Question

The graph of $\log \frac{x}{m}$ vs $\log p$ for an adsorption process is a straight line inclined at an angle of $45^{\circ}$ with intercept equal to $0.6020$. The mass of gas adsorbed per unit mass of adsorbent at the pressure of $0.4 \,atm$ is______ $\times 10^{-1}$ (Nearest integer).
Given: $\log 2=0.3010$

Answer 1

$ \log \frac{x}{m}=\log k +\frac{1}{n} \log P $
$ \frac{1}{ n }=\tan 45^{\circ}=1 $
$ \log k =0.6020=\log 4 $
$ \Rightarrow K =4 $
$ \therefore \frac{ x }{ m }= K \cdot P ^{1 / n } $
$ \frac{ x }{ m }=4(0.4)=1.6 $
$ \frac{ x }{ m }=1.6=16 \times 10^{-1}$