Q

Help me please, - Surface Chemistry - JEE Main

Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas adsorbed on mass m of the adsorbent. The plot of $\log\frac{x}{m}$ versus $\log P$ is shown in the given graph. $\frac{x}{m}$ is proportional to :

• Option 1)

$P^{2/3}$

• Option 2)

$P^{3}$

• Option 3)

$P^{2}$

• Option 4)

$P^{3/2}$

Views

$\frac{x}{m}\; \alpha \; \; P^{1/x}\; \; \; \; \; \; \; \; \; Let\; 1/x=a$

$\frac{x}{m}=kp^{1/x}\rightarrow \therefore \frac{x}{m}=kp^{a}$

$\log\frac{x}{m}=\log k +\frac{1}{x}\log p$

$slope=\frac{1}{x}=a=\frac{2}{3}$

$\frac{x}{m}\; \alpha \; p^{2/3}$

Option 1)

$P^{2/3}$

Option 2)

$P^{3}$

Option 3)

$P^{2}$

Option 4)

$P^{3/2}$

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