Perfect gas

The perfect gas is a Modèle Thermodynamique describing the behavior of all the real Gaz S with low Pression p .

This model was developed at the 19th century by noting that all the gases tend towards this same behavior to sufficiently low pressure, whatever the chemical nature of gas what expresses the Loi of Avogadro, discovered in 1811: the relation between the pressure, the Volume and the Température is, under these conditions, independent of the nature of gas. This appeared so extraordinary that it took time to believe in it (see the article Nombre of Avogadro ). This property is explained by the fact why when the pressure is low, the gas particles are sufficiently distant from/to each other to be able to neglect the interactions of an electrostatic nature which depend, they, of the physicochemical nature of the gas (polar molecules more or less ). Many real gases check with an excellent approximation the model of perfect gas, in the normal conditions. It is the case of principal gases of the air, the Diazote and the Dioxygène.

Description of a perfect gas

On the macroscopic levels, one calls perfect gas any simultaneously checking gas: ; Law of Boyle-Mariotte: at constant temperature, the product of the pressure p by volume V : statement is regarded as constant when the pressure is low. ; Law of Avogadro: all the gases have same the molar Volume under the same conditions of pressure and temperature.

On the microscopic level, the kinetic Théorie of the gases makes it possible to find this perfect gas reaction of: a perfect gas is a gas whose Molécule S do not interact between them apart from the shocks and whose size is negligible compared to the average intermolecular distance. The energy of perfect gas is thus the sum of the kinetic energy of the Center of mass of the molecules and the internal energy of each molecule (rotation, oscillation). When these two energies are proportional, there is the Perfect gas of Laplace.

Limits of the model and real gas

With low pressure, all the gases can be modelled by a perfect gas. When the pressure increases, one cannot neglect any more the interactions at short distance, in particular the effect of size of the molecules and the interactions of the type van der Waals.

To caricature, if the pressure further is increased, the gas liquefies and, obviously, a liquid cannot be regarded as a perfect gas.

A real gas has a behavior close to a perfect gas if the intermolecular distances are large compared to the size of the molecules and with the range of the forces of interaction. One calls perfect gas “associated with gas the real” perfect gas whose heat-storage capacity with constant pressure is that of real gas with null pressure $C\_p\; (p=0,\; T)$. One draws up the thermodynamic tables of a real gas starting from the corrections made to this associated perfect gas.

Equation of state and laws of Joule

As for any gas, the thermodynamic state of balance of a perfect gas is fixed for N mole S of molecules, by two macroscopic parameters, with the choice. The other parameters can be calculated starting from the two parameters chosen by the equation of state.

The equation most usually used is the equation of perfect gases, a thermoelastic equation. This one can be written in two forms, one whose vision is macroscopic (utilizing volume V gas), the other having a more microscopic approach where one considers the number of molecules contained in a unit of volume.

$pV\; =\; nRT\; =\; nN\_Ak\_BT\; \backslash \; or\; \backslash \; p\; =\; Nk\_BT$

where

• p is the Pression (in Pascal);
• V is the Volume occupied by gas (in Cubic meter);
• N is the Quantité of matter, in mole
• NR is the number of particles
• R is the constant of the gases parfaits
  R = 8,314  472 J·K^-1·mol^-1
  on has in fact R = N_A · k_B where N_A is the Nombre of Avogadro (6,022) and k_B is the Boltzmann constant (1,38);
• T is the absolute Température (in Kelvin).

This equation derives from other laws found before: the Law of Charles, the Law of Boyle-Mariotte and the Law of Gay-Lussac.

Numerical application:

• for a pressure of an atmosphere ( p = p ₀ = 1,013  25 Pa)
• and a temperature of 0°C ( T = T ₀ = 273,15  K, temperature of the melting ice under p ₀),

molar volume is

V ₀ = 22,413  996 (39) L/mol

One retains the approximate value 22,4 in general L/mol.

What gives a volume by molecule (“free” volume around the molecule, independently of its dimension):

$\backslash \; frac\; \{V\_0\}\; \{NR\}$

if one compares this free volume to a cube, then stops it of this cube is overall the average distance separating the molecules at every moment, that one calls “length of Loschmidt” D ₀. This value is the cubic Racine “free” volume:

D ₀ = 3,338  792 5 Nm

one uses the approximate value in general 3,33 Nm.

For a pressure being worth one millionth Atmospheric pressure ( p ₀/: 1000000), the interparticle distance is 333 Nm = 1/3 micrometer and is independent of the nature of gas.

One can also use an energy equation, the Loi of Joule-Thomson where the enthalpy $H\; (p,\; T)\; =\; U\; +\; p.V$ does not depend on the pressure.

From these two equations, it comes that energy interns U ( V , T ), depends only on the temperature and not on volume: a perfect gas follows the Loi of Joule.

The Heat-storage capacity with constant volume of a perfect gas is worth

$C\_V\; =\; NR\; \backslash \; left\; (\backslash \; frac\; \{\backslash \; partial\; U\}\; \{\backslash \; partial\; T\}\; \backslash \; \backslash \; right)\; \_V$

and is thus independent of volume.

In the same way, the heat-storage capacity with constant pressure of a perfect gas is worth:

$C\_p\; =\; NR\; \backslash \; left\; (\backslash \; frac\; \{\backslash \; partial\; H\}\; \{\backslash \; partial\; T\}\; \backslash \; \backslash \; right)\; \_p$

and is thus independent of the pressure.

Like, by definition, the enthalpy is worth

$H\; =\; U\; +\; pV~,$

one obtains the Relation of Mayer:

$C\_p\; -\; C\_V\; =\; R~$

for a mole.

If one applies the formula of Clapeyron giving the latent heat of expansion L , of latent transfer of heat of dilation (see Formules of thermodynamics), and one finds that the coefficient β relative increase in isochoric pressure checks:

$\backslash \; beta\; T\; =\; 1~$,

thus:

U (V, T) does not depend on volume: a perfect gas is a Gaz of Joule, $U\; =\; NR\; \backslash \; cdot\; U\; (T)\; +cste~$

and the enthalpy is worth:

$H\; (p,\; T)\; =\; U\; +\; pV~$

thus does not depend on p : a perfect gas is a Gaz of Joule-Thomson (see Loi of Joule).

Perfect gas of Laplace

Definition

It happens that $C\_p\; \backslash ,$ does not depend on $T\; \backslash ,$, therefore $\backslash \; gamma\; =\; C\_p/C\_V\; \backslash ,$ does not depend on it either. The perfect gas is known as of Laplace (the GP of Laplace). It often happens that one specifies the value of $\backslash \; gamma\; \backslash ,$.

By convention, if one chooses $\backslash \; gamma\; =\; \backslash \; frac\; \{5\}\; \{3\}$, the perfect gas of Laplace is known as “monoatomic perfect gas” (GPM). The behavior of the Argon is very close to a monoatomic perfect gas.

By convention, if one chooses $\backslash \; gamma\; =\; \backslash \; frac\; \{7\}\; \{5\}$ the perfect gas of Laplace is known as “diatomic perfect gas” (GPD). The behavior of the Diazote N₂ is close to a diatomic perfect gas.

It is of use to write for a perfect gas of Laplace:

$statement\; =\; (\backslash \; gamma\; -1)\; \backslash \; cdot\; U\; =\; (1\; -\; \backslash \; frac\; \{1\}\; \{\backslash \; gamma\})\; \backslash \; cdot\; H$

maybe, for a monoatomic perfect gas

statement = 2/3 U = 2/5 H

and for a diatomic perfect gas:

statement = 2/5 U = 2/7 H .

Law of Laplace

In an Adiabatic transformation reversible elementary, $dU\; =\; -\; p.dV\; \backslash ,$. This involves, by replacing U by its value according to p and V , $V.dp\; +\; \backslash \; gamma.p.dV\; =\; 0~$

Maybe while integrating, the Law of Laplace:

$statement\; ^\; \{\backslash \; gamma\}\; =\; cste\; =\; p\_0\; V\_0^\; \{\backslash \; gamma\}\; ~$

from where the name given to these perfect gases.

It results from it, in a relaxation, a considerable cooling, the fascinating gas on its energy interns the work, (- W ), that it provides .

See also: adiabatic Pressure drop

; Numerical application For a monoatomic perfect gas, one finds easily:

$\backslash \; ln\; \backslash \; frac\; \{T\}\; \{T^0\}\; =\; \backslash \; frac\; \{2\}\; \{5\}\; \backslash \; cdot\; \backslash \; ln\; \backslash \; frac\; \{p\}\; \{p^0\}$

Maybe for a report/ratio of pressure 1/10 (the pressure drops by 90%), a fall in absolute temperature of a factor 0,398; of 300  K the temperature goes down to 120  K is a fall of 180  K. This process is used in industry to obtain low temperatures. Unfortunately, the value obtained does not reflect reality, because the real gases are not perfect gases at low temperature.

For the orders of magnitude, one retains R · T ~ 2  500 J/mol to 300 K. the work recovered in this relaxation is 2500·(180/300) = 900 J/mol.

Experiment: relaxation in an empty container

Another remarkable law is the produced heating when one lets penetrate a perfect gas of Laplace in an empty bottle. The gas is engulfed and then very quickly all becomes again chaotic: the temperature is standardized and become

$T$ = $\backslash \; gamma\; T$₀

$T$₀ being the external temperature. For an external temperature of 300 K and a monoatomic perfect gas, one obtains:

$T$ = 500 K

that is to say a rise in 200 K. In the blower of Modane, it is well what one can observe.

Thus, there are two cases of expansion of gas. The second astonishes a little, if one did not reflect there before.

(See also the adiabatic article Compression .)

Cycle of Carnot of a perfect gas of Laplace

A cycle of driving Carnot of a perfect gas has like wants it the theorem of Carnot, the output of Carnot: $r\; =\; 1\; \backslash \; frac\; \{T\_2\}\; \{T\_1\}$.

In this case, all can be calculated easily. Let us describe the cycle - one will be able to easily draw it in coordinates of Clapeyron logarithmic curves (log V , log p ):

1. Transformation (has → B): isothermal compression at temperature low, T ₂, reversible: it is thus necessary, to prevent that the gas does not warm up, releasing a quantity of heat - Q ₂ >0 with the cold source which maintains the temperature T ₂.
2. Transformation (B → C): compression of Laplace: the temperature assembles T ₂ to T ₁.
3. Transformation (C → D): isothermal at high temperature T ₁, reversible relaxation: to prevent that the gas does not cool, it is necessary that the hot source provides a quantity of heat Q ₁>0.
4. Transformation (C → A) (the point a.c. be selected with the intersection of the reversible adiabat passing by A): the reversible adiabatic pressure drop brings back the temperature of T ₁ to T ₂, and gas in its initial state.

End of the cycle.

Since the gas returned in its initial state, the First principle of thermodynamics says to us that

W + Q ₁ + Q ₂ = 0

The output of the engine is recovered work - W (thus equal to Q ₁ + Q ₂) divided by the quantity of heat delivered by the hot source, is Q ₁:

$r\; =\; 1\; +\; \backslash \; frac\; \{Q\_2\}\; \{Q\_1\}$

It is shown that

$\backslash \; frac\; \{Q\_1\}\; \{T\_1\}\; +\; \backslash \; frac\; \{Q\_2\}\; \{T\_2\}\; =0$

from where

$r\; =\; -\; \backslash \; frac\; \{W\}\; \{Q\_1\}\; =\; 1\; \backslash \; frac\; \{T\_2\}\; \{T\_1\}$.

What gives us the announced formula.

The equality of Clausius

$\backslash \; frac\; \{Q\_1\}\; \{T\_1\}\; +\; \backslash \; frac\; \{Q\_2\}\; \{T\_2\}\; =\; 0$

comes owing to the fact that the cycle was reversible: the total entropy remained constant, that of gas is null because it returned in state A. source 1 saw its entropy varying - Q ₁/ T ₁, source 2 of - Q ₂/ T ₂, from where the equality.

numerical Application : even by taking a water of river to 300 K and an hot source to 600 K, the output would be only 50%. On a electric Giga Watt provided by a “reversible” power station, 2GW must be consumed (out of coal, oil, nuclear methane, or mox) of which 1 GW will go to the river (rise in temperature) or in the atmosphere (heat of smoke, vapor of a Tower cooling). If it is considered that all heat goes in the river and that this one has a flow 100 m/s: in 1 second, 10 9 J will heat 100 G of water: that is to say a rise in temperature of 10 (J) /4,18 (J/K) = 2,4 K.

Six power stations producing 6 GW would give a rise in 6·2,4 = 14,4 K. In France, it are interdict to exceed 27 °C in a river (legal requirement, for the survival of the aquatic life in moderate zone): the summer 2003 was very hot, it was necessary to stop certain power stations (see the article Canicule 2003 ). This thermal pollution ( Q ₂ < 0) calculated by the formula of Carnot is lowest possible; it is acted in fact of a minimum, real thermal pollution is higher. Still we took an output R = 0,5; reality is close to 0,42.

Experiment of Joule

Here a third example: this time the perfect gas is contained in a bottle of volume V ₁ and is brutally put in contact with an empty container of volume V ₂, where it is engulfed partially: which is the final temperature by admitting that the walls do not absorb any heat?

This experiment is called experiment of Joule and Gay-Lussac, or relaxation of Joule; in fact, it is Hirn in 1856 which really succeeds in implementing it in experiments.

One suspects a little what occurs: in the second container, the gas must be hotter; if V ₂ is much lower than V ₁, one must find T ₂ = γ· T ₀, according to experiment 2.

In container 1, the gas which remains there is slackened, it must be colder; it is what one notes. But which is the final temperature after return to thermal balance? Answer: as a GP is a gas of Joule, the temperature does not change (cf Loi of Joule).

Entropy of a perfect gas

For all to know gas, it any more but does not remain to calculate its entropy S; what is easy if the fact is admitted that T , absolute temperature, is the thermodynamic temperature (Thermométrie).

Then writing

$dS\; =\; \backslash \; frac\; \{1\}\; \{T\}\; \backslash \; cdot\; (dH\; -\; V\; \backslash \; cdot\; dp)\; =\; C\_p\; \backslash \; cdot\; \backslash \; frac\; \{dT\}\; \{T\}\; -\; R\; \backslash \; cdot\; \backslash \; frac\; \{dp\}\; \{p\}$

it comes immediately:

$S\; (p,\; T)\; =\; molar\; S\_\; \{\backslash \; rm\}\; =\; C\_p\; \backslash \; cdot\; \backslash \; ln\; T\; -\; R\; \backslash \; cdot\; \backslash \; ln\; p\; +\; \{\backslash \; rm\; constant\}$.

Example: in the experiment of Joule preceding, irreversible and adiabat the entropy increased:

$\backslash \; Delta\; S\; =\; R\; \backslash \; cdot\; \backslash \; ln\; \backslash \; left\; (1\; +\; \backslash \; frac\; \{V\_2\}\; \{V\_1\}\; \backslash \; right)$.

Only statistical mechanics (quantum) can give the value of the constant known as of Sackur-Tetrode (cf Third principle of thermodynamics).

For a monoatomic perfect gas, one prefers to retain the value of

$G\; =\; U\; +\; p\; \backslash \; cdot\; V\; -\; T\; \backslash \; cdot\; S$

called free Enthalpy:

$G\; (V,\; T,\; NR)\; =\; -\; NR\; \backslash \; cdot\; K\; \backslash \; cdot\; T\; \backslash \; ln\; \backslash \; frac\; \{p\}\; \{p^0\}$

with

$p^0\; (T)\; =\; \backslash \; frac\; \{kT\}\; \{\backslash \; lambda^3\}$

where

• λ is the Wavelength of Broglie
  $\backslash \; lambda\; =\; \backslash \; frac\; \{H\}\; \{p\}$
• H is the Constante of Planck
• p is the Quantité of movement
  $p\; =\; \backslash \; sqrt\; \{2\; \backslash \; pi\; \backslash \; cdot\; m\; \backslash \; cdot\; K\; \backslash \; cdot\; T\}$

One then finds all the values given in the tables (for example for argon, neon…). Calculations are hardly more complicated for diatomic gases.

This value is valid only if the interatomic distance is much higher than the wavelength of Broglie

$d\; =\; \backslash \; sqrt\; \{V/N\}\; >>\; \backslash \; lambda$

if not, one falls into the field from the Gaz from cold atoms (the study of the atoms ultra-colds is besides the object of the Nobel Prize of physics of Claude Cohen-Tannoudji).

Perfect gas and kinetic theory of gases

The perfect gases are the subject of a theory known as kinetic Théorie of the gas explaining the physical laws which govern them.

Mix Perfect gases

As regards a ideal mixture of perfect gases , one has the law of Dalton :

$p\; =\; \backslash \; sum\_\; \{I\}\; p\_\; \{I\}$

where $p\_\; \{I\}$ indicates the partial Pression gas I , i.e. the pressure which the gas I would have if it occupied only all volume.

That is to say a perfect gas of molecules has occupying a volume V ₁, and a perfect gas of molecules B of volume V ₂, separated by a partition they are in balance (even temperature T and even pressure P ₀). It is not at all obvious that for the same pressure and the same temperature, the mixture obtained by removing the partition is a of the same system pressure, of the same temperature and of volume V ₁+ V ₂. As comparison, if one mixes 1 water L and 1 alcohol L, one obtains 1,84 alcoholic water L; admittedly they are not perfect gases, but that shows that the properties are not always additive.

Let us consider that the gases are chemically neutral, i.e. the molecules of has and of B do not interact: according to the preceding paragraph,

• has then undergoes a relaxation of Joule of V ₁ with V ₁+ V ₂, since he “does not see” not B (not of interaction);
• symmetrically, B undergoes a relaxation of Joule of V ₂ with V ₁+ V ₂.

Thus the temperature did not change and the pressure on the walls is due to has and B , that is to say

$p\_A\; =\; p\_0\; \backslash \; left\; (\backslash \; frac\; \{V\_1\}\; \{V\_1+V\_2\}\; \backslash \; right)$ for the contribution of has , and

$p\_B\; =\; p\_0\; \backslash \; left\; (\backslash \; frac\; \{V\_2\}\; \{V\_1+V\_2\}\; \backslash \; right)$ for the contribution of B .

The total pressure is $p\_A\; +\; p\_B\; =\; p\_0$: the pressure did not change, the law of Mariotte thus remains true. On the other hand, there was “loss of information per mixing” (Paradoxe of Gibbs)).

If there is not interaction between has and B , internal energy is simply that of has more that of B : the law of Joule remains true. Thus the mixture behaves like a perfect gas.

The mixture of two perfect gases of Laplace is a perfect gas of Laplace, but whose factor γ is not the weighted average of the γ: they are $C\_V$ and $C\_p$ which is the weighted averages.

The entropy of a mixture is the sum of the entropies of each gas taken under its partial pressure (cf Paradoxe of Gibbs). This point is capital: it will be the key in chemistry of the Loi of action of masses.

See too

Point triples

Internal bonds

• Isobar Equation of Clausius

• Isochoric
• Isothermal
• real Gas
• Gas of Van der Waals
• Adiabatic
• Compression and adiabatic pressure drop
• Loi of Joule
• Loi of Joule-Thomson
• Expérience of Clement-Desormes
• Expérience of isothermal Rüchardt
• Atmosphère
• Atmosphère dries adiabatic

External bonds

• perfect gases and real, University of Strasbourg

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