JEE Main 24 Feb 2021 Shift 1 Solved Paper

Given graph is Freundlich adsorption isotherm.
According to equation, ‌

x

m

=kp‌

1

n

... (i)
where, ‌

x

m

= amount of gas
k and n= constants depending upon nature of adsorbate and absorbent.
Now, taking log both side in Eq. (i), we get log‌

x

m

=log‌k+‌

1

n

‌log‌p
For straight line, comparing above equation to y=mx+c
SO, m= slope =‌

1

n

hence, 0≤‌

1

n

≤1
It value lies in between 0 to 1 .
Hence, ‌

1

n

with (‌

1

n

=0 to 1).