Free Energy and Chemical Thermodynamics

Gibbs Free Energy and Chemical Potential

PHYS 4050: Thermodynamics and Statistical PhysicsProf. B. A. Foreman

Extensive quantities

• Previously we defined an extensive quantity as one that grows in proportion to the size of the system

𝑆 ( 𝜆𝑈 ,𝜆𝑉 ,𝜆𝑁 )=𝜆𝑆 (𝑈 ,𝑉 ,𝑁 )• In mathematical language, S is a

homogeneous function of degree one in U, V, and N

• For example, entropy is usually an extensive quantity:

Intensive quantities

• We defined an intensive quantity as one that is independent of the size of the system

1𝑇≡( 𝜕𝑆𝜕𝑈 )

𝑉 ,𝑁

,𝑃≡𝑇 ( 𝜕𝑆𝜕𝑉 )𝑈 ,𝑁

,𝜇≡−𝑇 ( 𝜕𝑆𝜕𝑁 )𝑈 ,𝑉

• In other words, they are homogeneous functions of degree zero in U, V, and N:

𝑇 ( 𝜆𝑈 ,𝜆𝑉 ,𝜆𝑁 )=𝑇 (𝑈 ,𝑉 ,𝑁 )

• Temperature, pressure, and chemical potential are all intensive:

Thermodynamic potentials

• The thermodynamic potentials are extensive too:

𝐻≡𝑈+𝑃𝑉 ,𝐹≡𝑈−𝑇𝑆 ,𝐺≡𝑈−𝑇𝑆+𝑃𝑉

• In terms of their natural independent variables, this means that

𝑈 (𝜆𝑆 ,𝜆𝑉 ,𝜆𝑁 )=𝜆𝑈 (𝑆 ,𝑉 ,𝑁 )𝐻 (𝜆𝑆 ,𝑃 ,𝜆𝑁 )=𝜆𝐻 (𝑆 ,𝑃 ,𝑁 )𝐹 (𝑇 ,𝜆𝑉 ,𝜆𝑁 )=𝜆𝐹 (𝑇 ,𝑉 ,𝑁 )𝐺 (𝑇 ,𝑃 ,𝜆𝑁 )=𝜆𝐺 (𝑇 ,𝑃 ,𝑁 )

Gibbs free energy

• Let’s examine the consequences of

𝐺+𝛼𝑁𝜕𝐺𝜕𝑁

=𝐺+𝛼𝐺⇒𝑁𝜕𝐺𝜕𝑁

=𝐺

• Setting and expanding in a Taylor series for small , we get

𝑑𝐺=−𝑆𝑑𝑇+𝑉 𝑑𝑃+𝜇𝑑𝑁⇒𝜇=( 𝜕𝐺𝜕𝑁 )𝑇 ,𝑃

• But this derivative is just the chemical potential:

Gibbs energy and chemical potential

• The fact that G is extensive therefore implies

𝐺=𝑁𝜇

• In other words, the chemical potential is the Gibbs free energy per particle:

𝜇=𝐺𝑁

• If we extend this argument to systems with more than one type of particle, we get

𝐺=𝑁 1𝜇1+𝑁2𝜇2+⋯=∑𝑖

𝑁 𝑖𝜇𝑖

Gibbs-Duhem relation

• Since , small changes in must satisfy

𝑑𝐺=𝑁𝑑𝜇+𝜇𝑑𝑁=−𝑆𝑑𝑇 +𝑉 𝑑𝑃+𝜇𝑑𝑁• After cancelling from both sides, we obtain the Gibbs-Duhem

relation:−𝑆𝑑𝑇 +𝑉 𝑑𝑃−𝑁 𝑑𝜇=0

• This tells us that the three intensive variables , , and cannot all be varied independently, because their differentials are related

Isothermal changes in m

• As an application of the Gibbs-Duhem relation, consider the special case of an isothermal process ():

𝑉 𝑑𝑃−𝑁 𝑑𝜇=0⇒( 𝜕𝜇𝜕𝑃 )𝑇

=𝑉𝑁

• In an ideal gas, this becomes

( 𝜕𝜇𝜕𝑃 )𝑇

=𝑘𝑇𝑃

Chemical potential

• Integrating from to , we get

𝜇 (𝑇 ,𝑃 )=𝜇 (𝑇 ,𝑃∘ )+𝑘𝑇 ln (𝑃 /𝑃∘ )

• This allows us to determine at a general pressure from the tabulated value at some reference pressure (usually 1 bar)

• The value of can be obtained from tables of , since