Editing Adiabatic process (section)

==Ideal gas (reversible process)==
{{Main|Reversible adiabatic process}}

[[Image:Adiabatic.svg|thumb|upright=1.2|For a simple substance, during an adiabatic process in which the volume increases, the [[internal energy]] of the working substance must decrease.]]
The mathematical equation for an [[ideal gas]] undergoing a reversible (i.e., no entropy generation) adiabatic process can be represented by the [[polytropic process]] equation<ref name="Bailyn 53"/>

<math display="block"> P V^\gamma = \text{constant}, </math>

where {{math|''P''}} is pressure, {{math|''V''}} is volume, and {{math|''γ''}} is the [[adiabatic index]] or heat capacity ratio defined as

<math display="block"> \gamma = \frac{C_P}{C_V} = \frac{f + 2}{f}. </math>

Here {{math|''C<sub>P</sub>''}} is the [[specific heat]] for constant pressure, {{math|''C<sub>V</sub>''}} is the specific heat for constant volume, and {{math|''f''}} is the number of [[Degrees of freedom (physics and chemistry)|degrees of freedom]] (3 for a monatomic gas, 5 for a diatomic gas or a gas of linear molecules such as carbon dioxide).

For a monatomic ideal gas, {{math|1=''γ'' = {{sfrac|5|3}}}}, and for a diatomic gas (such as [[nitrogen]] and [[oxygen]], the main components of air), {{math|1=''γ'' = {{sfrac|7|5}}}}.<ref>{{cite web |url=http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html |title=Adiabatic Process |website=HyperPhysics |publisher=Georgia State University}}</ref> Note that the above formula is only applicable to classical ideal gases (that is, gases far above absolute zero temperature) and not [[Bose–Einstein condensate|Bose–Einstein]] or [[Fermionic condensate|Fermi gases]].

One can also use the ideal gas law to rewrite the above relationship between {{math|''P''}} and {{math|''V''}} as <ref name="Bailyn 53"/>

<math display="block">\begin{align}
P^{1-\gamma} T^\gamma &= \text{constant},\\
TV^{\gamma - 1} &= \text{constant}
\end{align}</math>

where ''T'' is the absolute or [[thermodynamic temperature]].

===Example of adiabatic compression===
The compression stroke in a [[gasoline engine]] can be used as an example of adiabatic compression. The model assumptions are: the uncompressed volume of the cylinder is one litre (1&nbsp;L = 1000&nbsp;cm<sup>3</sup> = 0.001&nbsp;m<sup>3</sup>); the gas within is the air consisting of molecular nitrogen and oxygen only (thus a diatomic gas with 5 degrees of freedom, and so {{math|1=''γ'' = {{sfrac|7|5}}}}); the compression ratio of the engine is 10:1 (that is, the 1&nbsp;L volume of uncompressed gas is reduced to 0.1&nbsp;L by the piston); and the uncompressed gas is at approximately room temperature and pressure (a warm room temperature of ~27&nbsp;°C, or 300&nbsp;K, and a pressure of 1&nbsp;bar = 100&nbsp;kPa, i.e. typical sea-level atmospheric pressure).

<math display="block">\begin{align}
P_1 V_1^\gamma &= \mathrm{constant}_1 \\
 & = 100\,000~\text{Pa} \times (0.001~\text{m}^3)^\frac75 \\
 & = 10^5 \times 6.31 \times 10^{-5}~\text{Pa}\,\text{m}^{21/5} \\
 & = 6.31~\text{Pa}\,\text{m}^{21/5},
\end{align}</math>

so the adiabatic constant for this example is about {{nowrap|6.31 Pa&thinsp;m<sup>4.2</sup>.}}

The gas is now compressed to a 0.1&nbsp;L (0.0001&nbsp;m<sup>3</sup>) volume, which we assume happens quickly enough that no heat enters or leaves the gas through the walls. The adiabatic constant remains the same, but with the resulting pressure unknown

<math display="block">\begin{align} 
P_2 V_2^\gamma &= \mathrm{constant}_1 \\
&= 6.31~\text{Pa}\,\text{m}^{21/5} \\
&= P \times (0.0001~\text{m}^3)^\frac75,
\end{align}</math>

We can now solve for the final pressure<ref>{{cite book |last1=Atkins |first1=Peter |last2=de Paula |first2=Giulio |title=Atkins' Physical Chemistry |date=2006 |publisher=W. H. Freeman |isbn=0-7167-8759-8 |page=48 |edition=8th}}</ref>

<math display="block">\begin{align}
P_2 &= P_1\left (\frac{V_1}{V_2}\right)^\gamma \\
&= 100\,000~\text{Pa} \times \text{10}^{7/5} \\
&= 2.51 \times 10^6~\text{Pa}
\end{align}</math>

or 25.1&nbsp;bar. This pressure increase is more than a simple 10:1 compression ratio would indicate; this is because the gas is not only compressed, but the work done to compress the gas also increases its internal energy, which manifests itself by a rise in the gas temperature and an additional rise in pressure above what would result from a simplistic calculation of 10 times the original pressure.

We can solve for the temperature of the compressed gas in the engine cylinder as well, using the ideal gas law, ''PV''&nbsp;=&nbsp;''nRT'' (''n'' is amount of gas in moles and ''R'' the gas constant for that gas). Our initial conditions being 100&nbsp;kPa of pressure, 1&nbsp;L volume, and 300&nbsp;K of temperature, our experimental constant (''nR'') is:

<math display="block">\begin{align}
\frac{PV}{T} &= \mathrm{constant}_2 \\
&= \frac{10^5~\text{Pa} \times 10^{-3}~\text{m}^3}{300~\text{K}} \\
&= 0.333~\text{Pa}\,\text{m}^3\text{K}^{-1}.
\end{align}</math>

We know the compressed gas has {{mvar|V}}&nbsp;= 0.1&nbsp;L and {{mvar|P}}&nbsp;= {{val|2.51|e=6|u=Pa}}, so we can solve for temperature:

<math display="block">\begin{align}
T &= \frac{P V}{\mathrm{constant}_2} \\
&= \frac{2.51 \times 10^6~\text{Pa} \times 10^{-4}~\text{m}^3}{0.333~\text{Pa}\,\text{m}^3\text{K}^{-1}} \\
&= 753~\text{K}.
\end{align}</math>

That is a final temperature of 753&nbsp;K, or 479&nbsp;°C, or 896&nbsp;°F, well above the ignition point of many fuels. This is why a high-compression engine requires fuels specially formulated to not self-ignite (which would cause [[engine knocking]] when operated under these conditions of temperature and pressure), or that a [[supercharger]] with an [[intercooler]] to provide a pressure boost but with a lower temperature rise would be advantageous. A diesel engine operates under even more extreme conditions, with compression ratios of 16:1 or more being typical, in order to provide a very high gas pressure, which ensures immediate ignition of the injected fuel.

===Adiabatic free expansion of a gas===
{{See also|Free expansion}}
For an adiabatic free expansion of an [[ideal gas]], the gas is contained in an insulated container and then allowed to expand in a vacuum. Because there is no external pressure for the gas to expand against, the work done by or on the system is zero. Since this process does not involve any heat transfer or work, the first law of thermodynamics then implies that the net internal energy change of the system is zero. For an ideal gas, the temperature remains constant because the internal energy only depends on temperature in that case. Since at constant temperature, the entropy is proportional to the volume, the entropy increases in this case, therefore this process is irreversible.

===Derivation of ''P''–''V'' relation for adiabatic compression and expansion===
The definition of an adiabatic process is that heat transfer to the system is zero, {{math|1=''δQ'' = 0}}. Then, according to the first law of thermodynamics,

{{NumBlk||<math display="block"> d U + \delta W = \delta Q = 0, </math>|{{EquationRef|a1}}}}

where {{math|''dU''}} is the change in the internal energy of the system and {{math|''δW''}} is work done ''by'' the system. Any work ({{math|''δW''}}) done must be done at the expense of internal energy {{math|''U''}}, since no heat {{math|''δQ''}} is being supplied from the surroundings. Pressure–volume work {{math|''δW''}} done ''by'' the system is defined as

{{NumBlk||<math display="block"> \delta W = P \, dV. </math>|{{EquationRef|a2}}}}

However, {{math|''P''}} does not remain constant during an adiabatic process but instead changes along with {{math|''V''}}.

It is desired to know how the values of {{math|''dP''}} and {{math|''dV''}} relate to each other as the adiabatic process proceeds. For an ideal gas (recall ideal gas law {{math|1=''PV'' = ''nRT''}}) the internal energy is given by

{{NumBlk||<math display="block"> U = \alpha n R T = \alpha P V, </math>|{{EquationRef|a3}}}}

where {{math|''α''}} is the number of degrees of freedom divided by 2, {{math|''R''}} is the [[universal gas constant]] and {{math|''n''}} is the number of moles in the system (a constant).

Differentiating equation (a3) yields

{{NumBlk||<math display="block">\begin{align}
d U &= \alpha n R \, dT\\
                 & = \alpha \, d (P V)\\
                 & = \alpha (P \, dV + V \, dP).
\end{align}</math>|{{EquationRef|a4}}}}

Equation (a4) is often expressed as {{math|1=''dU'' = ''nC<sub>V</sub> dT''}} because {{math|1=''C<sub>V</sub>'' = ''αR''}}.

Now substitute equations (a2) and (a4) into equation (a1) to obtain

<math display="block"> -P \, dV = \alpha P \, dV + \alpha V \, dP,</math>

factorize {{math|−''P'' ''dV''}}:

<math display="block"> -(\alpha + 1) P \, dV = \alpha V \, dP,</math>

and divide both sides by {{math|''PV''}}:

<math display="block"> -(\alpha + 1) \frac{dV}{V} = \alpha \frac{dP}{P}. </math>

After integrating the left and right sides from {{math|''V''<sub>0</sub>}} to {{math|''V''}} and from {{math|''P''<sub>0</sub>}} to {{math|''P''}} and changing the sides respectively,

<math display="block"> \ln \left( \frac{P}{P_0} \right) = -\frac{\alpha + 1}{\alpha} \ln \left( \frac{V}{V_0} \right). </math>

Exponentiate both sides, substitute {{math|{{sfrac|''α'' + 1|''α''}}}} with {{math|''γ''}}, the heat capacity ratio

<math display="block"> \left( \frac{P}{P_0} \right) = \left( \frac{V}{V_0} \right)^{-\gamma}, </math>

and eliminate the negative sign to obtain

<math display="block"> \left( \frac{P}{P_0} \right) = \left( \frac{V_0}{V} \right)^\gamma. </math>

Therefore,

<math display="block"> \left( \frac{P}{P_0} \right) \left( \frac{V}{V_0} \right)^\gamma = 1,</math>

and

<math display="block"> P_0 V_0^\gamma = P V^\gamma = \mathrm{constant}. </math>

{{NumBlk||<math display="block"> \Delta U = \alpha R nT_2 - \alpha R nT_1 = \alpha Rn \Delta T. </math>|{{EquationRef|b1}}}}

At the same time, the work done by the pressure–volume changes as a result from this process, is equal to

{{NumBlk||<math display="block"> W = \int_{V_1}^{V_2}P \,dV. </math>|{{EquationRef|b2}}}}

Since we require the process to be adiabatic, the following equation needs to be true

{{NumBlk||<math display="block"> \Delta U + W = 0. </math>|{{EquationRef|b3}}}}

By the previous derivation,

{{NumBlk||<math display="block"> P V^\gamma = \text{constant} = P_1 V_1^\gamma. </math>|{{EquationRef|b4}}}}

Rearranging (b4) gives

<math display="block"> P = P_1 \left(\frac{V_1}{V} \right)^\gamma. </math>

Substituting this into (b2) gives

<math display="block"> W = \int_{V_1}^{V_2} P_1 \left(\frac{V_1}{V} \right)^\gamma \,dV. </math>

Integrating, we obtain the expression for work,

<math display="block">\begin{align}
W = P_1 V_1^\gamma \frac{V_2^{1-\gamma} - V_1^{1-\gamma}}{1 - \gamma} \\
&= \frac{P_2 V_2 - P_1 V_1}{1 - \gamma}.
\end{align}</math>

Substituting {{math|1=''γ'' = {{sfrac|''α'' + 1|''α''}}}} in the second term,

<math display="block"> W = -\alpha P_1 V_1^\gamma \left( V_2^{1-\gamma} - V_1^{1-\gamma} \right). </math>

Rearranging,

<math display="block"> W = -\alpha P_1 V_1 \left( \left( \frac{V_2}{V_1} \right)^{1-\gamma} - 1 \right). </math>

Using the ideal gas law and assuming a constant molar quantity (as often happens in practical cases),

<math display="block"> W = -\alpha n R T_1 \left( \left( \frac{V_2}{V_1} \right)^{1-\gamma} - 1 \right). </math>

By the continuous formula,

<math display="block"> \frac{P_2}{P_1} = \left(\frac{V_2}{V_1}\right)^{-\gamma}, </math>

or

<math display="block"> \left(\frac{P_2}{P_1}\right)^{-\frac{1}{\gamma}} = \frac{V_2}{V_1}. </math>

Substituting into the previous expression for {{math|''W''}},

<math display="block"> W = -\alpha n R T_1 \left( \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} - 1 \right). </math>

Substituting this expression and (b1) in (b3) gives

<math display="block"> \alpha n R (T_2 - T_1) = \alpha n R T_1 \left( \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} - 1 \right). </math>

Simplifying,

<math display="block">\begin{align}
T_2 - T_1 &=  T_1 \left( \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} - 1 \right), \\
\frac{T_2}{T_1} - 1 &=  \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} - 1, \\
T_2 &= T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}}.
\end{align}</math>

===Derivation of discrete formula and work expression===
The change in internal energy of a system, measured from state 1 to state 2, is equal to

<!-- equation missing here? TODO: check revisions -->At the same time, the work done by the pressure–volume changes as a result from this process, is equal to

{{NumBlk||<math display="block"> W = \int_{V_1}^{V_2}P \,dV. </math>|{{EquationRef|c2}}}}

Since we require the process to be adiabatic, the following equation needs to be true

{{NumBlk||<math display="block"> \Delta U + W = 0. </math>|{{EquationRef|c3}}}}

By the previous derivation,

{{NumBlk||<math display="block"> P V^\gamma = \text{constant} = P_1 V_1^\gamma. </math>|{{EquationRef|c4}}}}

Rearranging (c4) gives

<math display="block"> P = P_1 \left(\frac{V_1}{V} \right)^\gamma. </math>

Substituting this into (c2) gives

<math display="block"> W = \int_{V_1}^{V_2} P_1 \left(\frac{V_1}{V} \right)^\gamma \,dV. </math>

Integrating we obtain the expression for work,

<math display="block"> W = P_1 V_1^\gamma \frac{V_2^{1-\gamma} - V_1^{1-\gamma}}{1 - \gamma} = \frac{P_2 V_2 - P_1 V_1}{1 - \gamma}. </math>

Substituting {{math|1=''γ'' = {{sfrac|''α'' + 1|''α''}}}} in second term,

<math display="block"> W = -\alpha P_1 V_1^\gamma \left( V_2^{1-\gamma} - V_1^{1-\gamma} \right). </math>

Rearranging,

<math display="block"> W = -\alpha P_1 V_1 \left( \left( \frac{V_2}{V_1} \right)^{1-\gamma} - 1 \right). </math>

Using the ideal gas law and assuming a constant molar quantity (as often happens in practical cases),

<math display="block"> W = -\alpha n R T_1 \left( \left( \frac{V_2}{V_1} \right)^{1-\gamma} - 1 \right). </math>

By the continuous formula,

<math display="block"> \frac{P_2}{P_1} = \left(\frac{V_2}{V_1}\right)^{-\gamma}, </math>

or

<math display="block"> \left(\frac{P_2}{P_1}\right)^{-\frac{1}{\gamma}} = \frac{V_2}{V_1}. </math>

Substituting into the previous expression for {{math|''W''}},

<math display="block"> W = -\alpha n R T_1 \left( \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} - 1 \right). </math>

Substituting this expression and (c1) in (c3) gives

<math display="block"> \alpha n R (T_2 - T_1) = \alpha n R T_1 \left( \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} - 1 \right). </math>

Simplifying,

<math display="block">\begin{align}
T_2 - T_1 &=  T_1 \left( \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} - 1 \right), \\
\frac{T_2}{T_1} - 1 &=  \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}} - 1, \\
T_2 &= T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}}.
\end{align}</math>