CH _ 4 is adsorbed on 1g charcoal at 0 ° C following the Freundlich...

\dfrac{x}{m} = Kp\dfrac{1}{n}$</p><p>$x ∝ Vforthesamegas$<br />$\dfrac{10}{1} = K$\left(100\right)\dfrac{1}{n}$ ...(1)
$\dfrac{15}{1} = K$\left(200\right)\dfrac{1}{n}$....(2)<br />$\dfrac{V}{1} = K$\left(300\right)\dfrac{1}{n}$ ...(3)
Divide (2) to (1) $= \dfrac{15}{10} = 2\dfrac{1}{n}$
$\log \left(\dfrac{3}{2}\right) = \dfrac{1}{n}\log 2$
$\dfrac{1}{n} = \dfrac{0 . 4771 - 0 . 3010}{0 . 3010} = 0 . 585$
Divide (3) to (1) $= \dfrac{V}{10} = 3\dfrac{1}{n}$
$\log \left(\dfrac{V}{10}\right) = \dfrac{1}{n}\log 3$
$\log \left(\dfrac{V}{10}\right) = 0 . 585 \times 0 . 4771 = 0 . 2791$
$\dfrac{V}{10} = {10}^{0 . 2791}$
$V = 10 \times {10}^{0 . 2791} = {10}^{1 . 2791} = {10}^{x}$
$x = 1 . 2791 = 127 . 91 \times {10}^{- 2} \approx 128 \times {10}^{- 2}$