 Solution

At fixed temperature,Thus magnetic-field-dependent Langmuir adsorption is even in $B$ and increases with $|B|$. At fixed nonzero gas pressure, stronger field increases surface coverage; in an isotherm plot, the half-coverage pressure $p_{1/2}=1/b$ moves to lower pressure. The saturation coverage remains one because the number of sites is unchanged.

For weak fields, with ${\theta }_0=b(0)p/[1+b(0)p]$,There is no linear small-field change. At large $|x|$, $\cosh x\sim e^{|x|}/2$, and at fixed $p>0$ the coverage approaches one. The field lowers the free energy of the surface spin ensemble relative to the unsplit bulk; simply averaging the two surface energies to $-E_b$ would miss this equilibrium effect.