Gibbs Free Energy for Incompressible Substances
Theoretical Background
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Definition of Gibbs free energy: \(G = H - TS\)
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Differential form for a closed system of fixed composition: \(dG = V\,dp - S\,dT\)
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For an isothermal process ($dT = 0$): \(dG = V\,dp\)
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For a fixed amount of an incompressible substance (constant total volume $V$): \(\Delta G = V\,(p_2 - p_1)\)
For a fixed amount of an incompressible substance at constant temperature, the Gibbs free-energy change is therefore proportional to the pressure change.
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 $\Delta G = n V_m \Delta p$:
\(\Delta G
= (55.6\,\mathrm{mol})
(18.0 \times 10^{-6}\,\mathrm{m^3\,mol^{-1}})
(9.9 \times 10^6\,\mathrm{Pa})\)
Final Answer:
\(\Delta G \approx 9.9\, kJ\)
Notes
- The calculation shows that for liquids, $\Delta G$ is proportional to the pressure change.
- Increasing the pressure from 1 bar to 100 bar changes $G$ by about $9.9\,\text{kJ}$ for one litre of water; the change per mole is only about $0.18\,\text{kJ mol}^{-1}$ because the molar volume is small.
- For a pure ideal gas, the molar Gibbs free energy, or chemical potential, is $ \mu(T,p)=\mu^\circ(T)+RT\ln!\left(\frac{p}{p^\circ}\right). $ Its pressure dependence is logarithmic.
- Pressure enters gas chemical potentials explicitly. For liquids and solids, the pressure contribution is usually much smaller because their molar volumes are small.