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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?

Respuesta :

Explanation:

The given data is as follows.

      [tex]V_{1}[/tex] = 250 mL,     [tex]V_{2}[/tex] = 750 mL

      [tex]T_{1}[/tex] = [tex]35^{o}C[/tex] = 35 + 273 K = 308 K

      [tex]T_{2}[/tex] = 35 + 273 K = 308 K

      [tex]P_{1}[/tex] = 0.55 atm,    [tex]P_{2}[/tex] = 1.5 atm

               P = ? ,         V = 10.0 L

Since, temperature is constant.

So,    [tex]P_{1}V_{1} + P_{2}V_{2}[/tex] = PV

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

         [tex]P_{1}V_{1} + P_{2}V_{2}[/tex] = PV

         [tex]0.55 atm \times 250 mL + 1.5 atm \times 1.5 atm[/tex] = [tex]P \times 10.0 L[/tex]

                     P = 0.126 atm

As, 1 atm = 760 torr. So, [tex]0.126 atm \times \frac{760 torr}{1 atm}[/tex] = 95.76 torr.

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

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