Question

Exercise A mixture of 250 mL of methane, CH4, at 35 °C and 0.55 atm and 750 mL of propane, C3Hg, at 35° C and 1.5 atm, were introduced into a 10.0 L container. What is the final pressure, in torr, of the mixture?

Answer 1

Explanation:

The given data is as follows.

      $V_{1}$ = 250 mL,     $V_{2}$ = 750 mL

      $T_{1}$ = $35^{o}C$ = 35 + 273 K = 308 K

      $T_{2}$ = 35 + 273 K = 308 K

      $P_{1}$ = 0.55 atm,    $P_{2}$ = 1.5 atm

               P = ? ,         V = 10.0 L

Since, temperature is constant.

So,    $P_{1}V_{1} + P_{2}V_{2}$ = PV

Now, putting the given values into the above formula as follows.

         $P_{1}V_{1} + P_{2}V_{2}$ = PV

         $0.55 atm \times 250 mL + 1.5 atm \times 1.5 atm$ = $P \times 10.0 L$

                     P = 0.126 atm

As, 1 atm = 760 torr. So, $0.126 atm \times \frac{760 torr}{1 atm}$ = 95.76 torr.

Thus, we can conclude that the final pressure, in torr, of the mixture is 95.76 torr.