Gibbs Free Energy for Incompressible Substances
Theoretical Background
-
Definition of Gibbs free energy: \(G = H - TS\)
-
Differential form: \(dG = V\,dp - S\,dT\)
-
For an isothermal process ($dT = 0$): \(dG = V\,dp\)
-
If the substance is incompressible (constant $V$): \(\Delta G = V\,(p_2 - p_1)\)
š This shows that for liquids and solids, pressure has a linear effect on Gibbs free energy, unlike gases where the relation is logarithmic.
Exercise
For 1.00 L of liquid water at $25^\circ C$, calculate the change in Gibbs free energy when pressure increases from 1 bar to 100 bar, assuming water is incompressible with molar volume $V_m = 18.0 \times 10^{-6}\, m^3 mol^{-1}$.
Step-by-Step Solution
Step 1. Number of moles
From the total volume and the molar volume:
\(n = \frac{V}{V_m} = \frac{1.00 \times 10^{-3}\, m^3}{18.0 \times 10^{-6}\, m^3 mol^{-1}} = 55.6\, mol\)
Step 2. Pressure change
\(\Delta p = p_2 - p_1 = (100 - 1)\, bar = 99\, bar\)
Convert to SI units:
\(1\, bar = 10^5\, Pa \quad \Rightarrow \quad \Delta p = 99 \times 10^5 = 9.9 \times 10^6\, Pa\)
Step 3. Gibbs free energy change
Use $\Delta G = n V_m \Delta p$:
\(\Delta G = (55.6)(18.0 \times 10^{-6})(9.9 \times 10^6)\, J\)
Final Answer:
\(\Delta G \approx 9.9\, kJ\)
Notes
- The calculation shows that for liquids, $\Delta G$ is proportional to the pressure change.
- Even a large increase in pressure (from 1 bar to 100 bar) only changes $\Delta G$ by a few kJ because the molar volume is small.
- For gases, the relation is instead: \(G = G^\circ + RT \ln \frac{p}{p^\circ}\) which grows logarithmically with $p$.
- This distinction explains why pressure strongly influences gas equilibria, but has only minor effects on liquid or solid phases.