Third Law of Thermodynamics

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The third law of thermodynamics states that as the temperature approaches absolute zero, the entropy of a system reaches a constant value. Typically, this constant is zero, indicating that there is only one possible microscopic state for the system.

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The third law of thermodynamics states that as a system approaches absolute zero, the entropy of the system approaches a minimum value. At absolute zero, the entropy of a perfectly crystalline substance becomes zero. This law implies that it is impossible to reach absolute zero through a finite number of processes. The third law also provides a reference point for the determination of entropy, ensuring that entropy values are absolute and not relative. It has significant implications for the behavior of materials at very low temperatures, influencing phenomena such as superconductivity and the specific heat of substances. In essence, the third law establishes a baseline for entropy and highlights the unattainability of absolute zero, emphasizing the inherent limitations in cooling processes.

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Mechanisms

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Third law of thermodynamics

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The third law of thermodynamics [1] states that as the absolute temperature ($T$) of a system approaches absolute zero, the entropy in the zero temperature limit ($S$) of a perfect crystalline substance approaches a minimum value, typically zero. This implies that it is impossible to reach absolute zero through a finite number of processes because entropy would become constant. Essentially, the third law establishes that the entropy of a perfect crystal at absolute zero is zero, and it highlights the unattainability of absolute zero due to the infinite amount of energy removal required

$\lim_{T\rightarrow 0}S=0$

This law has significant implications for understanding the behavior of materials at very low temperatures and provides a fundamental reference point for calculating entropy values.

[1] "Über die Berechnung chemischer Gleichgewichte aus thermischen Messungen" (On the Calculation of Chemical Equilibria from Thermal Measurements), Walther Nernst, Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen, Mathematisch-Physikalische Klasse, 1906

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Entropy Calculation

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The relationship of the variation of heat ($\delta Q$) with the absolute temperature ($T$) and the entropy variation ($dS$) can be expressed as:

$ \delta Q = T dS $



When combined with the relationship between the heat supplied to liquid or solid ($\Delta Q$), the caloric Capacity ($C$), and the temperature variation in a liquid or solid ($\Delta T$):

$ \delta Q = C dT $



We obtain this relationship in the infinitesimal limit, where:

$\delta Q = C dT = T dS$



After integration, this leads us to the following equation:

$ S = S_0 + C \log\left(\displaystyle\frac{ T }{ T_0 }\right)$



with the condition that the base entropy ($S_0$) is less than the base temperature ($T_0$).

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Entropy and Phase Change

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If the entropy is estimated as a function of temperature, the following observations can be made:

• In each phase (solid, liquid, gas), entropy tends to slightly increase with temperature.
• During phase transitions, there is a significant jump in entropy.

This can be represented as:

In this way, entropy can be understood as an average measure of the degrees of freedom that a system possesses. In each phase, entropy gradually increases as a few additional degrees of freedom are \\"released\\". However, during phase transitions, the increase in entropy is dramatic. In a solid, multiple bonds restrict the movement of atoms, resulting in limited degrees of freedom. In a liquid, many bonds are broken, creating new freedoms that allow for relative movement, leading to numerous new degrees of freedom. Finally, in the transition to the gas phase, all bonds are lost, and each particle has its three degrees of freedom. As the temperature increases further, particles can rotate and oscillate, introducing new degrees of freedom and additional increases in entropy.

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Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$S_0$
S_0
Base entropy
J
$T_0$
T_0
Base temperature
K
$C$
C
Heat capacity
$T$
T
Temperature
K

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$T$
T
Absolute temperature
K
$\delta Q$
dQ
Differential inexact Heat
J
$S$
S
Entropy
J/K
$S$
S
Entropy in the zero temperature limit
J/K
$dS$
dS
Entropy variation
J/K
$dT$
dT
Temperature variation
K

Calculations


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

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$ \delta Q = C dT $

DQ = C * DT


$ \delta Q = T dS $

dQ = T * dS


$ S = S_0 + C \log\left(\displaystyle\frac{ T }{ T_0 }\right)$

S = S_0 + C *log( T / T_0 )


$\lim_{T\rightarrow 0}S=0$

S_T>0=0

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Third law of thermodynamics

Equation

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The third law of thermodynamics states that if a system is brought to absolute zero temperature, its entropy will approach zero. This can be expressed as follows:

$\lim_{T\rightarrow 0}S=0$

$S$
Entropy in the zero temperature limit
$J/K$
10374

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Second law of thermodynamics

Equation

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The differential inexact Heat ($\delta Q$) is equal to the absolute temperature ($T$) times the entropy variation ($dS$):

$ \delta Q = T dS $

$T$
Absolute temperature
$K$
5177
$\delta Q$
Differential inexact Heat
$J$
5220
$dS$
Entropy variation
$J/K$
5225

ID:(9639, 0)



Caloric content

Equation

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When the heat supplied to liquid or solid ($\Delta Q$) is added to a body, we observe a proportional increase of the temperature variation in a liquid or solid ($\Delta T$). Therefore, we can introduce a proportionality constant the heat capacity ($C$), known as heat capacity, which establishes the following relationship:

$ \delta Q = C dT $

$ \Delta Q = C \Delta T $

$C$
Heat capacity
$J/K$
8482
$\Delta Q$
$\delta Q$
Differential inexact Heat
$J$
5220
$\Delta T$
$dT$
Temperature variation
$K$
5217

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Entropy Calculation

Equation

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The entropy ($S$) is a function of the absolute temperature ($T$) with the base entropy ($S_0$) and the base temperature ($T_0$) according to:

$ S = S_0 + C \log\left(\displaystyle\frac{ T }{ T_0 }\right)$

$S_0$
Base entropy
$J$
8503
$T_0$
Base temperature
$K$
8504
$S$
Entropy
$J/K$
5227
$C$
Heat capacity
$J/K$
8482
$T$
Temperature
$K$
5393

The relationship of the variation of heat ($\delta Q$) with the absolute temperature ($T$) and the entropy variation ($dS$) can be expressed as:

$ \delta Q = T dS $



When combined with the relationship between the heat supplied to liquid or solid ($\Delta Q$), the caloric Capacity ($C$), and the temperature variation in a liquid or solid ($\Delta T$):

$ \delta Q = C dT $



We obtain this relationship in the infinitesimal limit, where:

$\delta Q = C dT = T dS$



After integration, this leads us to the following equation:

$ S = S_0 + C \log\left(\displaystyle\frac{ T }{ T_0 }\right)$



with the condition that the base entropy ($S_0$) is less than the base temperature ($T_0$).

ID:(11185, 0)