What will be the partial pressures of He and `O_(2)` respectively if 200 ml of He at 0.66 atm pressure and 400 ml of `O_(2)` at 0.52 atm pressure are mixed in 400 ml vessel at `20^(@)C` ?

Options

A. 0.33 and 0.56
B. 0.33 and 0.52
C. 0.38 and 0.52
D. 0.25 and 0.45

Answer

B

Explanation

`PV=nRT`
`n_(He)=(0.66xx200)/(RT )`
`n_(O_(2))=(0.52xx400)/(RT)`
`PV=nRT`
`P=((n_(He)+n_(O_(2)))RT)/(400)`
`P_(T)=(0.66xx200+400xx0.52)/(400)`
`:' X_(He)=(n_(He))/(n_(O_(2))+n_(He))`
`X_(He)=(0.66 xx 200)/(0.66 xx200+400xx0.52)=(132)/(132+208)`
`P_(He)=X_(He)P_(T)=(132)/(340)xx(340)/(400)implies P_(He)=0.33`
`X_(O_(2))=(208)/(400)`
So, `P_(O_(2))=0.52`