b$.$ Consider a vertical tube with a cross$-$sectional area of $1c{m}^2.$ The bottom of the tube is close with a semipermeable membrane, and $1$ gram of sugar is placed in the tube. The membrane$-$closed end of the tube is then put into an inexhaustible supply of pure water at $T=300K.$ What will be the height of the water in the tube at equilibrium?

Suppose two columns $($of equal cross$-$section$)$ of water are separated at the bottom by a rigid porous membrane. If $n$ molecules of sugar are dissolved in column one, what will be the height difference between the two columns after they achieve steady state? At steady state there is no flux between the two columns, so at the level of the membrane, $P_1-{}_s=P_2.$ Since $P_1$ and $P_2$ are related to the height of the column of water through $P=gh,$ where $$ is the density of the fluid, $g$ is the gravitational constant, and $h$ is the height of the column. We suppose that the density of the two columns is the same, unaffected lyy the presence of the dissolved molecule, so we have

$g{h}_2=g{h}_1-\frac{nkT}{h_{1}A}$

where $A$ is the cross$-$sectional area of the columns. Since fluid is conserved, $h_1+h_2=2h_0,$ where $h_0$ is the height of the two columns of water before the sugar was added. From these, we find a single quadratic equation for $h_1$ :

$h_1^2-h_0{h}_1-\frac{nkT}{2gA}=0$

The positive root of this equation is $h_1=\frac{h_0}{2}+\frac{1}{2}\sqrt[\phantom{2}]{h_0^2+\frac{2nkT}{gA}},$ so that

$h_1-h_2=\sqrt[\phantom{2}]{h_0^2+\frac{2nkT}{gA}}-h_0.$

When the solute is at a high enough concentration, physical solutions of $(2\mathrm{.}145)$ are not possible. Specifically, if the solute is too concentrated with $\frac{nkT}{gA}>4h_0^2,$ the weight of a column of water of height $2h_0$ is insufficient to balance the osmotic pressure, in which case there is not enough water to reach equilibrium.