Question

Use chemical potentials to explain the driving force for the spontaneous mixing of gases or liquids

Answer 1

Chemical potentials to explain the driving force for the spontaneous mixing of gases or liquid

We have seen that for pure A, the chemical potential is given by:

where p is the pressure of the gas. If another substance B is added to A, then the chemical potential of A in the mixture is given by:

where p_A is the partial pressure of A in the mixture.

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(1-2) The Gibbs energy, entropy and enthalpy of mixing of gases

The Gibbs free energy of mixing -

We have seen that for a binary system:

and that for pure A, (and similarly for B), the chemical potential is given by:

Thus let us assume that we start with two pure gases A and B (amounts n_A and n_B respectively) in two separate containers at the same temperature T and pressure p. At this stage, the gases have their 'pure' chemical potential, and hence, the total initial Gibbs energy whilst they are still in separate containers is given by:

When we mix the gases, the pressures 'p' have to be replaced by their partial pressures in the mixture, i.e. p_A and p_B where p_A = p_B = p. The total Gibbs energy then changes to:

and hence:

i.e.:

[why?]

But, (i) n_J=x_J n , and (ii) from Dalton's law, p_J / p = x_A . Hence we have:

Note that since x_J are always less than 1, ln(x_J) are always negative and hence for all perfect gasses, and in all proportions, D_mixG < 0 and hence gases mix spontaneously.

The entropy of mixing can be calculated from a corollary of the fundamental equation of thermodynamics, i.e.:

and hence:

i.e. D_mixS > 0 for all compositions since ln(x_A) and ln(x_B) are both negative since x_A and x_B are necessarily less that 1 (see graph of ln(x))

The enthalpy of mixing (isothermal and isobaric) of two perfect gases may be calculated from:

which gives:

as expected since the molecules of perfect gases do not interact.

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(2) Liquid mixtures

The basis for this discussion shall be that the chemical at equilibrium, the chemical potential of a substance present as a vapour must be equal to its chemical potential in the liquid, as illustrated in Fig. 1 below.
  
  

Fig. 1 - At equilibrium, the chemical potential of the gaseous form of a substance A is equal to the chemical potential of its condensed phase. The equality is preserved if a solute is also present. Because the chemical potential of A in the vapour depends on its partial vapour pressure, it follows that the chemical potential of liquid A can be related to its partial vapour pressure.
  

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(2-1) The chemical potential of ideal liquid solutions - Raoult's law

In this discussion we shall make use of the superscript * to denote that the chemical potential refers to the pure substance. The chemical potential of pure A is written as if we need to distinguish between the liquid phase or the vapour (gas) phase. The vapour pressure of the pure liquid A is hence written as .

The chemical potential of A (in liquid and in the vapour) is given by

where:

If another substance (a solute, B) is also present in the liquid, then we must replace the chemical potential and vapour pressure of the pure liquid by the partial chemical potential and the partial vapour pressure, i.e.:

We can combine these two equations to eliminate , to give:

In the case of an ideal solution, when both the solute and solvent obey Raoult's law (for closely related liquids, the ratio of the partial pressure of each component to its vapour pressure as a pure liquid is approximately equal to the mole fraction in the liquid mixture, see Fig. 2a/b for a graphical illustration), i.e.:

We have: