If x gram of gas is adsorbed by m gram of adsorbent at pressure P the plot of log $$\frac{x}{m}$$ versus log P is linear. The slope of the plot is: (m and k are constants and n > 1)

The behaviour of many adsorption systems at low pressure is described by the empirical Freundlich adsorption isotherm. Its mathematical form is stated first, because we will need it for the derivation:

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

Here $$\frac{x}{m}$$ is the amount of gas adsorbed per unit mass of the adsorbent, $$P$$ is the equilibrium pressure of the gas, while $$k$$ and $$n>1$$ are empirical constants characteristic of the particular adsorbate-adsorbent pair and the temperature.

Now we are told that a graph of $$\log\left(\dfrac{x}{m}\right)$$ versus $$\log P$$ is a straight line. To see the equation of that straight line, we take the common (base-10) logarithm of both sides of the Freundlich equation. Using the property of logarithms $$\log(a\,b)=\log a+\log b$$ and $$\log(a^{b})=b\log a$$, we proceed step by step:

First take the logarithm of the left side:

$$\log\left(\frac{x}{m}\right)=\log\left(k\,P^{1/n}\right)$$

Next separate the product inside the logarithm:

$$\log\left(k\,P^{1/n}\right)=\log k+\log\left(P^{1/n}\right)$$

Now apply the power rule of logarithms to the second term:

$$\log\left(P^{1/n}\right)=\frac{1}{n}\,\log P$$

Substituting this back we get:

$$\log\left(\frac{x}{m}\right)=\log k+\frac{1}{n}\,\log P$$

We can compare this equation with the standard equation of a straight line, which in the form $$y=mx+c$$ has slope $$m$$ and intercept $$c$$. Identifying the variables:

$$y=\log\left(\dfrac{x}{m}\right),\quad x=\log P,$$

we see that the coefficient of $$\log P$$—namely $$\dfrac{1}{n}$$—is the slope of the straight line.

Hence, the slope of the plot of $$\log\left(\dfrac{x}{m}\right)$$ versus $$\log P$$ is $$\dfrac{1}{n}$$.

Hence, the correct answer is Option B.