Question

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 4.80 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 36.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 845 mmHg ?

Express you answer numerically in millimeters of mercury.

Answer 1

Part-A: Given P = 4.80 atm, V = 10.0 L and T = 23 C = 23 + 273 = 296 K

Applying ideal gas equation

PV = nRT

=> n = PV/RT = (4.80 atm x 10.0 L) / (0.0821 L.atm.mol-1K-1 x 296K) = 1.975 mol

moles of methane = mass of methane / molecular mass = 8.00 g / 16.0 g/mol = 0.50 mol

moles of ethane = mass of ethane / molecular mass = 18.0 g / 30.0 g/mol = 0.60 mol

Hence moles of propane = 1.975 - 0.50 - 0.60 = 0.875 mol

Hence partial pressure of methane = mole fraction x P = (0.50 mol / 1.975 mol) x 4.80 atm = 1.2152 atm (answer)

partial pressure of ethane = mole fraction x P = (0.60 mol / 1.975 mol) x 4.80 atm = 1.4582 atm (answer)

partial pressure of propane = mole fraction x P = (0.875 mol / 1.975 mol) x 4.80 atm = 2.1266 atm (answer)

Part-B:

Given total pressure, P = 845 mmHg

Let's consider 100 g of the gaseous mixture.

mass of N2 = 36.8 g

Hence moles of N2 = 36.8 g / 28.0 g/mol = 1.3143 mol

mass of O2 = 100 - 36.8 = 63.2 mol

Hence moles of O2 = 63.2 g / 32.0 g/mol = 1.975 mol

Hence partial pressure of O2 = mole fraction x P = (1.975 mol / 3.2893 mol) x 845 mmHg = 507 mmHg (answer)