Statistical Molecular Thermodynamics

Christopher J. Cramer

Video 8.4

Maxwell Relations from G

Working With Gibbs Free Energy

PVTSUG +−=

using dU = TdS – PdV

compare with the formal derivative of G = G(P,T): dP

PGdT

TGdG

TP⎟⎠

⎞⎜⎝

⎛∂

∂+⎟

⎠

⎞⎜⎝

⎛∂

∂=

VdPPdVSdTTdSdUdG ++−−=differentiate

VdPSdTdG +−=

STG

P

−=⎟⎠

⎞⎜⎝

⎛∂

∂ VPG

T

=⎟⎠

⎞⎜⎝

⎛∂

∂and Thus

Equating Key Cross Derivatives

James Clerk Maxwell

€

∂G∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

=V

€

∂G∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

= −Sand

€

∂G∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

=V

€

∂∂T

∂G∂P⎛

⎝ ⎜

⎞

⎠ ⎟ =

∂V∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

€

∂G∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

= −S

€

∂∂P

∂G∂T⎛

⎝ ⎜

⎞

⎠ ⎟ = −

∂S∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

Another of many Maxwell relations

€

∂∂T

∂G∂P⎛

⎝ ⎜

⎞

⎠ ⎟ =

∂∂P

∂G∂T⎛

⎝ ⎜

⎞

⎠ ⎟ As equality of mixed

partial derivatives

€

∂V∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

= −∂S∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

Utility of The Maxwell Relation

€

∂V∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

= −∂S∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

From this Maxwell relation we can see how S changes with P given an equation of state

Integrate at constant T:

€

ΔS = −∂V∂T⎛

⎝ ⎜

⎞

⎠ ⎟

P1

P2∫P

dPNote that T is held

constant during integration over P

Get P dependence of S from P-V-T data.

Example: Ideal gas

€

∂V∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

=nRP

€

ΔS = −nR dPPP1

P2∫ = −nR ln P2P1

(isothermal)

Entropy of Ethane Again

€

ΔS = S T,P2( ) − S(P→ 0)id = −

∂V∂T⎛

⎝ ⎜

⎞

⎠ ⎟

P id

P2∫P

dP (constant T)

If P1 is chosen to be so small that a gas behaves ideally (P 0),

Ethane at 400 K

€

S (P→ 0)id 246.45 J • mol−1 • K−1 at 1 bar from Q!( )[ ]

€

S T,P( ) = S (P→ 0)id −

∂V∂T⎛

⎝ ⎜

⎞

⎠ ⎟

P id

P2∫P

dP

For real gases, i.e., those having no readily available, analytical equation of state, this requires data for how volume varies with temperature over a full range of pressures

Enthalpy Dependence on P Differentiating G = H – TS wrt P :

€

∂G∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

=∂H∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

−T ∂S∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

(isothermal)

using

€

∂V∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

= −∂S∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

Maxwell relation

and

€

∂G∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

=V

previously derived

€

∂H∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

=V −T ∂V∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

Ethane at 400 K

€

H (P→ 0)id 17.87 kJ • mol−1 from Q!( )[ ]

€

H T,P( ) = H (P→ 0)id + V −T ∂V

∂T⎛

⎝ ⎜

⎞

⎠ ⎟

P

⎡

⎣ ⎢

⎤

⎦ ⎥ P id

P2∫ dPFor real gases, i.e., those having no readily available, analytical equation of state, this again requires data for how volume varies with temperature over a full range of pressures

Pressure Dependence of G

Ideal gas example, V = nRT/P :

€

∂G∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

=V

€

ΔG = VdPP1

P2∫integrate (constant T)

€

ΔG = nRT 1PdP

P1

P2∫ = nRT ln P2P1

(constant T)

Compare this to a previous result for an ideal gas at constant T:

€

ΔS = nR lnV2V1

= −nR ln P2P1

As expected, ΔG = ΔH –TΔS is equal simply to –TΔS since ΔH = 0 at constant T for an ideal gas