Adiabatic processes

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In processes that occur rapidly, there is not enough time for the internal energy to vary significantly. In this case, any work done reduces the heat of the system, leading to modifications in the ideal gas equations.

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Mechanisms

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An adiabatic process is a thermodynamic process in which no heat is exchanged between the system and its surroundings. This means that all changes in the system's internal energy result solely from work done on or by the system. In an adiabatic expansion, the system does work on its surroundings, causing its temperature to drop. Conversely, in an adiabatic compression, work is done on the system, increasing its temperature. These processes are often idealized and occur in well-insulated systems where heat transfer is negligible.

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Concept

Mechanisms

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Adiabatic process

Concept

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When a gas expands rapidly, the water vapor molecules do not have enough time to exchange energy with the surroundings, so no heat is transferred, that is, the variation of heat ($\delta Q$) remains constant:

$\delta Q = 0$



The processes that are carried out under this condition are called adiabatic processes [1,2].

The expansion of the gas requires the system to do work or generate the differential inexact labour ($\delta W$). However, the energy needed for this cannot come from the internal energy ($U$), so it must be obtained from heat. As a result, the temperature of the system decreases, leading to a decrease in the variation of heat ($\delta Q$).

A typical example of this process is the formation of clouds. When air rises through convection, it expands, performs work, and cools down. The moisture in the air condenses, forming clouds.

Conversely, when work is done on the system, positive work the differential inexact labour ($\delta W$) is done. However, since the internal energy ($U$) cannot increase, the thermal energy in the variation of heat ($\delta Q$) increases, leading to an increase in the system's temperature.

A common example of this process is using a pump. If we try to inflate something rapidly, we do work on the system adiabatically, leading to an increase in ($$)5202

[1] "Réflexions sur la puissance motrice du feu" (Reflections on the Motive Power of Fire), Sadi Carnot, 1824

[2] "Über die bewegende Kraft der Wärme und die Gesetze, welche sich daraus für die Wärmelehre selbst ableiten lassen" (On the Moving Force of Heat and the Laws Which Can Be Deduced from It for the Theory of Heat Itself), Rudolf Clausius, Annalen der Physik und Chemie, 1850

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First law of Thermodynamics and Pressure

Concept

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Since the internal energy differential ($dU$) relates to the differential inexact Heat ($\delta Q$) and the differential inexact labour ($\delta W$) as shown below:

$ dU = \delta Q - \delta W $



And it is known that the differential inexact labour ($\delta W$) is related to the pressure ($p$) and the volume Variation ($dV$) as follows:

$ \delta W = p dV $



Therefore, we can conclude that:

$ dU = \delta Q - p dV $

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Caloric content of a gas at constant volume as a function of specific heat

Concept

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The variation of the internal energy ($dU$) in relation to the temperature variation in a liquid or solid ($\Delta T$) and the heat capacity at constant volume ($C_V$) is expressed as:

$ dU = C_V \Delta T $



Where the heat capacity at constant volume ($C_V$) can be replaced by the specific heat of gases at constant volume ($c_V$) and the mass ($M$) using the following relationship:

$ c_V =\displaystyle\frac{ C_V }{ M }$



Therefore, we obtain:

$ dU = c_V m \Delta T $

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Temperature and Volume Variation

Concept

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Since with the variation of the internal energy ($dU$), the variation of heat ($\delta Q$), and the differential inexact labour ($\delta W$) we have:

$dU = \delta Q - \delta W = 0 - \delta W = - \delta W$



the variation of the internal energy ($dU$) can be calculated from the specific heat of gases at constant volume ($c_V$), the mass ($M$) and the temperature variation in a liquid or solid ($\Delta T$) in the case of a constant volume:

$ dU = c_V m \Delta T $



Similarly, we can replace the differential inexact labour ($\delta W$) with the pressure ($p$) and the volume Variation ($dV$):

$ \delta W = p dV $



If we equate both expressions, we obtain the equation:

$c_VMdT=-pdV$



Which, with the inclusion of the volume ($V$), the universal gas constant ($R$), and número de Moles ($n$), leads to:

$ p V = n R T $



And with the mass ($M$) and the molar Mass ($M_m$):

$ n = \displaystyle\frac{ M }{ M_m }$



Finally, in the limit $\Delta T \rightarrow dt$, we obtain the relationship:

$\displaystyle\frac{ dT }{ T }=-\displaystyle\frac{ R }{ M_m c_V }\displaystyle\frac{ dV }{ V }$

ID:(15740, 0)



Adiabatic case relationship of temperature and volume

Concept

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In the adiabatic case, for absolute temperature ($T$) and the volume ($V$) with the universal gas constant ($R$), the molar Mass ($M_m$), the specific heat of gases at constant volume ($c_V$), the temperature variation ($dT$), and the volume Variation ($dV$), we have the following equation:

$\displaystyle\frac{ dT }{ T }=-\displaystyle\frac{ R }{ M_m c_V }\displaystyle\frac{ dV }{ V }$



By introducing the adiabatic index ($\kappa$), this equation can be expressed as:

$ \kappa \equiv1+\displaystyle\frac{ R }{ M_m c_V }$



This allows us to write the equation as:

$\displaystyle\frac{dT}{T}=-(\kappa - 1)\displaystyle\frac{dV}{V}$



If we integrate this expression between the volume in state i ($V_i$) and the volume in state f ($V_f$), as well as between the temperature in initial state ($T_i$) and the temperature in final state ($T_f$), we obtain:

$ T_i V_i ^{ \kappa -1}= T_f V_f ^{ \kappa -1}$

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Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$\kappa$
kappa
Adiabatic index
-
$N_A$
N_A
Avogadro's number
-
$M_m$
M_m
Molar Mass
kg/mol
$n$
n
Número de Moles
mol
$m$
m
Particle mass
kg
$c_V$
c_V
Specific heat of gases at constant volume
J/kg K
$R$
R
Universal gas constant
J/mol K
$\Delta T$
DT
Variación de 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
$p$
p
Pressure
Pa
$T_f$
T_f
Temperature in final state
K
$T_i$
T_i
Temperature in initial state
K
$dT$
dT
Temperature variation
K
$dU$
dU
Variation of the internal energy
J
$V$
V
Volume
m^3
$V_f$
V_f
Volume in state f
m^3
$V_i$
V_i
Volume in state i
m^3
$dV$
dV
Volume Variation
m^3

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 =0$

dQ =0


$\displaystyle\frac{ dT }{ T }=-\displaystyle\frac{ R }{ M_m c_V }\displaystyle\frac{ dV }{ V }$

dT / T =- R * dV /( M_m * c_V * V )


$ dU = c_V m dT $

dU = c_V * M * DT


$ dU = \delta Q - p dV $

dU = dQ - p * dV


$ \kappa \equiv1+\displaystyle\frac{ R }{ M_m c_V }$

k =1+ R /( M_m * c_V )


$ m =\displaystyle\frac{ M_m }{ N_A }$

m = M_m / N_A


$ p V = n R T $

p * V = n * R * T


$ T_i V_i ^{ \kappa -1}= T_f V_f ^{ \kappa -1}$

T_i * V_i ^( kappa -1)= T_f * V_f ^( kappa -1)

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Adiabatic condition

Equation

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In the adiabatic case, the system does not have the ability to alter the caloric Content ($Q$), meaning that the differential inexact Heat ($\delta Q$) must be zero:

$ \delta Q =0$

$\delta Q$
Differential inexact Heat
$J$
5220

ID:(4860, 0)



First law of thermodynamics and pressure

Equation

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With the first law of thermodynamics, it can be expressed in terms of the internal energy differential ($dU$), the differential inexact Heat ($\delta Q$), the pressure ($p$), and the volume Variation ($dV$) as:

$ dU = \delta Q - p dV $

$\delta Q$
Differential inexact Heat
$J$
5220
$p$
Pressure
$Pa$
5224
$dU$
Variation of the internal energy
$J$
5400
$dV$
Volume Variation
$m^3$
5223

Since the internal energy differential ($dU$) relates to the differential inexact Heat ($\delta Q$) and the differential inexact labour ($\delta W$) as shown below:

$ dU = \delta Q - \delta W $



And it is known that the differential inexact labour ($\delta W$) is related to the pressure ($p$) and the volume Variation ($dV$) as follows:

$ \delta W = p dV $



Therefore, we can conclude that:

$ dU = \delta Q - p dV $

ID:(3470, 0)



Caloric content of a gas at constant volume as a function of specific heat

Equation

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The relationship between the variation of the variation of the internal energy ($dU$) and the temperature variation in a liquid or solid ($\Delta T$) is with the specific heat of gases at constant volume ($c_V$) and the mass ($M$) equal to:

$ dU = c_V m \Delta T $

$ dU = c_V M \Delta T $

$M$
$m$
Particle mass
$kg$
5516
$c_V$
Specific heat of gases at constant volume
$J/kg K$
6662
$\Delta T$
Variación de Temperature
$K$
7510
$dU$
Variation of the internal energy
$J$
5400

The variation of the internal energy ($dU$) in relation to the temperature variation in a liquid or solid ($\Delta T$) and the heat capacity at constant volume ($C_V$) is expressed as:

$ dU = C_V \Delta T $



Where the heat capacity at constant volume ($C_V$) can be replaced by the specific heat of gases at constant volume ($c_V$) and the mass ($M$) using the following relationship:

$ c_V =\displaystyle\frac{ C_V }{ M }$



Therefore, we obtain:

$ dU = c_V M \Delta T $

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Particle mass and molar mass

Equation

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The particle mass ($m$) can be estimated from the molar Mass ($M_m$) and the avogadro's number ($N_A$) using

$ m =\displaystyle\frac{ M_m }{ N_A }$

$N_A$
Avogadro's number
6.02e+23
$-$
9860
$M_m$
Molar Mass
$kg/mol$
6212
$m$
Particle mass
$kg$
5516

ID:(4389, 0)



Temperature and Volume Variation

Equation

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In the adiabatic case it is given that the universal gas constant ($R$), the molar Mass ($M_m$), and the specific heat of gases at constant volume ($c_V$) vary in the temperature variation ($dT$) and ($$)5223< /var> according to:

$\displaystyle\frac{ dT }{ T }=-\displaystyle\frac{ R }{ M_m c_V }\displaystyle\frac{ dV }{ V }$

$T$
Absolute temperature
$K$
5177
$M_m$
Molar Mass
$kg/mol$
6212
$c_V$
Specific heat of gases at constant volume
$J/kg K$
6662
$dT$
Temperature variation
$K$
5217
$R$
Universal gas constant
8.4135
$J/mol K$
4957
$V$
Volume
$m^3$
5226
$dV$
Volume Variation
$m^3$
5223

Since with the variation of the internal energy ($dU$), the variation of heat ($\delta Q$), and the differential inexact labour ($\delta W$) we have:

$dU = \delta Q - \delta W = 0$



We can replace the variation of heat ($\delta Q$) with the infinitesimal version of the equation for the heat supplied to liquid or solid ($\Delta Q$) involving the specific heat at constant pressure ($c_p$), the mass ($M$), and the temperature variation in a liquid or solid ($\Delta T$) in the case of constant pressure, as shown below:

$ \Delta Q = c_p M \Delta T $



Similarly, we can replace the differential inexact labour ($\delta W$) with the pressure ($p$) and the volume Variation ($dV$):

$ \delta W = p dV $



If we equate both expressions, we obtain the equation:

$c_pMdT=-pdV$



Which, with the inclusion of the volume ($V$), the universal gas constant ($R$), and número de Moles ($n$), leads to:

$ p V = n R T $



And with the mass ($M$) and the molar Mass ($M_m$):

$ n = \displaystyle\frac{ M }{ M_m }$



Finally, in the limit $\Delta T \rightarrow dt$, we obtain the relationship:

$\displaystyle\frac{ dT }{ T }=-\displaystyle\frac{ R }{ M_m c_V }\displaystyle\frac{ dV }{ V }$

ID:(4861, 0)



Adiabatic Index

Equation

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Using the universal gas constant ($R$), the molar Mass ($M_m$), the specific heat of gases at constant volume ($c_V$), the temperature variation ($dT$), and the volume Variation ($dV$), the adiabatic index ($\kappa$) can be defined as follows:

$ \kappa \equiv1+\displaystyle\frac{ R }{ M_m c_V }$

$\kappa$
Adiabatic index
$-$
6661
$M_m$
Molar Mass
$kg/mol$
6212
$c_V$
Specific heat of gases at constant volume
$J/kg K$
6662
$R$
Universal gas constant
8.4135
$J/mol K$
4957

ID:(4864, 0)



General gas law

Equation

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The pressure ($p$), the volume ($V$), the absolute temperature ($T$), and the number of moles ($n$) are related by the following equation:

$ p V = n R T $

$T$
Absolute temperature
$K$
5177
$n$
Número de Moles
$mol$
6679
$p$
Pressure
$Pa$
5224
$R$
Universal gas constant
8.4135
$J/mol K$
4957
$V$
Volume
$m^3$
5226

The pressure ($p$), the volume ($V$), the absolute temperature ($T$), and the number of moles ($n$) are related through the following physical laws:

• Boyle's law

$ p V = C_b $



• Charles's law

$\displaystyle\frac{ V }{ T } = C_c$



• Gay-Lussac's law

$\displaystyle\frac{ p }{ T } = C_g$



• Avogadro's law

$\displaystyle\frac{ n }{ V } = C_a $



These laws can be expressed in a more general form as:

$\displaystyle\frac{pV}{nT}=cte$



This general relationship states that the product of pressure and volume divided by the number of moles and temperature remains constant:

$ p V = n R T $



where the universal gas constant ($R$) has a value of 8.314 J/K·mol.

ID:(3183, 0)



Adiabatic case relationship of temperature and volume

Equation

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From an initial state (i) with the volume in state i ($V_i$) and the temperature in initial state ($T_i$) it goes to a final state (f) with the volume in state f ($V_f$) and the temperature in final state ($T_f$) according to:

$ T_i V_i ^{ \kappa -1}= T_f V_f ^{ \kappa -1}$

$\kappa$
Adiabatic index
$-$
6661
$T_f$
Temperature in final state
$K$
5237
$T_i$
Temperature in initial state
$K$
5236
$V_f$
Volume in state f
$m^3$
5235
$V_i$
Volume in state i
$m^3$
5234

In the adiabatic case, for absolute temperature ($T$) and the volume ($V$) with the universal gas constant ($R$), the molar Mass ($M_m$), the specific heat at constant pressure ($c_p$), the temperature variation ($dT$), and the volume Variation ($dV$), we have the following equation:

$\displaystyle\frac{ dT }{ T }=-\displaystyle\frac{ R }{ M_m c_V }\displaystyle\frac{ dV }{ V }$



By introducing the adiabatic index ($\kappa$), this equation can be expressed as:

$ \kappa \equiv1+\displaystyle\frac{ R }{ M_m c_V }$



This allows us to write the equation as:

$\displaystyle\frac{dT}{T}=-(\kappa - 1)\displaystyle\frac{dV}{V}$



If we integrate this expression between the volume in state i ($V_i$) and the volume in state f ($V_f$), as well as between the temperature in initial state ($T_i$) and the temperature in final state ($T_f$), we obtain:

$ T_i V_i ^{ \kappa -1}= T_f V_f ^{ \kappa -1}$

ID:(4865, 0)