressure of the mixture

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