Question

Graph between log (X/m) Vs logP has a slope = 2 and intercept= 0.477 find (x/m) at pressure 4 atm
Given log3 = 0.477​

Answer 1

Graph between log(X/m) Vs logP has a slope = 1/2 and intercept = 0.477

given log3 = 0.477

To find : find (x/m) at pressure 4 atm

solution : we know the relation between rate of adsorption and pressure is given by,

x/m = kp^(1/n)

taking log both sides,

log(x/m) = log(k p^(1/n))

⇒log(x/m) = logK + log(P^(1/n))

⇒log(x/m) = logK + 1/n logP

here logK is intercept, 1/n is slope of equation

here, logK = 0.477

so, K = 3 [ as log3 = 0.477 ]

and slope of equation, 1/n = 1/2

and pressure, P = 4

now (x/m) = kp^(1/n)

= (3) × (4)½

= 3 × 2

= 6

Therefore the rate of adsorption (x/m) is 6

Answer 2

Answer:

48

Explanation:

k=3

1/n = slope = 2

(x/m)= k (p)^(1/n)

x/m = 3 × 4^2 =48