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The volume percentages of N(2),O(2) and ...

The volume percentages of `N_(2),O_(2)` and He in gas mixture are 25, 35 and 40 , respectively. At a given temperature, the pressure of the mixture is 760 mm Hg. Calculate the partial pressure of each gas at the same temperature.

Text Solution

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According to Amagat's law, at constant temperature and pressure, the partial volume of a component gas in a gas mixture=molee-fractino of the compound`xx`total volume of the mixture.
therefore, in the given mixture`,x_(N_(2))=(25)/(100)=0.25`,
`x_(O_(2))=(35)/(100)=0.35 and x_(He)=(40)/(100)=0.40`
`therefore `In mixture,
Partial pressure of `N_(2)=x_(N_(2))xxP=0.25xx760=190` mm Hg
Partial pressure of `O_(2)=x_(X_(2))xxP=0.35xx760`=266 mm Hg
and partial pressure of `He=x_(He)xxP=0.40xx760`=304 mm Hg
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