User:Dbrogioli/Calculation of consumed free energy

Calculation of osmotic pressure from Gibbs free energy
Take a volume V of salt water at concentration c; put an osmotic membrane; put a small volume v of fresh water on the other side. The osmotic pressure is $$\Pi\left(c\right)$$. When the volume v passes into the more concentrated solution, a work is done; in isothermal conditions, it is equal to the Gibbs free energy decrease:

$$ \Pi\left(c\right) v = V g\left(c\right) - \left(V+v\right) g\left(\frac{cV}{V+v}\right) $$

where $$g\left(c\right)$$ is the Gibbs free energy density (i.e. per unit volume of solution).

By taking the limit $$v/V \to 0$$:

$$ \Pi\left(c\right) = g\left(c\right) - c \frac{\partial}{\partial c} g\left(c\right) $$

Calculation of the Gibbs free energy density
Once $$\Pi\left(c\right)$$ is known, $$g\left(c\right)$$ can be calculated, but is defined up to a constant $$\delta$$ and a linear term $$\gamma c$$: any term $$\delta + \gamma c$$ can be added without changing the equality.

Any linear term is negligible when considering free energy changes in mixing (see below).

This allows to define the Gibbs free energy density $$g\left(c, c_0\right)$$ relative to a concentration $$c_0$$, that is the Gibbs free energy density up to a constant and linear term $$\delta + \gamma c$$, which moreover has the property that:

$$ g\left(c, c_0\right) = 0 $$

and

$$ \frac{\partial}{\partial c} g\left(c, c_0\right) = 0 $$

Free energy changes in mixing
Consider a Gibbs free energy density $$g\left(c\right)$$ defined up to a constant and linear term $$\delta + \gamma c$$. Consider two volumes $$V_A$$ and $$V_B$$ of solution at concentrations $$c_A$$ and $$c_B$$. The total free energy is:

$$ F_{initial} = V_A g\left(c_A\right) + \delta V_A + \gamma V_A c_A + V_B g\left(c_B\right) + \delta V_B + \gamma V_B c_B $$

Now, we mix the two solutions. For NaCl solutions, the volumes are summed:

$$ V_{total} = V_A + V_B $$

while the total concentration is:

$$ c_{total} = \frac{c_A V_A + c_B V_B}{V_A + V_B} $$

The total free energy becomes:

$$ F_{final} = \left(V_A +V_B\right) g\left(\frac{c_A V_A + c_B V_B}{V_A + V_B}\right) + \delta \left(V_A + V_B\right) \gamma \left(V_A + V_B\right) \frac{c_A V_A + c_B V_B}{V_A + V_B} $$

We immediately see that the difference $$\Delta F=F_{initial}-F_{final}$$ does not depend on $$\delta$$ and $$\gamma$$:

$$ \Delta F = V_A g\left(c_A\right) + V_B g\left(c_B\right) - \left(V_A +V_B\right) g\left(\frac{c_A V_A + c_B V_B}{V_A + V_B}\right) $$

The same calculations apply also to the Gibbs free energy density $$g\left(c, c_0\right)$$ relative to a concentration $$c_0$$; as we have seen, we can use any value of $$c_0$$ for calculations of free energy changes, since $$g\left(c, c_0\right)$$ differs from $$g\left(c\right)$$ only by a constant and linear term.

Free energy change when water is dispersed into the sea
This is a particular case of the previous section. In particular, $$V_A=v$$ is a small quantity with respect to the volume of the sea $$V_B$$. Taking the limit:

$$ \Delta F = v\left[ g\left(c_A\right) - g\left(c_B\right) - \left(c_A - c_B\right) \frac{\partial}{\partial c}g\left(c_B\right) \right] $$

It is useful to use the Gibbs free energy density $$g\left(c, c_0\right)$$ relative to a concentration $$c_B$$. In this case, the equation simplifies to:

$$ \Delta F = v g\left(c_A, c_0\right) $$

When we chose a value $$c_0$$, the function $$g\left(c, c_0\right)$$ defines the free energy density with respect to the solution disposed into a sea with concentration $$c_0$$.