Editing Adiabatic process (section)

===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>