Answer:
80.11 torr
Explanation:
Hello,
In this case, since the masses are equal, we could assume 1 gram for each gas, which result in the following moles:
[tex]n_{CH_4}=\frac{1g}{16.0g/mol}=0.0625molCH_4 \\n_{Ar}=\frac{1g}{40.0g/mol}=0.025molAr[/tex]
Thus, the molar fractions:
[tex]x_{CH_4}=\frac{0.0625}{0.0625+0.025}=0.714\\x_{Ar}=1- x_{CH_4}=1-0.714=0.286[/tex]
Thus, the total pressure is:
[tex]p=p_{CH_4}+p_{Ar}\\p=x_{CH_4}p+x_{Ar}p\\p=200torr+0.286p\\p=\frac{200torr}{1-0.286}\\ p=280.11torr[/tex]
Hence the pressure of methane:
[tex]p_{CH_4}=280.11torr-200torr=80.11torr[/tex]
Best regards.
