Question

Derive Gibb's Duhem equation.​

Answer 1

Answer:

Gibbs–Duhem equation

In thermodynamics, the Gibbs–Duhem equation describes the relationship between changes in chemical potential for components in a thermodynamic system: {\displaystyle \sum _{i=1}^{I}N_{i}\mathrm {d} ...

Answer 2

$\sf{{\displaystyle \sum _{i=1}^{I}N_{i}\mathrm {dμ_{i} = −S dT + V dP}}}$

Deriving the Gibbs–Duhem equation from the fundamental thermodynamic equation is straightforward. The total differential of the extensive Gibbs free energy

${\displaystyle \mathbf {G} }\mathbf {G} \: \sf \tiny{ in \: terms \: of \: its \: natural \: variables \: is}$

$\tiny</p><p>{\displaystyle \mathrm {d} \mathbf {G} =\left.{\frac {\partial \mathbf {G} }{\partial p}}\right|_{T,N} }\\ \tiny{\mathrm {d} p+\left.{\frac {\partial \mathbf {G} }{\partial T}}\right|_{p,N}\mathrm {d} T+\sum _{i=1}^{I}\left.{\frac {\partial \mathbf {G} }{\partial N_{i}}}\right|_{p,T,N_{j\neq i}}\mathrm {d} N_{i}.}$

Since the Gibbs free energy is the Legendre transformation of the internal energy, the derivatives can be replaced by its definitions transforming the above equation into:

$\tiny</p><p>{\displaystyle \mathrm {d} \mathbf {G} =V\mathrm {d} p-S\mathrm {d} T+\sum _{i=1}^{I}\mu _{i}\mathrm {d} N_{i}}$

The chemical potential is simply another name for the partial molar Gibbs free energy (or the partial Gibbs free energy, depending on whether N is in units of moles or particles). Thus the Gibbs free energy of a system can be calculated by collecting moles together carefully at a specified T, P and at a constant molar ratio composition (so that the chemical potential doesn't change as the moles are added together), i.e.

$</p><p>{{\displaystyle \mathbf {G} =\sum _{i=1}^{I}\mu _{i}N_{i}}} \\ </p><p></p><p> </p><p>$

The total differential of this expression is

$\small{ \mathrm {d} \mathbf {G} =\sum _{i=1}^{I}\mu _{i}\mathrm {d} N_{i}+\sum _{i=1}^{I}N_{i}\mathrm {d} \mu _{i}}$

Combining the two expressions for the total differential of the Gibbs free energy gives

$\tiny</p><p>{\displaystyle \sum _{i=1}^{I}\mu _{i}\mathrm {d} N_{i}+\sum _{i=1}^{I}N_{i}\mathrm {d} \mu _{i}=V\mathrm {d} p-S\mathrm {d} T+\sum _{i=1}^{I}\mu _{i}\mathrm {d} N_{i}}$

which simplifies to the Gibbs–Duhem relation:

$\tiny{\displaystyle \sum _{i=1}^{I}N_{i}\mathrm {d} \mu _{i}=-S\mathrm {d} T+V\mathrm {d} p}{\displaystyle \sum _{i=1}^{I}N_{i}\mathrm {d} \mu _{i}=-S\mathrm {d} T+V\mathrm {d} p}$

$\small\color{red}\star \: \sf{hope \: it \: helps}$