Gibbs Free Energy — Pressure and Incompressible Water

By Prof. Marco Ruzzi.

Theoretical background

The fundamental equation for Gibbs energy is:

\[dG=V\,dp-S\,dT+\sum_i\mu_i\,dn_i.\]

For a closed system with constant composition undergoing an isothermal change, dT=0 and dn_i=0. Therefore dG=Vdp.

Exercise 10 — Increasing the pressure on liquid water

The pressure on a 2 kg sample of water increases from 1.00 to 3.00 atm at constant temperature. Assume a constant liquid volume (incompressible water), with density 1 kg L⁻¹. Use 1 L atm=101.325 J.

Hint: use the fundamental thermodynamic equation.

Calculate the Gibbs-energy change.

1. Reduce the fundamental equation

The system is closed and its composition is constant. For water alone:

\[dG=V\,dp-S\,dT+\mu_{\mathrm{H_2O}}\,dn_{\mathrm{H_2O}}.\]

Since dT=0 and dn_H₂O=0:

\[dG=V\,dp.\]

2. Integrate at constant volume

\[\Delta G=V(p_f-p_i)=\frac{m}{\rho}(p_f-p_i).\] \[V=\frac{2\ \mathrm{kg}}{1\ \mathrm{kg\,L^{-1}}}=2\ \mathrm{L}.\] \[\Delta G=(2\ \mathrm{L})(2\ \mathrm{atm})=4\ \mathrm{L\,atm}.\] \[\Delta G=4(101.325)\ \mathrm{J}=+405.3\ \mathrm{J}.\]