The mass of gas adsorbed, $x$ , per unit mass of adsorbate, m, was measured at various pressures, $p$ . A graph between $\log [x / m]$ and $\log p$ gives a straight line with slope equal to $2$ and the intercept equal to $0.4771$ . The value of $[x / m]$ at a pressure of $4$ atm is: $(Given \log 3 = 0.4771)$

Open in App

Solution

Step 1 : Given data
$c = \log k$ , $x = 2, k = 3$
$c = \log 3 = 0.4771$
Step 2: Apply absorbed equation
$\frac{x}{m} = k p^{x} . . . . . (1)$
Take logarithm on both sides
$\log \frac{x}{m} = \log k + x \log p$
Substitute the given values in the pressure equation
$\frac{x}{m} = 3 \times (4)^{2}$
$\frac{x}{m} =$ 48
Therefore the value for $\frac{x}{m} = 48$

Suggest Corrections

Similar questions

Q. The Freundlich adsorption isotherm, a plot of

log (x / m) (along y-axis ) and log P (along x -axis )

gives a straight line. The intercept and slope for the line is

0.4771 and 2

respectively. The mass of gas adsorbed per gram of adsorbent is

x \times 10^{- 4} g

. If the initial pressure is

0.04 atm

. The value of

x

is.

(log 3 = 0.4771)

Q. A graph plotted between

l o g (\frac{x}{m})

l o g P

(

P

is pressure of gas in atm) shows a straight line with slope equal to

1

and intercept equal to

0.4771

. The extent of adsorption

(\frac{x}{m})

of gas at the pressure

2

atm is :

Q. The slope of the straight line graph between log x/m and log P for the adsorption of a gas on solid at constant temperature is:
(Where, x = mass of the gas adsorbed, m = mass of adsorbent, P = pressure)

For Freundlich adsorption isotherm, a plot of $\log (x / m) (y - a x i s)$ and $\log p (x - axis)$ gives a straight line. The intercept and slope for the line is $0.4771$ . and $2$ , respectively. The mass of gas, adsorbed per gram of adsorbent if the initial pressure is $0.04 a t m$ s……..… $\times 10^{- 4} g . (\log 3 = 0.4771)$

Q. Graph between log $\left ( \dfrac{x}{m} \right )$ and $log ~p$ is a straight line at an angle $45^\circ $ with intercept on y-axis at 0.3010. Calculate the amount of gas adsorbed in gram per gram of the adsorbent when pressure is $0.2 ~atm$.

Join BYJU'S Learning Program

The mass of gas adsorbed, x, per unit mass of adsorbate, m, was measured at various pressures, p. A graph between log[x/m]and logp gives a straight line with slope equal to 2and the intercept equal to 0.4771. The value of[x/m]at a pressure of 4atm is:(Givenlog3=0.4771)

Step 1 : Given datac=logk, x=2,k=3c=log3=0.4771Step 2: Apply absorbed equation xm=kpx.....(1)Take logarithm on both sideslogxm=logk+xlogpSubstitute the given values in the pressure equationxm=3×(4)2xm=48Therefore the value for xm=48