CH4 is adsorbed on 1g charcoal at 0ºC following the Freundlich adsorption 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 10x mL. The value of x is____________ x 10

Engineering

Chemistry

Adsorption

Question

CH₄ is adsorbed on 1g charcoal at 0ºC following the Freundlich adsorption 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 10x mL. The value of x is____________ x 10^–2

[Use log₁₀2 = 0.3010 log₁₀3 = 0.4771]

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We know

$\frac{x}{m}$ = KP^1/n ; using (x ∝ V)

$\frac{10}{1}$ = K × (100)^1/n .....(1)

$\frac{15}{1}$ = K × (200)^1/n .....(2)

$\frac{V}{1}$ = K × (300)^1/n .....(3)

Divide
(2) / (1)

$\frac{15}{10} = 2^{1 / n}$

$\log (\frac{3}{2}) = \frac{1}{n} \log^{2}$

$\frac{1}{n} = \frac{\log^{3} - \log^{2}}{\log^{2}} = \frac{0.4771 - 0.3010}{0.3010}$

$\frac{1}{n} = 0.585$

Divide
(3) / (1)

$\frac{V}{10} = 3^{1 / n}$

$\log (\frac{V}{10}) = \frac{1}{n} \log^{3}$

$\log (\frac{V}{10})$ = 0.585 × 0.4771 = 0.2791

$\frac{V}{10}$ = 10^0.279 ⇒ V = 10 × 10^0.279

⇒ V = 10^1.279 = 10^x

⇒ x = 1.279

⇒ x = 128 × 10^–² (Nearest integer)

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