JEE Main · ChemistryHard

$Consider the Freundlich adsorption isotherm given by \frac{x}{m} = k P^{1/ n} . If a plot of lo g (\frac{x}{m}) versus lo g (P) yields a straight line with an intercept of 0.301 and a slope of 0.5, what is the amount of gas adsorbed per gram of adsorbent at a pressure of 4 atm ?$

A. $2 units$
B. $4 units$
C. $8 units$
D. $1 unit$

Show correct answer & step-by-step solution

Correct answer: B — $4 units$

Solution

$The equation is lo g (\frac{x}{m}) = lo g (k) + \frac{1}{n} lo g (P) .$
$The intercept lo g (k) = 0.301, so k = 1 0^{0.301} \approx 2 .$
$The slope \frac{1}{n} = 0.5 .$
$At P = 4, \frac{x}{m} = 2 \times (4)^{0.5} = 2 \times 2 = 4 .$
$Hence the answer is (B).$

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