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Thermodynamics of Mixing

The molar Gibbs free energy of mixing is

\[\Delta G_{m,mix} = RT(x_i) \ln(x_i )\]

or for a two component solution

\[ \Delta G_{m,mix} = nRT(x_A\ln x_A + x_B \ln x_B) \]

and for more than two components

\[ \Delta G_{m,mix} = nRT(x_A \ln x_A + x_B \ln x_B+ x_c \ln x_c+ ...)\]

For the first equation, \(m\) denotes Gibbs free energy per mole of solution, and \(x_i\) is the mole fraction of component \(i\) in the liquid state. Note that when calculating the free energy of mixing in an ideal solution is always negative, since each \(\ln x_i\) term becomes be negative value. Thus showing how ideal solutions are always completely miscible.

How the Equations Calculate Gibbs Free Energy For Ideal Solutions

The origins of the m value comes from the equation of partial molar quantities which is the combination of the first and second laws of thermodynamics's equations for an open system. Note that these equations are expressed in terms of chemical potentials of their individual components.

First law's:

\[ dU = (dq) - p(dV) = (dq) + w\]

Second law's:

\[dq_{rev}= T(dS)\]

\(dq\) represents the change in enthalpy or heat, \(w\) is work, \(dS\) is the change in entropy, \(T\) is temperature, and \(dU\) is the changes of internal energy. By combining these two equations, integrating them, and placing in the number of moles for a solution, the new formula becomes

\[dU = TdS = pdV + m_1\,dn_1+m_2\,dn_2\]

where \(m\) came from was the replacement of dU divided by dn (the change of moles). However it can also be \(dG\) (the change in Gibbs Free energy) divided by \(dn\). By using the integrated form of \(G\):

\[G = m_1n_1+m_2n_2+...\]

one can find the Gibbs free energy of the ideal mixture for multiple components: 

\[ \Delta G_{m,mix} = nRT(x_A \ln x_A + x_B \ln x_B+x_c \ln x_c+...)\]

Ideal Vapor Pressure Relationship

Raoult's Law says that \(P_A\),the experimental partial pressure of the solvent, \(x_A\) is the mole fraction of the solvent, and \(P^*_A\) is the vapor pressure of the pure solvent at the given temperature.

\[ P_A = x_AP^*_A\]

For ideal solutions, the entire composition range of temperature and concentration and shows no internal energy change on mixing and no attractive force between components. The equilibrium vapor pressure of the pure component is the mole fraction of the component in solution. It can also be shown that volumes are strictly additive for ideal solutions.

Why is Raoult's law applicable for ideal solutions?

  1. mi = m*i+ RT ln Xi                        (formula for chemical potentials only)
  2. ml = mg           (the partial gibbs energy in both liquid and gas form is equal*)
  3. m*l+ RT ln Xl = m^g + RT ln (P1/1 atm) (for the second, ideal conditions in gas form)
  4. In liquid form, the ideal solution is pure, thus its vapor pressure is also pure. Replace m*l with m^g + RT ln (P*1/1 atm)
  5. Subtract: [m^g + RT ln (P1/1 atm)] - [m^g + RT ln (P*1/1 atm)]
  6. ln Xl = ln (P1/P*1)                    (get rid of natural logs)
  7. p1 = XlP*1        (Raoult's law)


  1. Petrucci, Harwood, Herring, Madura. General Chemistry: Principles & Modern Applications, Ninth Ed. Upper Saddle River, NJ: Pearson Education, Inc., 2007.
  2. Salzman, W.R. "Mixtures; Partial Molar Quantities; Ideal Solutions". Chemical Thermodynamics. Last updated: 21 Oct 04.
  3. Petrucci, Herring, Madura, Bissonette. General Chemistry: Principles and Modern Applications, Tenth Ed. Upper Saddle River, NJ: Pearson Education, Inc., 2011.


  • Ryan Woods (UCD), Shelley Chu (UCD)