Are there fundamentally two ways for volume of total solution might not be preserved?

Under the assumption that outside factors, such as pressure and temperature, remain invariant…

Chemical reaction:

Suppose that the reaction $\ce{aA + bB -> cC + dD}$ takes place in a solvent in which solvent-solute interaction is not present:

$$\begin{align}\Delta\mathcal{\bar{V}}&= (c\mathcal{\bar{V}}_C+d\mathcal{\bar{V}}_D)-(a\mathcal{\bar{V}}_a + b\mathcal{\bar{V}}_B) \\&= c\left (\dfrac{\partial V}{\partial n_C} \right)_{p,T,n_{\neq C}} + d\left (\dfrac{\partial V}{\partial n_D} \right)_{p,T,n_{\neq D}}\nonumber-\biggr(a\left (\dfrac{\partial V}{\partial n_A} \right)_{p,T,n_{\neq A}}+b\left (\dfrac{\partial V}{\partial n_B} \right)_{p,T,n_{\neq B}}\biggr)\end{align}$$

Where $\mathcal{\bar{V}}$ denotes the partial molar volume of a substance.

Physical reaction:

Molar volumes are (not) preserved, but (also) final products react with the solvent and causes volume contraction or expansion. Which is the case in the dissolution of $\text{NaCI}$ in $\ce{H_2O}$ solvent:

$$\ce{NaCl <=> Na+ + Cl-}$$

  • $\begingroup$ It would be a matter of molar volumes, if it was not in solution. $\endgroup$
    – Mithoron
    Commented May 28 at 17:51
  • $\begingroup$ Usually chemical reactions take place in rather dilute solution milli molar for example so any volume change will be small as the volume of solvent dominates. But if you form a polymer, for example, there will be a huge change, see video of making nylon66 from solution for example. Mixing liquids often leads to change in volume, alcohols and water for example but there are many other examples. $\endgroup$
    – porphyrin
    Commented May 29 at 6:50
  • $\begingroup$ Imagine dissolving a compound in a solvent, or mixing two miscible liquids. There is no reason to expect that the mixture density has incidentally such a value that the mixture volume is the sum of the component volumes. $\endgroup$
    – Poutnik
    Commented May 30 at 9:02

2 Answers 2


The second case is correct -- even without ionization. For example, 50 ml $\ce{H2O}$ + 50 ml $\ce{C2H5OH}$ mix to ~90 ml of solution, not 100 ml.

However, the first case you propose does not take into account precipitation or evolution of a gas or the density of the products. The resultant solution might have the same volume or lesser or greater volume compared to the reactants; there is no way to tell without specifying the reactants. The volume might change, or might not.

Consider a solution of glucose in water, with density greater than that of water. Add zymase, and the solution changes to ethanol in water, with evolution of $\ce{CO2}$, both less dense than water, with greater volume.

There is also this example of a double-displacement reaction, where all ions precipitate: $\ce{CuSO4(aq) + Ba(OH)2(aq) -> Cu(OH)2\downarrow + BaSO4\downarrow}$ . Would the volume increase or decrease,or stay the same? My SWAG: the volume of the liquid is about the same, and the total volume, including the precipitate, increases. One might need to experiment to get an answer.


By thermodynamics, the molar volume within a phase can change because of three things (in almost all cases): changes in pressure $p$, temperature $T$, and compositions of the species $x_1,x_2,...,x_n$ comprising the phase. As you said that $p$ and $T$ are invariant, in those two processes you have the same phenomenon, namely the change of compositions.

In my opinion, it would be not wise to classify the molar volume regarding the type of process. Firstly, this would lead to a possible confusion that two different processes under the same $p,T,\{\mathbf{x}\}$ exhibit different molar volumes, but this is not true. Secondly, there are many, many physical processes and chemical processes. For example, when we go up in height, the temperature and pressure decreases, and thus molar volume experiences a change. You would have to add a subcategory there, because this is a physical process but not a dissolution. Digging even more, if you take into account electromagnetic radiation, you must add to the physical process (gravity), the radical-based mechanism that changes the constituents of the atmosphere (ozone and oxygen), and again the molar volume changes at a certain height. Likewise, you would have to add another subcategory in the chemical process, because not all homogeneous phase reactions are induced by the interaction with photons.

Thermodynamics is a fine way to skip all these exceptions, in one way. So, I would simply say:

  • The first case represents a change in molar volume due to the change of the composition of the substances as the reaction progress. I would like to say that the formula you have is not true. The molar volume is in fact \begin{equation} V = \overline{V}_\ce{A}(p,T,\{x_j\})x_\ce{A} + \overline{V}_\ce{B}(p,T,\{x_j\})x_\ce{B} + \overline{V}_\ce{C}(p,T,\{x_j\})x_\ce{C} + \overline{V}_\ce{D}(p,T,\{x_j\})x_\ce{D} \end{equation} where the $\overline{V}$'s are the partial molar volumes of the species. They are not constants, and in the general case they are function of pressure, temperature, and also the molar fractions of other species. Almost always they must be obtained by experimental data. They could also be calculated with statistical thermodynamics under some assumptions, and postulating a form of the Helmholtz free energy.
  • The second case represents a change in molar volume also due to the change of the composition of the phase. Initially we only have solvent, i.e. $\ce{H2O}$, and after some time some solute enters the liquid phase so we have $\ce{H2O}$, $\ce{Na^+}$, and $\ce{Cl^-}$.

Both cases respond to the same explanation: change in composition $\{\mathbf{x}\}$.


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