Question

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 5.00 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 30.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 685 mmHg ? Express you answer numerically in millimeters of mercury.

Answer 1

Ans). According to Dalton’s Law of Partial Pressures (P) :

P_total = P_CH3 + P_C2H4 + P_C3H8

We know that for an ideal gas, PV = nRT and P = nRT / V

Therefore,

P_total =(nRT/V)_CH3 + (nRT/V)_C2H6 + (nRT/V)_C3H8

Provided that,  P_total = 5.00 atm, V = 10.00 L, T = 273 + 23 = 296 K, R = 0.0821 atm L mol^-1 K^-1

no. of moles of CH₃, n_CH3 = given mass / molar mass = 8 / 16 = 0.5 moles

no. of moles of C₂H₆, n_C2H6  = 18 / 30 = 0.6 moles.

Thus we have,

P_CH3 = (nRT/V)_CH3 = (0.5 x 0.0821 x 296) / 10 = 1.125 atm

P_C2H6 = (nRT/V)_C2H6 = (0.6 x 0.0821 x 296) / 10 = 1.458 atm

P_C3H8 = P_total - (P_{CH3 +} P_C2H6) = 5.00 - (1.125 + 1.148) = 5.00 - 2.583 = 2.417 atm

Hene, calculated the Partial pressure of each gas in the container.

Part B.Ans). We know that,

Partial pressure = Mole fraction x P_total

For a 100 g gaseous mixture of O₂ and N₂,

N₂ = 30.8 % nitrogen by mass = 30.8 g. Hence O₂ is = 69.2 g

no. of moles of N₂ , n_N2 = 30.8/28 = 1.1 moles

no. of moles of O₂ , n_O2 = 69.2/32 = 2.163 moles

Mole fraction of O₂ = n_O2 / (n_N2 + n_O2) = 2.163 / (2.163 + 1.1) = 2.163 / 3.263 = 0.663

Hence,

Partial Pressure of  O₂ = 0.663 x 685 = 454.155 mm of Hg.