Adiabatic process

For a closed system, one may write the first law of thermodynamics as : ΔU = Q − W, where ΔU denotes the change of the system's internal energy, Q the quantity of energy added to it as heat, and W the work done by the system on its surroundings.

• If the system has such rigid walls that work cannot be transferred in or out (W = 0), and the walls are not adiabatic and energy is added in the form of heat (Q > 0), and there is no phase change, then the temperature of the system will rise.
• If the system has such rigid walls that pressure–volume work cannot be done, but the walls are adiabatic (Q = 0), and energy is added as isochoric (constant volume) work in the form of friction or the stirring of a viscous fluid within the system (W < 0), and there is no phase change, then the temperature of the system will rise.
• If the system walls are adiabatic (Q = 0) but not rigid (W ≠ 0), and, in a fictive idealized process, energy is added to the system in the form of frictionless, non-viscous pressure–volume work (W < 0), and there is no phase change, then the temperature of the system will rise. Such a process is called an isentropic process and is said to be "reversible". Ideally, if the process were reversed the energy could be recovered entirely as work done by the system. If the system contains a compressible gas and is reduced in volume, the uncertainty of the position of the gas is reduced, and seemingly would reduce the entropy of the system, but the temperature of the system will rise as the process is isentropic (ΔS = 0). Should the work be added in such a way that friction or viscous forces are operating within the system, then the process is not isentropic, and if there is no phase change, then the temperature of the system will rise, the process is said to be "irreversible", and the work added to the system is not entirely recoverable in the form of work.
• If the walls of a system are not adiabatic, and energy is transferred in as heat, entropy is transferred into the system with the heat. Such a process is neither adiabatic nor isentropic, having Q > 0, and ΔS > 0 according to the second law of thermodynamics.

Naturally occurring adiabatic processes are irreversible (entropy is produced).

The transfer of energy as work into an adiabatically isolated system can be imagined as being of two idealized extreme kinds. In one such kind, no entropy is produced within the system (no friction, viscous dissipation, etc.), and the work is only pressure-volume work (denoted by P dV). In nature, this ideal kind occurs only approximately because it demands an infinitely slow process and no sources of dissipation.

The other extreme kind of work is isochoric work (dV = 0), for which energy is added as work solely through friction or viscous dissipation within the system. A stirrer that transfers energy to a viscous fluid of an adiabatically isolated system with rigid walls, without phase change, will cause a rise in temperature of the fluid, but that work is not recoverable. Isochoric work is irreversible.[7] The second law of thermodynamics observes that a natural process, of transfer of energy as work, always consists at least of isochoric work and often both of these extreme kinds of work. Every natural process, adiabatic or not, is irreversible, with ΔS > 0, as friction or viscosity are always present to some extent.

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 L = 1000 cm³ = 0.001 m³); the gas within is the air consisting of molecular nitrogen and oxygen only (thus a diatomic gas with 5 degrees of freedom, and so γ = 7/5); the compression ratio of the engine is 10:1 (that is, the 1 L volume of uncompressed gas is reduced to 0.1 L by the piston); and the uncompressed gas is at approximately room temperature and pressure (a warm room temperature of ~27 °C, or 300 K, and a pressure of 1 bar = 100 kPa, i.e. typical sea-level atmospheric pressure).

${\displaystyle {\begin{aligned}&P_{1}V_{1}^{\gamma }=\mathrm {constant} _{1}=100\,000~{\text{Pa}}\times (0.001~{\text{m}}^{3})^{\frac {7}{5}}\\&=10^{5}\times 6.31\times 10^{-5}~{\text{Pa}}\,{\text{m}}^{21/5}=6.31~{\text{Pa}}\,{\text{m}}^{21/5},\end{aligned}}}$

so the adiabatic constant for this example is about 6.31 Pa m^4.2.

The gas is now compressed to a 0.1 L (0.0001 m³) 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

${\displaystyle P_{2}V_{2}^{\gamma }=\mathrm {constant} _{1}=6.31~{\text{Pa}}\,{\text{m}}^{21/5}=P\times (0.0001~{\text{m}}^{3})^{\frac {7}{5}},}$

We can now solve for the final pressure[12]

${\displaystyle P_{2}=P_{1}\left({\frac {V_{1}}{V_{2}}}\right)^{\gamma }=100\,000~{\text{Pa}}\times {\text{10}}^{7/5}=2.51\times 10^{6}~{\text{Pa}}}$

or 25.1 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 = nRT (n is amount of gas in moles and R the gas constant for that gas). Our initial conditions being 100 kPa of pressure, 1 L volume, and 300 K of temperature, our experimental constant (nR) is:

${\displaystyle {\frac {PV}{T}}=\mathrm {constant} _{2}={\frac {10^{5}~{\text{Pa}}\times 10^{-3}~{\text{m}}^{3}}{300~{\text{K}}}}=0.333~{\text{Pa}}\,{\text{m}}^{3}{\text{K}}^{-1}.}$

We know the compressed gas has V = 0.1 L and P = 2.51×10⁶ Pa, so we can solve for temperature:

${\displaystyle T={\frac {PV}{\mathrm {constant} _{2}}}={\frac {2.51\times 10^{6}~{\text{Pa}}\times 10^{-4}~{\text{m}}^{3}}{0.333~{\text{Pa}}\,{\text{m}}^{3}{\text{K}}^{-1}}}=753~{\text{K}}.}$

That is a final temperature of 753 K, or 479 °C, or 896 °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 temperature, which ensures immediate ignition of the injected fuel.

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.

The definition of an adiabatic process is that heat transfer to the system is zero, δQ = 0. Then, according to the first law of thermodynamics,

${\displaystyle dU+\delta W=\delta Q=0,}$

(a1)

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

${\displaystyle \delta W=P\,dV.}$

(a2)

However, P does not remain constant during an adiabatic process but instead changes along with V.

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

${\displaystyle U=\alpha nRT=\alpha PV,}$

(a3)

where α is the number of degrees of freedom divided by 2, R is the universal gas constant and n is the number of moles in the system (a constant).

Differentiating equation (a3) yields

${\displaystyle dU=\alpha nR\,dT=\alpha \,d(PV)=\alpha (P\,dV+V\,dP).}$

(a4)

Equation (a4) is often expressed as dU = nC_V dT because C_V = αR.

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

${\displaystyle -P\,dV=\alpha P\,dV+\alpha V\,dP,}$

factorize −P dV:

${\displaystyle -(\alpha +1)P\,dV=\alpha V\,dP,}$

and divide both sides by PV:

${\displaystyle -(\alpha +1){\frac {dV}{V}}=\alpha {\frac {dP}{P}}.}$

After integrating the left and right sides from V₀ to V and from P₀ to P and changing the sides respectively,

${\displaystyle \ln \left({\frac {P}{P_{0}}}\right)=-{\frac {\alpha +1}{\alpha }}\ln \left({\frac {V}{V_{0}}}\right).}$

Exponentiate both sides, substitute α + 1/α with γ, the heat capacity ratio

${\displaystyle \left({\frac {P}{P_{0}}}\right)=\left({\frac {V}{V_{0}}}\right)^{-\gamma },}$

and eliminate the negative sign to obtain

${\displaystyle \left({\frac {P}{P_{0}}}\right)=\left({\frac {V_{0}}{V}}\right)^{\gamma }.}$

Therefore,

${\displaystyle \left({\frac {P}{P_{0}}}\right)\left({\frac {V}{V_{0}}}\right)^{\gamma }=1,}$

and

${\displaystyle P_{0}V_{0}^{\gamma }=PV^{\gamma }=\mathrm {constant} .}$

${\displaystyle \Delta U=\alpha RnT_{2}-\alpha RnT_{1}=\alpha Rn\Delta T.}$

(b1)

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

${\displaystyle W=\int _{V_{1}}^{V_{2}}P\,dV.}$

(b2)

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

${\displaystyle \Delta U+W=0.}$

(b3)

By the previous derivation,

${\displaystyle PV^{\gamma }={\text{constant}}=P_{1}V_{1}^{\gamma }.}$

(b4)

Rearranging (b4) gives

${\displaystyle P=P_{1}\left({\frac {V_{1}}{V}}\right)^{\gamma }.}$

Substituting this into (b2) gives

${\displaystyle W=\int _{V_{1}}^{V_{2}}P_{1}\left({\frac {V_{1}}{V}}\right)^{\gamma }\,dV.}$

Integrating we obtain the expression for work,

${\displaystyle W=P_{1}V_{1}^{\gamma }{\frac {V_{2}^{1-\gamma }-V_{1}^{1-\gamma }}{1-\gamma }}={\frac {P_{2}V_{2}-P_{1}V_{1}}{1-\gamma }}.}$

Substituting γ = α + 1/α in second term,

${\displaystyle W=-\alpha P_{1}V_{1}^{\gamma }\left(V_{2}^{1-\gamma }-V_{1}^{1-\gamma }\right).}$

Rearranging,

${\displaystyle W=-\alpha P_{1}V_{1}\left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right).}$

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

${\displaystyle W=-\alpha nRT_{1}\left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right).}$

By the continuous formula,

${\displaystyle {\frac {P_{2}}{P_{1}}}=\left({\frac {V_{2}}{V_{1}}}\right)^{-\gamma },}$

or

${\displaystyle \left({\frac {P_{2}}{P_{1}}}\right)^{-{\frac {1}{\gamma }}}={\frac {V_{2}}{V_{1}}}.}$

Substituting into the previous expression for W,

${\displaystyle W=-\alpha nRT_{1}\left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right).}$

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

${\displaystyle \alpha nR(T_{2}-T_{1})=\alpha nRT_{1}\left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right).}$

Simplifying,

${\displaystyle T_{2}-T_{1}=T_{1}\left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right),}$

${\displaystyle {\frac {T_{2}}{T_{1}}}-1=\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1,}$

${\displaystyle T_{2}=T_{1}\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}.}$

The change in internal energy of a system, measured from state 1 to state 2, is equal to

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

${\displaystyle W=\int _{V_{1}}^{V_{2}}P\,dV.}$

(c2)

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

${\displaystyle \Delta U+W=0.}$

(c3)

By the previous derivation,

${\displaystyle PV^{\gamma }={\text{constant}}=P_{1}V_{1}^{\gamma }.}$

(c4)

Rearranging (c4) gives

${\displaystyle P=P_{1}\left({\frac {V_{1}}{V}}\right)^{\gamma }.}$

Substituting this into (c2) gives

${\displaystyle W=\int _{V_{1}}^{V_{2}}P_{1}\left({\frac {V_{1}}{V}}\right)^{\gamma }\,dV.}$

Integrating we obtain the expression for work,

${\displaystyle W=P_{1}V_{1}^{\gamma }{\frac {V_{2}^{1-\gamma }-V_{1}^{1-\gamma }}{1-\gamma }}={\frac {P_{2}V_{2}-P_{1}V_{1}}{1-\gamma }}.}$

Substituting γ = α + 1/α in second term,

${\displaystyle W=-\alpha P_{1}V_{1}^{\gamma }\left(V_{2}^{1-\gamma }-V_{1}^{1-\gamma }\right).}$

Rearranging,

${\displaystyle W=-\alpha P_{1}V_{1}\left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right).}$

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

${\displaystyle W=-\alpha nRT_{1}\left(\left({\frac {V_{2}}{V_{1}}}\right)^{1-\gamma }-1\right).}$

By the continuous formula,

${\displaystyle {\frac {P_{2}}{P_{1}}}=\left({\frac {V_{2}}{V_{1}}}\right)^{-\gamma },}$

or

${\displaystyle \left({\frac {P_{2}}{P_{1}}}\right)^{-{\frac {1}{\gamma }}}={\frac {V_{2}}{V_{1}}}.}$

Substituting into the previous expression for W,

${\displaystyle W=-\alpha nRT_{1}\left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right).}$

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

${\displaystyle \alpha nR(T_{2}-T_{1})=\alpha nRT_{1}\left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right).}$

Simplifying,

${\displaystyle T_{2}-T_{1}=T_{1}\left(\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1\right),}$

${\displaystyle {\frac {T_{2}}{T_{1}}}-1=\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}-1,}$

${\displaystyle T_{2}=T_{1}\left({\frac {P_{2}}{P_{1}}}\right)^{\frac {\gamma -1}{\gamma }}.}$