If the initial pressure of a gas is 0.03 , at | vedclass

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(C) The Freundlich adsorption isotherm is given by $\frac{x}{m} = k \, P^{\frac{1}{n}}$.Taking logarithm on both sides,we get $\log \frac{x}{m} = \log

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$12$

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(C) The Freundlich adsorption isotherm is given by $\frac{x}{m} = k \, P^{\frac{1}{n}}$.Taking logarithm on both sides,we get $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$.Comparing this with the equation of a straight line $y = mx + c$,the slope is $\frac{1}{n}$ and the intercept is $\log k$.From the graph,the slope is $1$,so $\frac{1}{n} = 1 \Rightarrow n = 1$.The intercept is $\log k = 0.602$. Since $\log 4 \approx 0.602$,we have $k = 4$.Now,substitute the values into the equation: $\frac{x}{m} = 4 	imes (0.03)^1$.$\frac{x}{m} = 0.12 \, g$.To express this in the form $...... 	imes 10^{-2} \, g$,we write $0.12 = 12 	imes 10^{-2} \, g$.

Gas	$H_2$	$CH_4$	$CO_2$	$SO_2$
Critical Temp $/ K$	$33$	$190$	$304$	$630$

If the initial pressure of a gas is $0.03 \, atm$,the mass of the gas adsorbed per gram of the adsorbent is $...... \, \times 10^{-2} \, g$. Based on the provided graph,determine the value.

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