JEE PYQ: Surface Chemistry Question 12
Question 12 - 2020 (02 Sep Shift 1)
Type: Numerical
The mass of gas adsorbed, $x$, per unit mass of adsorbate, $m$, was measured at various pressures, $p$. A graph between $\log \frac{x}{m}$ and $\log p$ gives a straight line with slope equal to 2 and the intercept equal to 0.4771. The value of $\frac{x}{m}$ at a pressure of 4 atm is: (Given $\log 3 = 0.4771$)
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Answer: 6
Solution
$\left(\frac{x}{m}\right) = k(p)^n$
$\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log p$
Slope $= \frac{1}{n} = 2$, so $n = \frac{1}{2}$
Intercept $= \log k = 0.477$. So $k = \text{Antilog}(0.477) = 3$
$\frac{x}{m} = k(p)^n = 3(4)^{1/2} = 3 \times 2 = 6$
BETA
