Question

Dalton's law states that the total pressure, Ptotal, of a mixture of gases in a container equals the sum of the pressures of each individual gas: Ptotal=P1+P2+P3+… The partial pressure of the first component, P1, is equal to the mole fraction of this component, X1, times the total pressure of the mixture: P1=X1×Ptotal The mole fraction, X, represents the concentration of the component in the gas mixture, so X1=moles of component 1total moles in mixture

Part A 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 5.40 atm . Calculate the partial pressure of each gas in the container. Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.

Part B A gaseous mixture of O2 and N2 contains 33.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 825 mmHg ? Express you answer numerically in millimeters of mercury.

Answer 1

Part A

Molar mass of methane, CH₄ = (1*12 + 4*1) g/mol = 16 g/mol.

Molar mass of ethane, C₂H₆ = (2*12 + 6*1) g/mol = 30 g/mol.

Molar mass of propane, C₃H₈ = (3*12 + 8*1) g/mol = 44 g/mol.

Moles of methane = (8.00 g)/(16 g/mol) = 0.500 mole.

Moles of ethane = (18.00 g)/(30 g/mol) = 0.600 mole.

Let us have x g of propane; therefore, moles of propane = (x g/44 g/mol) = x/44 mole.

Mole fraction of methane = (moles of methane)/(total number of moles) = (0.500 mole)/[(0.500 + 0.600 + x/44) mole] = 0.500/(1.100 + x/44)

Mole fraction of ethane = (moles of ethane)/(total number of moles) = (0.600 mole)/[(0.500 + 0.600 + x/44) mole] = 0.600/(1.100 + x/44)

Mole fraction of propane = (moles of propane)/(total number of moles) / (x/44 mole)/[(0.500 + 0.600 + x/44) mole] = (x/44)/(1.100 + x/44)

The temperature of the gas is T = 23.0°C = (23.0 + 273) K = 296 K; the volume of the gas is V = 10.0 L while the pressure of the gas is 5.40 atm. Use the ideal gas law to find out the total number of moles of gases as

(5.40 atm)*(10.0 L) = n*(0.082 L-atm/mol.K)*(296 K)

====> 54.00 L-atm = n*24.272 L-atm/mol

====> n = 54.00/24.272 mol = 2.22478 mole ≈ 2.225 mole.

As per the problem,

(0.500 + 0.600 + x/44) mole = 2.225 mole

====> 1.100 + x/44 = 2.225

====> x/44 = 1.125

====> x = 1.125*44 = 49.5

The mass of propane taken is 49.5 g and the moles of propane taken = (49.5 g)/(44 g/mol) = 1.125 mole.

The partial pressures are as below:

Methane = (0.500/2.225)*(5.40 atm) = 1.20 atm.

Ethane = (0.600/2.25)*(5.40 atm) = 1.44 atm.

Propane = (1.125/2.25)*(5.40 atm) = 2.70 atm (ans).