Editing Adiabatic process (section)

===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.