Question

A gas mixture consists of equal masses of methane (molecular weight 16.0) and argon (atomic weight 40.0). If the partial pressure of argon is 200. torr, what is the pressure of methane, in torr? Hint: what is the mole fraction of each gas?

Answer 1

The partial pressure of the methane is 500 Torr.

Mole fraction of methane is 0.7142 and mole fraction of argon is 0.2857.

Step-by-step explanation:

Equal masses of methane and argon. Suppose 1 gram of methane and argon.

Moles of methane =
`n_1=(1 g)/(16.0 g/mol)=0.0625 mol`

Moles of argon =
`n_2=(1 g)/(40 g/mol)=0.025 mol`

Mole fraction of methane =
`\chi_1=(n_1)/(n_1+n_2)`

`\chi_1=(0.0625 mol)/(0.0625 mol+0.025 mol)=0.7142`

Mole fraction of argon=
`\chi_2=(n_2)/(n_1+n_2)`

`\chi_2=(0.025 mol)/(0.0625 mol+0.025 mol)=0.2857`

Total pressure of the gases = P

Let the partial pressure methane and argon be
`p_1 \& p_2`.

`p_2=200 Torr`

According Dalton's law of partial pressure :

`p_1=P* \chi_1=P* (n_1)/(n_1+n_2)`

`p_2=P* \chi_2=P* (n_2)/(n_1+n_2)`

`200 Torr=P* (0.025 mol)/(0.0625 mol+0.025 mol)`

P = 700 Torr

`p_1=700* (0.0625 mol)/(0.0625 mol+0.025 mol)=500 Torr`

The partial pressure of the methane is 500 Torr.

Mole fraction of methane is 0.7142 and mole fraction of argon is 0.2857.