Question

Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.

What is the mole fraction of O2 in the mixture?

___ atm

What is the mole fraction of Ar in the mixture?

___atm

Answer 1

Mole fraction of
`O_(2)` = 0.42

Moles fraction of Ar = 0.037

Step-by-step explanation:

Let's assume all the four gases and their mixture behave ideally.

According to Dalton's law of partial pressure for mixture of ideal gases-

`P_(i)=x_(i)P_(total)`

where
`P_(i)` and
`x_(i)` are partial pressure and mole fraction of "i"-th gas respectively.
`P_(total)` is total pressure of gaseous mixture.

Here
`P_(total)` =
`P_{N_(2)}+P_{O_(2)}+P_(Ar)+P_(He)`

= (3.5+2.8+0.25+0.15) atm

= 6.7 atm

So
`x_{O_(2)}=\frac{P_{O_(2)}}{P_(total)}` =
`(2.8atm)/(6.7atm)` = 0.42

`x_(Ar)=(P_(Ar))/(P_(total))` =
`(0.25atm)/(6.7atm)` = 0.037