Potenciales termodinámicos

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Thermodynamic potentials

Description

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The thermodynamic potentials correspond to different variations of the internal energy $U$, which can include the energy required to form the system equivalent to the work $pV$, as well as the energy that cannot be used to perform work, which is $TS$.

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Internal energy

Concept

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If the absolute temperature ($T$) and the pressure ($p$) are kept constant, the variation of the internal energy ($dU$), which depends on the entropy variation ($dS$) and the volume Variation ($dV$), is expressed as:

$ dU = T dS - p dV $



Integrating this results in the following expression in terms of the internal energy ($U$), the entropy ($S$), and the volume ($V$):

$ U = T S - p V $

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Internal Energy: Differential ratio

Equation

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The dependency of the internal energy differential ($dU$) on the pressure ($p$) and the volume Variation ($dV$), in addition to the absolute temperature ($T$) and the entropy variation ($dS$), is given by:

$ dU = T dS - p dV $

$T$
Absolute temperature
$K$
5177
$dS$
Entropy variation
$J/K$
5225
$p$
Pressure
$Pa$
5224
$dU$
Variation of the internal energy
$J$
5400
$dV$
Volume Variation
$m^3$
5223

As the internal energy differential ($dU$) depends on the differential inexact Heat ($\delta Q$), the pressure ($p$), and the volume Variation ($dV$) according to the equation:

$ dU = \delta Q - p dV $



and the expression for the second law of thermodynamics with the absolute temperature ($T$) and the entropy variation ($dS$) as:

$ \delta Q = T dS $



we can conclude that:

$ dU = T dS - p dV $

.

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Internal Energy

Equation

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The internal energy ($U$) is with the absolute temperature ($T$), the pressure ($p$), the entropy ($S$) and the volume ($V$) equal to:

$ U = T S - p V $

$T$
Absolute temperature
$K$
5177
$S$
Entropy
$J/K$
5227
$U$
Internal energy
$J$
5228
$p$
Pressure
$Pa$
5224
$V$
Volume
$m^3$
5226

If the absolute temperature ($T$) and the pressure ($p$) are kept constant, the variation of the internal energy ($dU$), which depends on the entropy variation ($dS$) and the volume Variation ($dV$), is expressed as:

$ dU = T dS - p dV $



Integrating this results in the following expression in terms of the internal energy ($U$), the entropy ($S$), and the volume ($V$):

$ U = T S - p V $

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Enthalpy

Concept

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The enthalpy ($H$) refers to the energy contained within a system, including any energy required to create it. It is composed of the internal energy ($U$) and the work necessary to form the system, which is represented as $pV$ where the pressure ($p$) and the volume ($V$) are involved.

It is a function of the entropy ($S$) and the pressure ($p$), allowing it to be expressed as $H = H(S,p)$ and it follows the following mathematical relation:

$ H = U + p V $

An article that can be considered as the origin of the concept, although it does not include the definition of the name, is:

[1] "Memoir on the Motive Power of Heat, Especially as Regards Steam, and on the Mechanical Equivalent of Heat," written by Benoît Paul Émile Clapeyron (1834).

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Enthalpy $H(S,p)$

Equation

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If we need to take into account the energy required to form the system in addition to the internal energy, we must consider the enthalpy ($H$).

the enthalpy ($H$) [1] is defined as the sum of the internal energy ($U$) and the formation energy. The latter corresponds to the work done in the formation, which is equal to $pV$ with the pressure ($p$) and the volume ($V$).

Therefore, we have:

$ H = U + p V $

$H$
Enthalpy
$J$
5229
$U$
Internal energy
$J$
5228
$p$
Pressure
$Pa$
5224
$V$
Volume
$m^3$
5226



the enthalpy ($H$) is a function of the entropy ($S$) and the pressure ($p$).

An article that can be considered as the origin of the concept, although it does not include the definition of the name, is:

[1] "Memoir on the Motive Power of Heat, Especially as Regards Steam, and on the Mechanical Equivalent of Heat," written by Benoît Paul Émile Clapeyron (1834).

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Differential Enthalpy Relationship

Equation

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Since the enthalpy ($H$) is a function of the internal energy ($U$), the pressure ($p$), and the volume ($V$) according to the equation:

$ H = U + p V $



and this equation depends solely on the entropy ($S$) and the pressure ($p$), we can show that its partial derivative with respect to the differential enthalpy ($dH$) is equal to:

$ dH = T dS + V dp $

$T$
Absolute temperature
$K$
5177
$dH$
Differential enthalpy
$J$
5171
$dS$
Entropy variation
$J/K$
5225
$dp$
Pressure Variation
$Pa$
5240
$V$
Volume
$m^3$
5226

If we differentiate the enthalpy function:

$ H = U + p V $



we obtain:

$dH = dU + Vdp + pdV$



With the differential of the internal energy:

$ dU = T dS - p dV $



we can conclude:

$ dH = T dS + V dp $



where the entropy variation ($dS$), the pressure Variation ($dp$), and the absolute temperature ($T$) are also considered.

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Enthalpy Ratio

Equation

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The enthalpy,

$ H = U + p V $

,

can be rewritten in terms of the internal energy as

$ H = T S $

$T$
Absolute temperature
$K$
5177
$H$
Enthalpy
$J$
5229
$S$
Entropy
$J/K$
5227

If we replace the internal energy in the expression:

$ H = U + p V $



with the internal energy:

$ U = T S - p V $



we obtain:

$ H = T S $

,

where $T$ is the temperature and $S$ is the entropy.

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Helmholtz free energy

Concept

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The helmholtz free fnergy ($F$) [1] refers to the energy contained within a system, but excludes the energy that cannot be used to perform work. In this sense, it represents the energy available to do work as long as it does not include the energy required to form the system. It is composed, therefore, of the internal energy ($U$), from which the thermal energy $ST$, where the entropy ($S$) and the absolute temperature ($T$) are involved, is subtracted.

This function depends on the absolute temperature ($T$) and the volume ($V$), allowing it to be expressed as $F = F(V,T)$, and it satisfies the following mathematical relation:

$ F = U - T S $

[1] "Über die Thermodynamik chemischer Vorgänge" (On the thermodynamics of chemical processes.), Hermann von Helmholtz, Dritter Beitrag. Offprint from: ibid., 31 May, (1883)

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Helmholtz free energy $F(V,T)$

Equation

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The helmholtz free fnergy ($F$) is defined as the difference between the internal energy ($U$) and the energy that cannot be utilized to perform work. The latter corresponds to $ST$ with the entropy ($S$) and the absolute temperature ($T$).

Therefore, we obtain:

$ F = U - T S $

$T$
Absolute temperature
$K$
5177
$S$
Entropy
$J/K$
5227
$F$
Helmholtz free fnergy
$J$
5230
$U$
Internal energy
$J$
5228

[1] "Über die Thermodynamik chemischer Vorgänge" (On the thermodynamics of chemical processes.), Hermann von Helmholtz, ritter Beitrag. Offprint from: ibid., 31 May, (1883)

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Differential relation Helmholtz Free Energy

Equation

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Since the Helmholtz free energy depends on the temperature $T$ and volume $V$, the differential is obtained as:

$ F = U - T S $



where:

$ dF =- S dT - p dV $

$dF$
Differential Helmholtz Free Energy
$J$
5251
$S$
Entropy
$J/K$
5227
$p$
Pressure
$Pa$
5224
$dT$
Temperature variation
$K$
5217
$dV$
Volume Variation
$m^3$
5223

If we differentiate the definition of Helmholtz free energy:

$ F = U - T S $



we obtain:

$dF = dU - TdS - SdT$



With the differential of internal energy:

$ dU = T dS - p dV $



we can conclude that:

$ dF =- S dT - p dV $

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Helmholtz Free Energy Relation

Equation

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Similar to the helmholtz free fnergy ($F$), with the inclusion of the internal energy ($U$), the absolute temperature ($T$), and the entropy ($S$), it is defined by the equation:

$ F = U - T S $



If we replace the equation with the definition of the internal energy ($U$), we obtain the following when considering the pressure ($p$) and the volume ($V$):

$ F = - p V $

$F$
Helmholtz free fnergy
$J$
5230
$p$
Pressure
$Pa$
5224
$V$
Volume
$m^3$
5226

Expressing the helmholtz free fnergy ($F$) in terms of the internal energy ($U$), the absolute temperature ($T$), and the entropy ($S$), we have the following equation:

$ F = U - T S $



By substituting the internal energy ($U$), which is a function of the pressure ($p$) and the volume ($V$), we obtain:

$ U = T S - p V $



This leads to the following expression:

$ F = - p V $

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Gibbs free energy

Concept

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The gibbs free energy ($G$) refers to the energy contained within a system, including the energy required for its formation, but excludes the energy that cannot be used to do work. In this sense, it represents the energy available to do work in a process that includes the energy required for its formation. It is composed, therefore, of the enthalpy ($H$), from which the thermal energy $ST$, where the entropy ($S$) and the absolute temperature ($T$) are involved, is subtracted.

This function depends on the absolute temperature ($T$) and the pressure ($p$), allowing it to be expressed as $G = G(T,p)$ and it satisfies the following mathematical relation:

$ G = H - T S $

[1] "On the Equilibrium of Heterogeneous Substances," J. Willard Gibbs, Transactions of the Connecticut Academy of Arts and Sciences, 3: 108-248 (October 1875 May 1876)

[2] "On the Equilibrium of Heterogeneous Substances," J. Willard Gibbs, Transactions of the Connecticut Academy of Arts and Sciences, 3: 343-524 (May 1877 July 1878)

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Gibbs and Helmholtz free energy

Equation

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The gibbs free energy ($G$) [1,2] represents the total energy, encompassing both the internal energy and the formation energy of the system. It is defined as the enthalpy ($H$), excluding the portion that cannot be used to perform work, which is represented by $TS$ with the absolute temperature ($T$) and the entropy ($S$). This relationship is expressed as follows:

$ G = H - T S $

$T$
Absolute temperature
$K$
5177
$H$
Enthalpy
$J$
5229
$S$
Entropy
$J/K$
5227
$G$
Gibbs free energy
$J$
5231



Where the absolute temperature ($T$) and the entropy ($S$) play a significant role.

[1] "On the Equilibrium of Heterogeneous Substances," J. Willard Gibbs, Transactions of the Connecticut Academy of Arts and Sciences, 3: 108-248 (October 1875 May 1876)

[2] "On the Equilibrium of Heterogeneous Substances," J. Willard Gibbs, Transactions of the Connecticut Academy of Arts and Sciences, 3: 343-524 (May 1877 July 1878)

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Gibbs free energy as differential

Equation

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As the gibbs free energy ($G$) [1,2] depends on the enthalpy ($H$), the entropy ($S$), and the absolute temperature ($T$):

$ G = H - T S $



The dependence on the gibbs free energy ($G$) with respect to the pressure ($p$) is obtained, and from the absolute temperature ($T$), we obtain the differential:

$ dG =- S dT + V dp $

$S$
Entropy
$J/K$
5227
$dp$
Pressure Variation
$Pa$
5240
$dT$
Temperature variation
$K$
5217
$dG$
Variation of Gibbs Free Energy
$J$
5402
$V$
Volume
$m^3$
5226

The gibbs free energy ($G$) as a function of the enthalpy ($H$), the entropy ($S$), and the absolute temperature ($T$) is expressed as:

$ G = H - T S $



The value of the differential of the Gibbs free energy ($dG$) is determined using the differential enthalpy ($dH$), the temperature variation ($dT$), and the entropy variation ($dS$) through the equation:

$dG=dH-SdT-TdS$



Since the differential enthalpy ($dH$) is related to the volume ($V$) and the pressure Variation ($dp$) as follows:

$ dH = T dS + V dp $



It follows that the differential enthalpy ($dH$), the entropy variation ($dS$), and the pressure Variation ($dp$) are interconnected in the following manner:

$ dG =- S dT + V dp $

[1] "On the Equilibrium of Heterogeneous Substances," J. Willard Gibbs, Transactions of the Connecticut Academy of Arts and Sciences, 3: 108-248 (October 1875 May 1876)

[2] "On the Equilibrium of Heterogeneous Substances," J. Willard Gibbs, Transactions of the Connecticut Academy of Arts and Sciences, 3: 343-524 (May 1877 July 1878)

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Gibbs Free Energy Ratio

Equation

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The gibbs free energy ($G$) with the internal energy ($U$), the entropy ($S$), the absolute temperature ($T$), the pressure ($p$), and the volume ($V$) is expressed as:

$ G = U - S T + p V $



If we replace the internal energy ($U$), the expression reduces to:

$ G = 0$

$G$
Gibbs free energy
$J$
5231

The gibbs free energy ($G$) with the internal energy ($U$), the entropy ($S$), the absolute temperature ($T$), the pressure ($p$), and the volume ($V$) is represented as:

$ G = U - S T + p V $



And with the substitution of the internal energy ($U$),

$ U = T S - p V $



We obtain:

$ G = 0$

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