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A mixture of N(2) and O(2) at 1 bar pres...

A mixture of `N_(2) and O_(2)` at 1 bar pressure contains 80% `N_(2)` by weight. Calculate the partial pressure of `N_(2)` in the mixture.

Text Solution

Verified by Experts

As 80 g `N_(2)` is present in 100 g of mixture, the amount of `O_(2)` in the mixture is 20 g.
Therefore, the number of moles of `N_(2)` in the mixture `(n_(N_(2)))=(80)/(28)=2.857` mol and the number of moles of `O_(2)` in the mixture `(n_(O_(2)))=(20)/(32)=0.625` mol.
`therefore`Total moles of the mixture `(n_("total"))=n_(N_(2))=n_(O_(2))`
`=2.857+0.625=3.482` mol
Mole fraction of `N_(2)` gas `(x_(N_(2)))=(n_(1))/(n)=(2.857)/(3.482)=0.82`
and that of `O_(2)` gas `(x_(O_(2)))=(n_(2))/(n)=(0.625)/(3.482)=0.18`
In the mixture, the partial pressure of `N_(2)` gas `=x_(N_(2))xx`total pressure of the mixture, `=0.82xx1` bar =0.82 bar and that of `O_(2)` gas `=x_(O_(2))xx`total pressure of the mixture `=0.18xx1`bar=0.18 bar.
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