Adsorption of gas on a surface follows Freundlich adsorption isotherm. The plot of log $$\frac{x}{m}$$ versus log(P) gives a straight line with slope equal to 0.5, then:
($$\frac{x}{m}$$ is the mass of the gas adsorbed per gram of adsorbent)

We start with the empirical Freundlich adsorption isotherm, which relates the amount of gas adsorbed to the equilibrium pressure of the gas. The mathematical form is stated first:

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

Here $$\frac{x}{m}$$ is the mass of gas adsorbed per gram of adsorbent, $$P$$ is the equilibrium pressure, $$k$$ is the Freundlich constant, and $$n$$ is another constant that characterises the adsorption system.

To obtain a straight-line relationship we take the logarithm of both sides (using common logarithm, i.e. base 10):

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

This is of the form $$y = c + m\,x$$, where $$y=\log\!\left(\frac{x}{m}\right)$$, $$x=\log(P)$$, $$c=\log(k)$$ is the intercept, and the slope $$m=\dfrac{1}{n}$$.

The question states that the plot of $$\log\!\left(\dfrac{x}{m}\right)$$ versus $$\log(P)$$ is a straight line with slope $$0.5$$. Therefore:

$$\frac{1}{n}=0.5 \qquad\Longrightarrow\qquad n=\frac{1}{0.5}=2$$

Substituting $$n=2$$ back into the original Freundlich equation gives:

$$\frac{x}{m}=k\,P^{\,\frac{1}{2}}$$

Thus, at constant temperature, the amount of gas adsorbed per gram of adsorbent is proportional to $$P^{1/2}$$, i.e. to the square root of the pressure.

Hence, the correct answer is Option A.