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JEE Main 202330 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
30 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Surface Chemistry

JEE Main 2023 · 30 January · Evening Shift · Chemistry · Surface Chemistry

The graph of log(x)/(m)vslogp for an adsorption process is a...

The graph of logxmvslogplog\dfrac{x}{m}vslogp for an adsorption process is a straight line inclined at an angle of 45°45 ° with intercept equal to 0.60200 . 6020. The mass of gas adsorbed per unit mass of adsorbent at the pressure of 0.4atm0 . 4atm is _____×10−1\times {10}^{- 1} (Nearest integer)

Given : log⁡2=0.3010\log 2 = 0 . 3010

The graph of log(x)/(m)vslogp for an adsorption process is a straight line inclined at an angle of 45 ° with intercept

Correct answer

0

Step-by-step solution

The graphical representation of logarithmic form of Freundilich adsorption isotherm is as follows,

The graph of log(x)/(m)vslogp for an adsorption process is a straight line inclined at an angle of 45 ° with intercept

The logarithmic form of Freundilich adsorption isotherm is

logxm=logk+1nlogPlog\dfrac{x}{m} = logk + \dfrac{1}{n}logP

1n=tan⁡45°=1\dfrac{1}{n} = \tan 45 ° = 1

logk=0.6020=log⁡4logk = 0 . 6020 = \log 4

⇒K=4⇒ K = 4

∴xm=K⋅P1/n∴ \dfrac{x}{m} = K \cdot {P}^{1 / n}

xm=4(0.4)=1.6\dfrac{x}{m} = 4\left(0 . 4\right) = 1 . 6

xm=1.6=16×10−1\dfrac{x}{m} = 1 . 6 = 16 \times {10}^{- 1}

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