91Ó°ÊÓ

Expression 5.68 for the Gibbs free energy of a dilute solution and expression 5.61 for the Gibbs free energy of an ideal mixture.

Dilute solution's Gibbs energy is given by

$G=N_A·{\mathrm{μ}}_0(T,P)+N_B·f(T,P)-N_{A}kT·\ln \left(N_A\right)+N_{B}kT·\ln \left(N_B\right)-N_{B}kT$

Ideal mixture's Gibbs free energy is given by:

$G=(1-x)G_A+x{G}_B+RT\left(x\ln \right(x)+(1-x\left)\ln \right(1-x\left)\right)$

The goal is to figure out when these two expressions are in agreement.

The Gibbs free energy is equal to the chemical potential multiplied by N_A and N_B.

$$\begin{gathered} G_A=N_A·{\mathrm{μ}}_A(T,P) \\ {G}_B=N_B·{\mathrm{μ}}_B(T,P) \end{gathered}$$

In the limit of N_A >> N_Bwe have:

$$\begin{gathered} x≈\frac{N_B}{N_A} \\ 1-x≈1 \end{gathered}$$

Now consider the Gibbs free energy of an ideal mixture using these approximations:

$G=N_A·{\mathrm{μ}}_A(T,P)+\frac{N_B}{N_A}·N_B·{\mathrm{μ}}_B(T,P)+RT\left(\frac{N_B}{N_A}\right)·\ln \left(\frac{N_B}{N_A}\right)$

Solvent chemical potential is given by:

${\mathrm{μ}}_A={\left(\frac{∂G}{∂N_A}\right)}_{T,P,N_A}={\mathrm{μ}}_0(T,P)-\frac{N_B}{N_A}·kT$

Solute chemical potential is given by:

${\mathrm{μ}}_B={\left(\frac{∂G}{∂N_B}\right)}_{T,P,N_B}=f(T,P)-kT\ln \left(\frac{N_B}{N_A}\right)$

On both sides of the membrane, the solvent's chemical potential must be the same:

${\mathrm{μ}}_0\left(T,P_1\right)={\mathrm{μ}}_0\left(T,P_2\right)-\frac{N_B}{N_A}·kT$

Because the two pressures aren't too dissimilar, we may approximate:

${\mathrm{μ}}_0\left(T,P_2\right)={\mathrm{μ}}_0\left(T,P_1\right)+\left(P_2-P_1\right)·\frac{∂{\mathrm{μ}}_0}{∂P}$

Putting the two equations together,

$\left(P_2-P_1\right)·\frac{∂{\mathrm{μ}}_0}{∂P}=\frac{N_B}{N_A}·kT$

It's important to remember that a pure substance's chemical potential is simply the Gibbs free energy per particle:

$\frac{∂{\mathrm{μ}}_0}{∂P}=\frac{1}{N_A}·\frac{∂G}{∂P}=\frac{V}{N_A}$

Therefore,

$\left(P_2-P_1\right)·\frac{V}{N_A}=\frac{N_B}{N_A}·kT$

The osmotic pressure of a dilute solution can be written as:

$P_2-P_1=\frac{N_B·k·T}{V}$

For every molecule of whatever else, there are around 200 water molecules in a typical cell. We can compute $\frac{N_B}{V}$ because a mole of water has a mass of 18 g and a volume of 18 cm³.

$\frac{N_B}{V}=\frac{{100}^3}{200×18}=278\frac{\mathrm{mol}}{{\mathrm{m}}^3}$

To find the pressure difference, substitute the values

$P_2-P_1=\frac{N_B·k·T}{V}=278×8.314×300=7\mathrm{atm}$