CH₄ is absorbed on 1g charcoal at 0$^\circ$C following the Freundlich adsorpiton isotherm 10.0 mL of CH₄ is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of CH₄ adsorbed at 300 mm of Hg is 10^x mL. The value of x is _________ $\times$ 10^-2.

(Nearest integer) [Use log₁₀2 = 0.3010, log₁₀3 = 0.4771]

We know<br><br>${x \over m} = K{P^{1/n}}$; using (x $\propto$ V)<br><br>$\Rightarrow {{10} \over 1} = K \times {(100)^{1/n}}$ ..... (1)<br><br>${{15} \over 1} = K \times {(200)^{1/n}}$ ..... (2)<br><br>${V \over 1} = K \times {(300)^{1/n}}$ ..... (3)<br><br>Divide<br><br>(2) / (1)<br><br>${{15} \over {10}} = {2^{1/n}}$<br><br>$\log \left( {{3 \over 2}} \right) = {1 \over n}\log 2$<br><br>$${1 \over n} = {{\log 3 - log2} \over {\log 2}} = {{0.4771 - 0.3010} \over {0.3010}}$$<br><br>${1 \over n} = 0.585$<br><br>Divide<br><br>(3) / (1)<br><br>${V \over {10}} = {3^{1/n}}$<br><br>$\log \left( {{V \over {10}}} \right) = {1 \over n}\log 3$<br><br>$\log \left( {{V \over {10}}} \right) = 0.585 \times 0.4771 = 0.2791$<br><br>${V \over {10}} = {10^{0.279}} \Rightarrow V = 10 \times {10^{0.279}}$<br><br>$\Rightarrow V = {10^{1.279}} = {10^x}$<br><br>$\Rightarrow$ x = 1.279<br><br>$\Rightarrow$ x = 128 $\times$ 10^$-$2 (Nearest integer)