JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Surface Chemistry

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

lo g \frac{x}{m}

versus

lo g p

is shown in the given graph,

\frac{x}{m}

is proportional to:

Answer: (A)

Solution

Freundlich's isotherm

\frac{x}{m} = K p^{1/ n}

becomes a straight line on taking logarithms,

lo g \frac{x}{m} = lo g K + \frac{1}{n} lo g p

, so the slope of the plotted line is

\frac{1}{n}

. The line rises two units for every three along, a slope of

\frac{2}{3}

, so

\frac{1}{n} = \frac{2}{3}

and

\frac{x}{m} \propto p^{2/3}

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option