1.

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

Answer»

0.33 and 0.56
0.33 and 0.52
0.38 and 0.52
0.25 and 0.45

Solution :`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`


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