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1

### AIPMT 2012 Prelims

In Freundlich adsorption isotherm, the value of 1/n is
A
between 0 and 1 in all cases
B
between 2 and 4 in all cases
C
1 in case of physical absorption
D
1 in case of chemisorption.

## Explanation

$${x \over m} = k.{p^{{1 \over n}}}$$

Where $${x \over m}$$ is the ratio of amount of adsorbent to the amount of adsorbate.

The value of n is always greater than 1. So, the value of $${{1 \over n}}$$ lies between 0 and 1 in all cases.
2

### AIPMT 2011 Prelims

If x is amount of adsorbate and m is amount of adsorbent, which of the following relations is not related to adsorption process?
A
x/m = $$f$$(p) at constant T
B
x/m = f(T) at constant p
C
p = f(T) at constant (x/m)
D
$${x \over m} = p \times T$$

## Explanation

$${x \over m} = p \times T$$ is incorrect.

3

### AIPMT 2007

The Langmuir adsorption isotherm is deduced using the assumption
A
the adsorption sites are equivalent in their ability to adsorb the particles
B
the heat of adsorption varies with coverage
C
the adsorbed molecules interact with each other
D
the adsorption takes place in multilayers.

## Explanation

Langmuir adsorption isotherm is based on the assumption that every adsorption site is equivalent and that the ability of a particle to bind there is independent of whether nearby sites are occupied or not occupied.
4

### AIPMT 2006

A plot of log(x/m) versus log p for the adsorption of a gas on a solid gives a straight line with slope equal to
A
log K
B
$$-$$ log K
C
n
D
1/n

## Explanation

According to Freundlich adsorption isotherm, in the median range of pressure

$${x \over m} \propto {P^{{1 \over n}}}$$

$$\Rightarrow$$  $${x \over m}$$ = kP$$^{{1 \over n}}$$

taking log both sides, we get,

log$${x \over m}$$ = logk + $${1 \over n}$$ logP

Here in graph between log$${x \over m}$$ and logP, slope is $${1 \over n}$$ and intercepts = log k.

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