Helmholtz free energy

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The Helmholtz free energy represents the portion of a system's internal energy that is available to perform work.

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Mechanisms

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The Helmholtz free energy represents the maximum amount of work that a thermodynamic system can perform at constant temperature and volume. It encompasses both the internal energy of the system and the entropy. By balancing the internal energy, which includes kinetic and potential energies at the microscopic level, and the entropy term, which accounts for energy dispersal due to temperature, Helmholtz free energy essentially measures the portion of the internal energy that can be converted into work, considering entropy-related energy losses.

Helmholtz free energy is crucial for analyzing systems at constant temperature and volume. It determines the spontaneity of processes, with a decrease in Helmholtz free energy indicating a spontaneous process. At equilibrium, the Helmholtz free energy of a system is minimized, thereby identifying the equilibrium state. It represents the maximum work that can be extracted from a system at constant temperature and volume, excluding pressure-volume work. In chemical reactions occurring at constant volume, Helmholtz free energy predicts the direction of reactions and the conditions under which they will proceed.

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Concept

Mechanisms

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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 differential

Concept

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The helmholtz free fnergy ($F$) explains how this behaves under variations in the absolute temperature ($T$) and the volume ($V$), expressed as:

$ dF =- S dT - p dV $



In response to changes in the absolute temperature ($T$), a positive slope equal to the entropy ($S$) is observed.

In response to changes in the volume ($V$), a negative slope equal to the pressure ($p$) is produced.

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Helmholtz Free Energy and Equation of State at Constant Temperature

Concept

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The differential Helmholtz Free Energy ($dF$) is a function of the variations of the absolute temperature ($T$) and the volume ($V$), as well as the slopes the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) and the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$), which is expressed as:

$ dF = DF_{T,V} dT + DF_{V,T} dV $



Comparing this with the equation for the differential Helmholtz Free Energy ($dF$):

$ dF =- S dT - p dV $



and with the first law of thermodynamics, it follows that the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$) is equal to negative the pressure ($p$):

$ DF_{V,T} =- p $

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Helmholtz Free Energy and Equation of State at Constant Volume

Concept

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The differential Helmholtz Free Energy ($dF$) is a function of the variations of the absolute temperature ($T$) and the volume ($V$), as well as the slopes the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) and the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$), expressed as:

$ dF = DF_{T,V} dT + DF_{V,T} dV $



Comparing this with the equation for the differential Helmholtz Free Energy ($dF$):

$ dF =- S dT - p dV $



and with the first law of thermodynamics, it follows that the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) is equal to negative the entropy ($S$):

$ DF_{T,V} =- S $

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Helmholtz Free Energy and its Relation of Maxwell

Concept

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Since the differential Helmholtz Free Energy ($dF$) is an exact differential, we should note that the helmholtz free fnergy ($F$) with respect to the absolute temperature ($T$) and the volume ($V$) must be independent of the order in which the function is derived:

$D(DF_{T,V})_{V,T}=D(DF{V,T})_{T,V}$



Using the relationship between the slope the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) and the entropy ($S$)

$ DF_{T,V} =- S $



and the relationship between the slope the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$) and the pressure ($p$)

$ DF_{V,T} =- p $



we can conclude that:

$ DS_{V,T} = Dp_{T,V} $

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Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$DS_{V,T}$
DS_VT
Partial derivative of entropy with respect to volume at constant temperature
$Dp_{T,V}$
Dp_TV
Partial derivative of pressure with respect to temperature at constant volume
$DF_{T,V}$
DF_TV
Partial derivative of the Helmholtz free energy with respect to temperature at constant volume
$DF_{V,T}$
DF_VT
Partial derivative of the Helmholtz free energy with respect to volume at constant temperature

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$T$
T
Absolute temperature
K
$dF$
dF
Differential Helmholtz Free Energy
J
$S$
S
Entropy
J/K
$F$
F
Helmholtz free fnergy
J
$U$
U
Internal energy
J
$p$
p
Pressure
Pa
$dT$
dT
Temperature variation
K
$V$
V
Volume
m^3
$dV$
dV
Volume Variation
m^3

Calculations


First, select the equation: to , then, select the variable: to

Calculations

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Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

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Equation

$ dF = DF_{T,V} dT + DF_{V,T} dV $

dF = DF_TV * dT + DF_VT * dV


$ dF =- S dT - p dV $

dF =- S * dT - p * dV


$ DF_{T,V} =- S $

DF_TV =- S


$ DF_{V,T} =- p $

DF_VT =- p


$ DS_{V,T} = Dp_{T,V} $

DS_VT = Dp_TV


$ F = - p V $

F = - p * V


$ F = U - T S $

F = U - T * S


$ U = T S - p V $

U = T * S - p * V

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Helmholtz free Energy

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

ID:(14047, 0)



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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Helmholtz Free Energy Relation

Equation

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The helmholtz free fnergy ($F$) is reduced with the pressure ($p$) and the volume ($V$) to:

$ 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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Differential relation Helmholtz Free Energy

Equation

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The dependency of the differential Helmholtz Free Energy ($dF$) on the entropy ($S$) and the temperature variation ($dT$), in addition to the pressure ($p$) and the volume Variation ($dV$), is given by:

$ 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

The helmholtz free fnergy ($F$) is defined using the internal energy ($U$), the absolute temperature ($T$), and the entropy ($S$) as:

$ F = U - T S $



When we differentiate this equation, we obtain with the differential Helmholtz Free Energy ($dF$), the variation of the internal energy ($dU$), the entropy variation ($dS$), and the temperature variation ($dT$):

$dF = dU - TdS - SdT$



With the differential of internal energy and the variables the pressure ($p$) and the volume Variation ($dV$),

$ dU = T dS - p dV $



we finally obtain:

$ dF =- S dT - p dV $

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Helmholtz Free Energy and Equation of State at Constant Volume

Equation

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Comparing this with the first law of thermodynamics, it turns out that the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) is equal to minus the entropy ($S$):

$ DF_{T,V} =- S $

$S$
Entropy
$J/K$
5227
$DF_{T,V}$
Partial derivative of the Helmholtz free energy with respect to temperature at constant volume
$J/K$
9321

The differential Helmholtz Free Energy ($dF$) is a function of the variations of the absolute temperature ($T$) and the volume ($V$), as well as the slopes the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) and the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$), expressed as:

$ dF = DF_{T,V} dT + DF_{V,T} dV $



Comparing this with the equation for the differential Helmholtz Free Energy ($dF$):

$ dF =- S dT - p dV $



and with the first law of thermodynamics, it follows that the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) is equal to negative the entropy ($S$):

$ DF_{T,V} =- S $

ID:(3550, 0)



Helmholtz Free Energy and Equation of State at Constant Temperature

Equation

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Comparing this with the first law of thermodynamics, it turns out that the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$) is equal to minus the pressure ($p$):

$ DF_{V,T} =- p $

$DF_{V,T}$
Partial derivative of the Helmholtz free energy with respect to volume at constant temperature
$J/m^3$
9320
$p$
Pressure
$Pa$
5224

The differential Helmholtz Free Energy ($dF$) is a function of the variations of the absolute temperature ($T$) and the volume ($V$), as well as the slopes the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) and the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$), which is expressed as:

$ dF = DF_{T,V} dT + DF_{V,T} dV $



Comparing this with the equation for the differential Helmholtz Free Energy ($dF$):

$ dF =- S dT - p dV $



and with the first law of thermodynamics, it follows that the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$) is equal to negative the pressure ($p$):

$ DF_{V,T} =- p $

ID:(3551, 0)



Differential of Helmholtz Free Energy

Equation

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The differential Helmholtz Free Energy ($dF$) is a function of the variations of the absolute temperature ($T$) and the volume ($V$), as well as the slopes the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) and the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$), which is expressed as:

$ dF = DF_{T,V} dT + DF_{V,T} dV $

$dF$
Differential Helmholtz Free Energy
$J$
5251
$DF_{T,V}$
Partial derivative of the Helmholtz free energy with respect to temperature at constant volume
$J/K$
9321
$DF_{V,T}$
Partial derivative of the Helmholtz free energy with respect to volume at constant temperature
$J/m^3$
9320
$dT$
Temperature variation
$K$
5217
$dV$
Volume Variation
$m^3$
5223

Given that the helmholtz free fnergy ($F$) depends on the absolute temperature ($T$) and the volume ($V$), the differential Helmholtz Free Energy ($dF$) can be calculated using:

$dF = \left(\displaystyle\frac{\partial F}{\partial T}\right)_V dT + \left(\displaystyle\frac{\partial F}{\partial V}\right)_T dV$



To simplify this expression, we introduce the notation for the derivative of the helmholtz free fnergy ($F$) with respect to the absolute temperature ($T$) while keeping the volume ($V$) constant as:

$DF_{T,V} \equiv \left(\displaystyle\frac{\partial F}{\partial T}\right)_V$



and for the derivative of the helmholtz free fnergy ($F$) with respect to the volume ($V$) while keeping the absolute temperature ($T$) constant as:

$DF_{V,T} \equiv \left(\displaystyle\frac{\partial F}{\partial V}\right)_T$



thus we can write:

$ dF = DF_{T,V} dT + DF_{V,T} dV $

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Helmholtz Free Energy and its Relation of Maxwell

Equation

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With the entropy ($S$), the volume ($V$), the absolute temperature ($T$) and the pressure ($p$) we obtain one of the so-called Maxwell relations:

$ DS_{V,T} = Dp_{T,V} $

$DS_{V,T}$
Partial derivative of entropy with respect to volume at constant temperature
$J/m^3$
9324
$Dp_{T,V}$
Partial derivative of pressure with respect to temperature at constant volume
$Pa/K$
9325

Since the differential Helmholtz Free Energy ($dF$) is an exact differential, we should note that the helmholtz free fnergy ($F$) with respect to the absolute temperature ($T$) and the volume ($V$) must be independent of the order in which the function is derived:

$D(DF_{T,V})_{V,T}=D(DF{V,T})_{T,V}$



Using the relationship between the slope the partial derivative of the Helmholtz free energy with respect to temperature at constant volume ($DF_{T,V}$) and the entropy ($S$)

$ DF_{T,V} =- S $



and the relationship between the slope the partial derivative of the Helmholtz free energy with respect to volume at constant temperature ($DF_{V,T}$) and the pressure ($p$)

$ DF_{V,T} =- p $



we can conclude that:

$ DS_{V,T} = Dp_{T,V} $

ID:(3554, 0)