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NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Surface Chemistry

NTA Abhyas 2020 · Chemistry · Surface Chemistry

Which of the following option is correct for given curve?

Which of the following option is correct for given curve?

Which of the following option is correct for given curve?

Options

A\dfrac{x}{m} ∝ { $\left(P\right)}^{}$
B\dfrac{x}{m} ∝ { $\left(P\right)}^{2}$
C $\dfrac{x}{m} ∝ {\left(P\right)}^{1 / 2}$
D\dfrac{x}{m} ∝ { $\left(P\right)}^{0}$

Correct answer

C. $\dfrac{x}{m} ∝ {\left(P\right)}^{1 / 2}$

Step-by-step solution

According to the Freundlich adsorption isotherm,

\dfrac{x}{m} = Kp^{\dfrac{1}{n}}

Taking log on bot sides.

\log \left(\dfrac{x}{m}\right) = logK + \dfrac{1}{n}log P

Slope

= \dfrac{1}{n}

Slope from graph

= \dfrac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \dfrac{2}{4} = \dfrac{1}{2}

or,

\dfrac{1}{n}=\dfrac{1}{2}

∴ \dfrac{x}{m} = Kp^{\left(\dfrac{1}{2}\right)}

or

\dfrac{x}{m} ∝ p^{\left(\dfrac{1}{2}\right)}

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