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In the adsorption of acetic acid vapours...

In the adsorption of acetic acid vapours by 1 g of charcoal, the following data was abtained :
`{:(x(cm^(3)),0.726,0.478),(p(cmof Hg),0.570,0.210):}`

Text Solution

Verified by Experts

According to Freudlich adsorptio siotherm :
`x//m=Kp^(1//n)`
Talking log, we get log x/m= log K + 1/n log p
For observation (I) `log"(0.726)/(1)=log K+(1)/(n)log (0.576)`
For observation (II) `log"(0.438)/(1)=logK +(1)/(n)log (0.210)`
Subtracting eqn. (iii) froe eqn. (ii)
`log"((0.726)/(0438))=1/nlog ((0.576)/(0.210))or log (1.675)=1/nlog (2.742)`
`(0.219)=1/n(0.428)or " "n=(0.438)/(0.219)=2`
Substituting the values of x, m and n in eqn. (i)
`(0.726)/(1)=K(0.576)^(1//n)=K(0.576)^(1//2)`
`0.726=K(0759)or K=(0.726)/(0759)=9.956~~1.`
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