# Equilibrium criteria for pure substances

(Difference between revisions)
 Revision as of 00:01, 29 January 2010 (view source)← Older edit Revision as of 06:51, 3 February 2010 (view source)Newer edit → Line 37: Line 37: (71) (71) - Assuming that the only work present in this closed system is of type $pV$, the work exchange between the system and the surroundings must be zero, i.e., $W = 0$. The fundamental thermodynamic relationship, eq. (54), simplifies as + Assuming that the only work present in this closed system is of type $pV$, the work exchange between the system and the surroundings must be zero, i.e., $W = 0$. The fundamental thermodynamic relationship, eq. $\Delta E \le T\Delta S - W$ from [[Thermodynamic property relations]], simplifies as
$\Delta E - T\Delta S \le 0\qquad \qquad(9)$
$\Delta E - T\Delta S \le 0\qquad \qquad(9)$
Line 73: Line 73: Once again, the goal is to determine the equilibrium criteria for such a system, with the assumption that the only work is of the $pV$ type. The work exchange between the system and its surroundings is $W = p\Delta V.$ Once again, the goal is to determine the equilibrium criteria for such a system, with the assumption that the only work is of the $pV$ type. The work exchange between the system and its surroundings is $W = p\Delta V.$ - The fundamental thermodynamic relationship, eq. (54), can be written as + The fundamental thermodynamic relationship, eq. $\Delta E \le T\Delta S - W$ from [[Thermodynamic property relations]], can be written as
$\Delta E - T\Delta S + p\Delta V \le 0\qquad \qquad(16)$
$\Delta E - T\Delta S + p\Delta V \le 0\qquad \qquad(16)$

## Constant-Volume Isolated System

For a constant-volume isolated system that exchanges neither heat nor work with its surroundings,

$Q = 0\qquad \qquad(1)$

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$\Delta V = 0\qquad \qquad(2)$

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$W = 0\qquad \qquad(3)$

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It follows from the first law of thermodynamics that a system that has no heat or work interaction with the surroundings also has no change in internal energy. Thus

$\Delta E = 0\qquad \qquad(4)$

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Applying eqs. (1) – (4) with the fundamental thermodynamic relation, eq. (54), gives

$\Delta {S_{E,V}} \ge 0\qquad \qquad(5)$

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Equation (5) asserts that system entropy always increases for a spontaneous and irreversible finite process occurring in a system with constant internal energy, E, and constant volume V. These spontaneous processes continuously move the system toward an equilibrium state where the entropy will reach a maximum value. When the system reaches an equilibrium state, any infinitesimal change in the system will result in a zero change of entropy, i.e.,

$d{S_{E,V}} = 0\qquad \qquad(6)$

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## Constant-Temperature and Volume System

Since the temperature and volume of the closed system are constants, we have

$\Delta T = 0\qquad \qquad(7)$

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$\Delta V = 0\qquad \qquad(8)$

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Assuming that the only work present in this closed system is of type pV, the work exchange between the system and the surroundings must be zero, i.e., W = 0. The fundamental thermodynamic relationship, eq. $\Delta E \le T\Delta S - W$ from Thermodynamic property relations, simplifies as

$\Delta E - T\Delta S \le 0\qquad \qquad(9)$

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Recalling the well known Helmholtz free energy function, F,

$F = E - TS\qquad \qquad(10)$

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and expanding eq. (10) to define a finite change in the system yields

$\Delta F = \Delta E - T\Delta S - S\Delta T\qquad \qquad(11)$

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Substituting eqs. (11) and (7) into eq. (9), a second equilibrium criterion is obtained:

$\Delta {F_{T,V}} \le 0\qquad \qquad(12)$

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Therefore, for a closed system at constant temperature and volume, the Helmholtz free energy must decrease with any spontaneous system change and be minimal at equilibrium. At equilibrium conditions any infinitesimal change from constant-temperature, constant-volume equilibrium must result in zero change in the Helmholtz free energy.

$d{F_{T,V}} = 0\qquad \qquad(13)$

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## Constant Temperature and Pressure System

For a closed system with constant temperature and constant pressure,

$\Delta T = 0\qquad \qquad(14)$

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$\Delta p = 0\qquad \qquad(15)$

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Once again, the goal is to determine the equilibrium criteria for such a system, with the assumption that the only work is of the pV type. The work exchange between the system and its surroundings is W = pΔV. The fundamental thermodynamic relationship, eq. $\Delta E \le T\Delta S - W$ from Thermodynamic property relations, can be written as

$\Delta E - T\Delta S + p\Delta V \le 0\qquad \qquad(16)$

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To determine a useful equilibrium criterion for such a system, another common thermodynamic property, the Gibbs free energy, is recalled:

$G = E - TS + pV\qquad \qquad(17)$

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Expanding eq. (17) for a system undergoing a finite system change results in

$\Delta G = \Delta E - T\Delta S - S\Delta T + p\Delta V + V\Delta p\qquad \qquad(18)$

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Substituting eqs. (18) and (14) – (15) into eq. (16) results in the well known criterion of equilibrium at constant temperature and pressure:

$\Delta {G_{T,p}} \le 0\qquad \qquad(19)$

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Thus, for a closed system at constant temperature and pressure, the Gibbs free energy of the system must decrease with any spontaneous finite system change and will be at its minimum value at equilibrium. Finally, if a system of constant temperature and pressure is at equilibrium, any infinitesimal system change will result in zero change in the system’s Gibbs free energy.

$d{G_{T,p}} \le 0\qquad \qquad(20)$

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## Summary of the Equilibrium Criteria

Other sets of inequality constraints exist for system equilibrium and can be found in a manner similar to the one detailed above, but have limited applications. In summary, the equilibrium constraints for a system undergoing a finite change as determined above are as follows:

$\Delta {S_{E,V}} \ge 0\qquad \qquad(21)$

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$\Delta {F_{T,V}} \le 0\qquad \qquad(22)$

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$\Delta {G_{T,p}} \le 0\qquad \qquad(23)$

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$\Delta {E_{S,V}} \le 0\qquad \qquad(24)$

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$\Delta {S_{H,p}} \ge 0\qquad \qquad(25)$

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$\Delta {H_{S,p}} \le 0\qquad \qquad(26)$

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where H = E + pV is the enthalpy of the system. During an infinitesimal change in the system, this change can be assumed to be reversible (dS = 0). Therefore, the criteria of equilibrium for a system undergoing an infinitesimal change from equilibrium conditions become

$d{S_{E,V}} = 0\qquad \qquad(27)$

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$d{F_{T,V}} = 0\qquad \qquad(28)$

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$d{G_{T,p}} = 0\qquad \qquad(29)$

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$d{U_{S,V}} = 0\qquad \qquad(30)$

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$d{S_{H,p}} = 0\qquad \qquad(31)$

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$d{H_{S,p}} = 0\qquad \qquad(32)$

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$\Delta E \le T\Delta S\qquad \qquad(33)$

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Since entropy is constant, i.e., ΔS = 0, it follows that

$\Delta {E_{S,V}} \le 0\qquad \qquad(34)$

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