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

Step-by-step 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 * 250 mL + 1.5 atm * 1.5 atm` =
`P * 10.0 L`

P = 0.126 atm

As, 1 atm = 760 torr. So,
`0.126 atm * (760 torr)/(1 atm)` = 95.76 torr.

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