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At 2 atm pressure the value of x/m will ...

At 2 atm pressure the value of `x/m` will be :(log 2=0.3010)

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The correct Answer is:
4
`x/m=K P^(1//n)`
`"log"x/m="log"K+1/n"log"P`
`:. K=2 & 1/n=1 or n=1`
On putting in `eq^n(1)`
`x/m=2xx2=4`
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