Respuesta :
Answer:
The partial pressure of the methane is 500 Torr.
Mole fraction of methane is 0.7142 and mole fraction of argon is 0.2857.
Explanation:
Equal masses of methane and argon. Suppose 1 gram of methane and argon.
Moles of methane = [tex]n_1=\frac{1 g}{16.0 g/mol}=0.0625 mol[/tex]
Moles of argon =[tex]n_2=\frac{1 g}{40 g/mol}=0.025 mol[/tex]
Mole fraction of methane =[tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]
[tex]\chi_1=\frac{0.0625 mol}{0.0625 mol+0.025 mol}=0.7142[/tex]
Mole fraction of argon=[tex]\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\chi_2=\frac{0.025 mol}{0.0625 mol+0.025 mol}=0.2857[/tex]
Total pressure of the gases = P
Let the partial pressure methane and argon be [tex]p_1 \& p_2[/tex].
[tex]p_2=200 Torr[/tex]
According Dalton's law of partial pressure :
[tex]p_1=P\times \chi_1=P\times \frac{n_1}{n_1+n_2}[/tex]
[tex]p_2=P\times \chi_2=P\times \frac{n_2}{n_1+n_2}[/tex]
[tex]200 Torr=P\times \frac{0.025 mol}{0.0625 mol+0.025 mol}[/tex]
P = 700 Torr
[tex]p_1=700\times \frac{0.0625 mol}{0.0625 mol+0.025 mol}=500 Torr[/tex]
The partial pressure of the methane is 500 Torr.
Mole fraction of methane is 0.7142 and mole fraction of argon is 0.2857.
