Surface Chemistry Question 6
Question 6 - 30 January - Shift 2
The graph of $\log \frac{X}{m}$ vs $\log p$ for an adsorption process is a straight line inclined at an angle of $45^{\circ}$ with intercept equal to 0.6020 . The mass of gas adsorbed per unit mass of adsorbent at the pressure of $0.4 atm$ is $\times 10^{-1}$ (Nearest integer)
Given : $\log 2=0.3010$
Show Answer
Answer: (16)
Solution:
$\log \frac{x}{m}=\log k+\frac{1}{n} \log P$
$m n$
$\underline{1}=\tan 45^{\circ}=1$
n
$\log k=0.6020=\log 4$
$\Rightarrow K=4$
$\therefore \frac{x}{m}=K \cdot P^{1 / n}$
$\frac{x}{m}=4(0.4)=1.6$
$m$
$\underline{X}=1.6=16 \times 10^{-1}$
$m$
