A mixture containing 1.6 g of `O_(2)`, 1.4 g of `N_(2)` and 0.4 g of He occupies a volume of 10 litres at `27^(@)C`. Calculate the total pressure of the mixutre and partial pressure of each compound.

Explanation

PV = nRT V=10 litre
`T=27^(@) C = 300K `
`n_(He)= (0.4)/( 4) = 0.1, n_(N_(2)) =(1.4)/( 28) = 0.05,n_(D_(2)) = ( 1.6)/( 32) =0.05 `
Total no. of moles `= 0.1 +0.005 + 0.05 rArr0.2`
`PV =nRT rArrP xx 10 = 0.2 xx0.082xx 300 = 4.926`
`rArr P=0.492 ` atm
Partial pressure = Total pressure ` xx` mole fraction
`P_(He) =0.492 xx (0.1)/(0.2)= 0.246 ` atm
`P_(N_(2)) = 0.492 xx (0.05)/( 0.2) = 0.123 atm`
`P_(O_(2)) xx( 0.05)/( 0.2) = 0.123 ` atm