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

Applying Boyle.s law (since the temperature remains constant)
`P_(He) V._(He)` (after mixing ) = `p_(He) V_(He)` (before mixing )
`P._(He) = (P_(He) V_(He))/(V._(He)) = (200 xx 0.66)/(400) = 0.33` atm
Similarly , for `O_(2)` ,
`p._(O_(2)) V._(O_(2))` (after mixing ) = `p_(O_(2)) , V_(O_(2))` (before mixing )
`p._(O_(2)) = (400 xx 0.52)/(400) = 0.52` atm .