Question

Compare expression 5.68 for the Gibbs free energy of a dilute solution to expression 5.61 for the Gibbs free energy of an ideal mixture. Under what circumstances should these two expressions agree? Show that they do agree under these circumstances, and identify the function f(T, P) in this case.

Short answer

Therefore,

$P_2-P_1=7\mathrm{atm}$

Step-by-step solution

Given information

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.

Explanation

Dilute solution's Gibbs energy is given by

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

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.

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

In the limit of N_A >> N_Bwe have:

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

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

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

Solvent chemical potential is given by:

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

Solute chemical potential is given by:

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

Explanation

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

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

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

${\mu }_0\left(T,P_2\right)={\mu }_0\left(T,P_1\right)+\left(P_2-P_1\right)·\frac{\partial {\mu }_0}{\partial P}$

Putting the two equations together,

$\left(P_2-P_1\right)·\frac{\partial {\mu }_0}{\partial P}=\frac{N_B}{N_A}·kT$

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

$\frac{\partial {\mu }_0}{\partial P}=\frac{1}{N_A}·\frac{\partial G}{\partial P}=\frac{V}{N_A}$

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:

$P_2-P_1=\frac{N_B·k·T}{V}$

Calculations

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.

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

To find the pressure difference, substitute the values

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