Mixture of gases w/ 1.32g of CO, 1.61g of N2, and 8.77g of SO2. What is the partial pressure in atm of each component if held in a 16.6-L vessel at 37.5 oC? R = 0.0821 L-atm/mol-K for gas constant.

Answer 1

There are a couple of ways to solve this problem. Maybe the easiest for the student to understand is the mole fraction approach. It says that the mol fraction of a gas times the total pressure = the partial pressure of that gas. The mole fraction of the gas is simply the moles of that gas / total moles of all gases. The calculations look like this:

moles CO = 1.32 g CO x 1 mol CO/28 g = 0.0471 moles CO

moles N₂ = 1.61 g N₂ x 1 mol N2 / 28 g = 0.0575 moles N₂

moles SO₂ = 8.77 g SO₂ x 1 mol SO2 / 64 g = 0.137 moles SO₂

TOTAL MOLE OF GAS = 0.0471 + 0.0575 + 0.137 = 0.242 total moles of gas

Now to calculate mole fractions of each gas:

mole fraction CO = 0.0471 / 0.242 = 0.195

mole fraction N₂ = 0.0575 / 0.242 = 0.238

mole fraction SO₂ = 0.137 / 0.242 = 0.566

partial pressures of each gas = mol fraction x total pressure. To find total pressure, use the ideal gas law

PV = nRT

P = nRT/V = (0.242 moles)(0.0821 Latm/Kmol)(310.5K) / 16.6 L

P = 0.372 atm

partial pressure CO = 0.195 x 0.372 atm = 0.0725 atm

partial pressure N₂ = 0.238 x 0.372 atm = 0.0885 atm

partial pressure SO₂ = 0.566 x 0.372 atm = 0.210 atm

=======================================================

An alternative way to solve the problem is to find the moles of each gas and then use the idea gas law for each gas and solve for pressure of each gas individually.