Question

A gas mixture contains 0.150 mol of O2 gas, 0.116 mol of N2 gas, and 0.211 mol of Ar gas in a 0.500 L flask at 298 K. What is the partial pressure of N2 the mixture?

Answer 1

Explanation:

The given data is as follows.

Moles of oxygen gas = 0.150 mol,    Moles of nitrogen gas = 0.116 mol

Moles of argon gas = 0.211 mol,        Volume of the flask = 0.5 L

Temperature = 298 K

So, total number of moles will be calculated as follows.

             $n_{total} = n_{O_{2}} + n_{N_{2}} + n_{Ar}$

                         = (0.150 + 0.116 + 0.211) mol

                         = 0.477 mol

Now, using ideal gas equation we will calculate the total pressure of given gas mixture as follows.

                      PV = nRT

    $P \times 0.5 L = 0.477 mol \times 0.0821 Latm/mol K \times 298 K$

                   P = 23.34 atm

Mole fraction of $N_{2}$ gas will be calculated as follows.

           Mole fraction = $\frac{\text{moles of nitrogen}}{\text{total no. of moles}}$

                                 = $\frac{0.116 mol}{0.477 mol}$

                                 = 0.243

Using Dalton's law,  $P_{i} = P_{total} \times x_{i}$

where,    $P_{i}$ = partial pressure of a gas

              $P_{total}$ = total pressure

              $x_{i}$ = mole fraction of gas

Therefore, putting the values into Dalton's law to calculate the partial pressure of nitrogen gas as follows.

                   $P_{i} = P_{total} \times x_{i}$

                              = $23.34 atm \times 0.243$

                              = 5.67 atm

Thus, we can conclude that the partial pressure of $N_{2}$ into the given mixture is 5.67 atm.