Question

A graph plotted between $log\left(\frac{x}{m}\right)$ vs $logP$ ($P$ is pressure of gas in atm) shows a straight line with slope equal to $1$ and intercept equal to $0.4771$. The extent of adsorption $\left(\frac{x}{m}\right)$ of gas at the pressure $2$ atm is :

Answer 1

We know,

$log\frac{x}{m}=logK+\frac{1}{n}logP$

$\therefore logK=0.4771$ or $K=3$

$\therefore log\frac{x}{m}=log3+\frac{1}{n}\times logP(\because P=1)$

$Slope=\frac{1}{n}=1$

$\therefore n=1$

Hence, $\frac{x}{m}=K{P}^{1/n}=3\times (2{)}^1=6$.