Question

When a graph is plotted between $log\frac{x}{m}$ and $logP$, it is a straight line with an angle ${45}^0$ and intercept at $0.3010$ on the y-axis. If the initial pressure is $0.3$ atm, then the amount of gas which will be adsorbed per gram of adsorbent is $x\times {10}^{-1}$. The value of $x$ is ______

Answer 1

According to Freundlich adsorption isotherm,

$\frac{x}{m}=k{P}^{1/n}$.

Here, as the angle is ${45}^o$, the value of $n$ is equal to $1$.

$log\frac{x}{m}=logk+logP$

$log\frac{x}{m}=0.3+log0.3$

Therefore, $\frac{x}{m}=0.6$.

Thus, the amount of gas adsorbed per gram is $6\times {10}^{-1}$.

Therefore, the value of $x$ is $6$.