Question

$CH _{4}$ is adsorbed on $1 g$ charcoal at $0^{\circ} C$ following the Freundlich adsorption isotherm. $10.0\, mL$ of $CH _{4}$ is adsorbed at $100 \,mm$ of $Hg$, whereas $15.0$ $mL$ is adsorbed at $200\, mm$ of $Hg$. The volume of $CH _{4}$ adsorbed at $300 \,mm$ of $Hg$ is $10^{ x } mL$. The value of $x$ is _______ $\times 10^{-2}$ (Nearest integer) $\left[\right.$ Use $\left.\log _{10} 2=0.3010, \log _{10} 3=0.4771\right]$

Answer 1

We know
$\frac{x}{m}=K P^{1 / n} ; \text { using }(x \propto V)$
$\Rightarrow \frac{10}{1} = K \times(100)^{1 / n } ....$ (1)
$\frac{15}{1} = K \times(200)^{1 / n } ....$ (2)
$\frac{ V }{1} = K \times(300)^{1 / n } ....$ (3)
Divide
(2) / (1)
$\frac{15}{10}=2^{1 / n} $
$\log \left(\frac{3}{2}\right)=\frac{1}{n} \log 2$
$\frac{1}{n}=\frac{\log 3-\log 2}{\log 2}=\frac{0.4771-0.3010}{0.3010} $
$\frac{1}{n}=0.585$
Divide
(3) / (1)
$\frac{ V }{10}=3^{1 / n } $
$\log \left(\frac{ V }{10}\right)=\frac{1}{ n } \log 3 $
$\log \left(\frac{ V }{10}\right)=0.585 \times 0.4771=0.2791$
$\frac{ V }{10}=10^{0.279} \Rightarrow V =10 \times 10^{0.279}$
$\Rightarrow V =10^{1.279}=10^{ x }$
$\Rightarrow x =1.279$
$\Rightarrow x =128 \times 10^{-2}$ (Nearest integer)