1.14.51: Partial Molar Properties: General
The laws of thermodynamics and the associated treatment of the thermodynamic properties of closed systems concentrate attention on macroscopic properties. Although we may define the composition of a closed system in terms of the amounts of each chemical substance in a system, general thermodynamic treatments direct our attention to macroscopic properties such as, volume \(\mathrm{V}\), Gibbs energy \(\mathrm{G}\), enthalpy \(\mathrm{H}\) and entropy \(\mathrm{S}\). We need to ‘tell’ these thermodynamic properties that a given system probably comprises different chemical substances. In this development the analysis is reasonably straightforward if we define the system under consideration by the ‘Gibbsian ‘set of independent variables; i.e. \(\mathrm{T}, \mathrm{~p}\) and amounts of each chemical substance. Thus,
\[\mathrm{G}=\mathrm{G}\left[\mathrm{T}, \mathrm{p}, \mathrm{n}_{1}, \mathrm{n}_{2}, \ldots \mathrm{n}_{\mathrm{i}}\right] \nonumber \]
In equation (a), the Gibbs energy is defined by intensive variables \(\mathrm{T}\) and \(\mathrm{p}\) together with extensive composition variables. In many cases the task of a chemist is to assay a system to determine the number and amounts of each chemical substance in the system.
The analysis leads to the definition of the chemical potential for each substance \(\mathrm{j}\), \(\mu_{\mathrm{j}}\) in a closed system.
Consider a solution comprising \(\mathrm{n}_{1}\) moles of solvent, liquid chemical substance 1, and \(\mathrm{n}_{\mathrm{j}}\) moles of solute, chemical substance \(\mathrm{j}\). We ask—what contributions are made by the solvent and by the solute to the volume of the solution at defined \(\mathrm{T}\) and \(\mathrm{p}\)? In fact we can only guess at these contributions [1]. This is disappointing. The best we can do is to probe the sensitivity of the volume of a given solution to the addition of small amounts of either solute or solvent. This approach leads to a set of properties called partial molar volumes. Here we explore the definition of these properties. The starting point is the Gibbs energy of a solution. We develop the argument in a way which places the Gibbs energy at the centre from which all other thermodynamic variables develop. For a closed system containing i-chemical substances,
\[\mathrm{G}=\sum_{\mathrm{j}=1}^{\mathrm{j}=\mathrm{i}} \mathrm{n}_{\mathrm{j}} \, \mu_{\mathrm{j}} \nonumber \]
The later equation signals that the total Gibbs energy is given by the sum of products of amounts and chemical potentials of each chemical substance in the system. For an aqueous solution containing \(\mathrm{n}_{\mathrm{j}}\) moles of solute \(\mathrm{j}\) and \(\mathrm{n}_{1}\) moles of solvent 1 ( water),
\[G(a q)=n_{1} \, \mu_{1}(a q)+n_{j} \, \mu_{j}(a q) \nonumber \]
We do not have to attach to equation (c) the condition ‘at fixed \(\mathrm{T} and \mathrm{p}\)’. Similarly the volume of the solution is given by equation (d).
\[V(a q)=n_{1} \, V_{1}(a q)+n_{j} \, V_{j}(a q) \nonumber \]
The same argument applies in the case of a system prepared using \(\mathrm{n}_{1}\) moles of water, \(\mathrm{n}_{\mathrm{X}}\) moles of solute \(\mathrm{X}\) and \(\mathrm{n}_{\mathrm{Y}}\) moles of solute \(\mathrm{Y}\).
\[\mathrm{G}(\mathrm{aq})=\mathrm{n}_{1} \, \mu_{1}(\mathrm{aq})+\mathrm{n}_{\mathrm{X}} \, \mu_{\mathrm{X}}(\mathrm{aq})+\mathrm{n}_{\mathrm{Y}} \, \mu_{\mathrm{Y}}(\mathrm{aq}) \nonumber \]
Complications emerge however if solute \(\mathrm{X}\) and \(\mathrm{Y}\) are in chemical equilibrium; e.g. \(X(a q) \Leftrightarrow Y(a q)\). Then account must be taken of the fact that \(\mathrm{n}_{\mathrm{x}}^{\mathrm{eq}}\) and \(\mathrm{n}_{\mathrm{x}}^{\mathrm{eq}}\) depend on \(\mathrm{T}\) and \(\mathrm{p}\).
Footnote
[1] I have on my desk a flask containing water (\(\ell ; 100 \mathrm{~cm}^{3}\)) and 20 small round steel balls, each having a volume of \(0.1 \mathrm{~cm}^{3}\). I add a steel ball to the flask and the volume of the system, water + steel ball, is \(100.1 \mathrm{~cm}^{3}\). I add one more steel ball and the volume of the system increases by \(0.1 \mathrm{~cm}^{3}\). So in this simple case I can equate directly the volume of the pure steel balls \(\mathrm{V}^{*}\)(balls) with the partial molar volume of the balls in the system, water + balls.
I have on my desk an empty egg carton designed to hold six eggs. The volume of the carton is represented as \(\mathrm{V}(\mathrm{c})\) as judged by the volume occupied in a food store.. The volume of one egg is \(\mathrm{V}^{*}\)(egg), the superscript * indicating that we are discussing the property of pure eggs. I now ‘add’ one egg to the egg carton which does not change its volume ----again as judged by the volume occupied in a food store. In other words the partial molar volume of eggs in the egg carton \(\mathrm{V}(\mathrm{egg})\) is zero;
\[V(\text { egg })=\left(\frac{\partial V(\text { system })}{\partial n(\text { egg })}\right)=\text { zero. Or, } V(\text { egg })-V^{*}(\text { egg })<0 \nonumber \]