1.

Two containers are connected by stopcock as shown. If initially P_(O_(2))=P_(He)=P when stopcock is closed. Then after opening the stopcock (alter a long time keeping initial temperature in each same as initially).

Answer»

<P>`P_("final")=P`
`P_(O_(2))=(P)/(2)`
`P_(He)=P`
`T=600K`

Solution :After opening the stopcock (for long TIME)
`n'_(1)+n'_(2)=n_(1)+n_(2)`
`(P_(f)xx10)/(Rxx300)+(P_(f)xx20)/(Rxx600)=(Pxx10)/(300R)+(Pxx20)/(600R)`
For `O_(2) & He` before MIXING
`(P_(O_(2))V_(O_(2)))/(n_(O_(2))T_(O_(2)))=(P_(He)V_(He))/(n_(He)T_(He))`
`T=T_(He)=600K`
For `O_(2)` in `1^(st)` container and `2^(nd)` container its MOLE distribution
`(Pxx10)/(Pxx20)=((n_(O_(2)))_(1)xxRxx300)/((n_(O_(2)))_(2)xxRxx600)`
`(n_(O_(2)))_(1)=(n_(O_(2)))_(2)=(X)/(2)`
Similarily `(n_(O_(2)))_(1)=(X)/(2)`
Partial pressure of `O_(2)` in any container `P_(O_(2))=x_(O_(2))P_("total")`
`P_(O_(2))=(X//2)/(X)P=(P)/(2)`
Partial pressure of He in any container `P_(He)=P-P_(O_(2))=(P)/(2)`
T=600K
A,B,D


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