Question

An apparatus consists of a 4 L flask containing nitrogen gas at 27◦C and 875 kPa, joined by a valve to a 13 L flask containing argon gas at 27◦C and 54.9 kPa. The valve is opened and the gases mix. What is the partial pressure of nitrogen after mixing? Answer in units of kPa.

Answer 1

Explanation:

Given data is as follows.

      $V_{1}$ = 4 L,        $P_{1}$ = 875 kPa

     $T_{1} = 27^{o}C$ = (27 + 273) K = 300 K,      $V_{2}$ = 13 L

     $P_{2}$ = 54.9 kPa,       $T_{2} = 27^{o}C$

Total volume will be 4 L + 13 L = 17 L

Hence, calculate the total pressure as follows.

                    $P_{1}V_{1} = P_{total}V_{total}$     (as T is same so it will cancel out)

                    $875 kPa \times 4 L = P_{total} \times 17 L$

                    $P_{total}$ = 205.8 kPa

Now, using ideal gas equation we will calculate the moles of nitrogen as follows.

               PV = nRT

  $875 kPa \times 4 L = n \times 8.314 L kPa/mol K \times 300 K$

     3500 = $n \times 2494.2$

          n = 1.4 mol

Moles of argon will be calculated as follows.

             PV = nRT

 $54.9 \times 13 L = n \times 8.314 L kPa/mol K \times 300 K$

        713.7 = $n \times 2494.2$

           n = 0.28 mol

Therefore, mole fraction of nitrogen will be as follows.

       Mole fraction of $N_{2} = \frac{moles of N_{2}}{\text{total moles}}$

                                       = $\frac{1.4 mol}{1.4 + 0.28}$

                                       = $\frac{1.4}{1.68}$

                                       = 0.833

Relation between partial pressure and total pressure is as follows.

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

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

             $x_{i}$ = mole fraction of the gas

Thus, we will calculate the partial pressure of nitrogen after mixing as follows.

          $P_{total} = P_{i}x_{i}$  

            205.8 kPa = $P_{i} \times 0.833$

              $P_{i} = 247.05 kPa$

Thus, we can conclude that the partial pressure of nitrogen after mixing is 247.05 kPa.