Two vessels connected by a valve of negligible volume. One container (I) has 2.8 g of `N_(2)` at temperature `T_(1) (K)`. The other container (II) is completely evacuated. The container (I) is heated to `T_(2)(K)`while container (II) is maintained at `T_(2)//3(K)`. volume of vessel (I) is half that of vessel (II). If the valve is opened then what is the mass ratio of `N_(2)` is both vessel `(W_(I)//W_(II))`?
A
`1:2`
B
`1:3`
C
`1:6`
D
`3:1`
Text Solution
Verified by Experts
The correct Answer is:
c
(c) Let x mole of `N_(2)` present into vessel II and P is final pressure of `N_(2)` `P(2V)=xR(T_(2)//3)and P(V)=(0.1-x)RT_(2)` `implies " "2=(x)/(3(0.1-x))` `implies" "x=0.6//7"moles",` `(0.6)/(7)xx28implies2.4 g N_(2)` II has 2.4 g `N_(2)` and I has 0.4 g of `N_(2),` `(W_(I))/(W_(II))=(0.4)/(2.4)implies1:6`
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