A 0.500 L of gases is at 307 K and contains N2 at a pressure of 1.40 atm and O2 at a pressure of 0.24 atm. What is the mole fraction of each of the two gases? (Molar mass: N=14 g/mol, O=16 g/mol)

Answer 1

First, we need to calculate the total pressure of the mixture:

Ptotal = PN2 + PO2 = 1.40 atm + 0.24 atm = 1.64 atm

Next, we can use the ideal gas law to calculate the number of moles of each gas:

nN2 = (PN2 * V) / (R * T) = (1.40 atm * 0.500 L) / (0.0821 L·atm/mol·K * 307 K) = 0.0273 mol
nO2 = (PO2 * V) / (R * T) = (0.24 atm * 0.500 L) / (0.0821 L·atm/mol·K * 307 K) = 0.00413 mol

The total number of moles is then:

ntotal = nN2 + nO2 = 0.0273 mol + 0.00413 mol = 0.0314 mol

Finally, we can calculate the mole fraction of each gas:

XN2 = nN2 / ntotal = 0.0273 mol / 0.0314 mol = 0.869
XO2 = nO2 / ntotal = 0.00413 mol / 0.0314 mol = 0.131

Therefore, the mole fraction of N2 is 0.869 and the mole fraction of O2 is 0.131.