# A plot of x/m versus log p for the adsorption of a gas on a solid gives a straight line with slope equal to

$(a)\;\large\frac{1}{n}\qquad(b)\;log\;k\qquad(c)\;-log\;k\qquad(d)\;n$

Answer : $\;\large\frac{1}{n}$
Explanation :
For the adsorption of a gas on a solid $\;\large\frac{x}{m}=kp^{\large\frac{1}{n}}\;$
Therefore , $\;log \large\frac{x}{m}=log\;k+\large\frac{1}{n}\;log\;p\;$ (ontaking log can both sides )
Therefore , plot of $\;\large\frac{x}{m}\;$ vs $\;log\;p\;$ will be linear with slope = $\;\large\frac{1}{n}$