Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p. (x/m) is proportional to

Question

Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p. (x/m) is proportional to <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-eynelokzvb2gndql.jpg" /> (A) P(1/4) (B) P2 (C) P (D) P(1/2)

Tardigrade Admin · Accepted Answer

(x/m)=K× P1 / n log(x/m)=logK+(1/n)logP on compairing with y=mx+c m=(1/n)=(2/4)=(1/2)⇒ n=2 So, (x/m)=K× P1 / 2

Q. Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p. $\frac{x}{m}$ is proportional to

$P^{\frac{1}{4}}$

$P^{2}$

P

$P^{\frac{1}{2}}$

$\frac{x}{m} = K \times P^{1/ n}$
$l o g \frac{x}{m} = l o g K + \frac{1}{n} l o g P$
on compairing with $y = m x + c$
$m = \frac{1}{n} = \frac{2}{4} = \frac{1}{2} \Rightarrow n = 2$
So, $\frac{x}{m} = K \times P^{1/2}$

Q. Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p. mx​ is proportional to

P41​

P2

P

P21​

mx​=K×P1/n logmx​=logK+n1​logP on compairing with y=mx+c m=n1​=42​=21​⇒n=2 So, mx​=K×P1/2

Q. Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p. $\frac{x}{m}$ is proportional to

$P^{\frac{1}{4}}$

$P^{2}$

$P^{\frac{1}{2}}$

$\frac{x}{m} = K \times P^{1/ n}$
$l o g \frac{x}{m} = l o g K + \frac{1}{n} l o g P$
on compairing with $y = m x + c$
$m = \frac{1}{n} = \frac{2}{4} = \frac{1}{2} \Rightarrow n = 2$
So, $\frac{x}{m} = K \times P^{1/2}$