Isotherms of anideal gas for different temperatures. The curved lines arerectangular hyperbolae of the form y = a/x. They represent the relationship betweenpressure (on the vertical axis) andvolume (on the horizontal axis) for an ideal gas at differenttemperatures: lines that are farther away from theorigin (that is, lines that are nearer to the top right-hand corner of the diagram) correspond to higher temperatures.
Molecular collisions within a closed container (a propane tank) are shown (right). The arrows represent the random motions and collisions of these molecules. The pressure and temperature of the gas are directly proportional: As temperature increases, the pressure of the propane gas increases by the same factor. A simple consequence of this proportionality is that on a hot summer day, the propane tank pressure will be elevated, and thus propane tanks must be rated to withstand such increases in pressure.
Thestate of an amount ofgas is determined by its pressure, volume, and temperature. The modern form of the equation relates these simply in two main forms. The temperature used in the equation of state is an absolute temperature: the appropriateSI unit is thekelvin.[4]
How much gas is present could be specified by giving the mass instead of the chemical amount of gas. Therefore, an alternative form of the ideal gas law may be useful. The chemical amount,n (in moles), is equal to total mass of the gas (m) (in kilograms) divided by themolar mass,M (in kilograms per mole):
By replacingn withm/M and subsequently introducingdensityρ =m/V, we get:
This form of the ideal gas law is very useful because it links pressure, density, and temperature in a unique formula independent of the quantity of the considered gas. Alternatively, the law may be written in terms of thespecific volumev, the reciprocal of density, as
It is common, especially in engineering and meteorological applications, to represent thespecific gas constant by the symbolR. In such cases, theuniversal gas constant is usually given a different symbol such as or to distinguish it. In any case, the context and/or units of the gas constant should make it clear as to whether the universal or specific gas constant is being used.[5]
Instatistical mechanics, the following molecular equation (i.e. the ideal gas law in its theoretical form) is derived from first principles:
wherep is the absolutepressure of the gas,n is thenumber density of the molecules (given by the ration =N/V, in contrast to the previous formulation in whichn is thenumber of moles),T is theabsolute temperature, andkB is theBoltzmann constant relating temperature and energy, given by:
whereNA is theAvogadro constant.The form can be further simplified by defining the kinetic energy corresponding to the temperature:
so the ideal gas law is more simply expressed as:
From this we notice that for a gas of massm, with an average particle mass ofμ times theatomic mass constant,mu, (i.e., the mass isμDa) the number of molecules will be given by
and sinceρ =m/V =nμmu, we find that the ideal gas law can be rewritten as
In SI units,p is measured inpascals,V in cubic metres,T in kelvins, andkB =1.38×10−23 J⋅K−1 inSI units.
Combining the laws of Charles, Boyle, and Gay-Lussac gives thecombined gas law, which can take the same functional form as the ideal gas law. This form does not specify the number of moles, and the ratio of to is simply taken as a constant:[6]
where is thepressure of the gas, is thevolume of the gas, is theabsolute temperature of the gas, and is a constant. More commonly, when comparing the same substance under two different sets of conditions, the law is written as:
According to the assumptions of the kinetic theory of ideal gases, one can consider that there are no intermolecular attractions between the molecules, or atoms, of an ideal gas. In other words, itspotential energy is zero. Hence, all the energy possessed by the gas is the kinetic energy of the molecules, or atoms, of the gas.
This corresponds to the kinetic energy ofn moles of amonoatomic gas having 3degrees of freedom:x,y,z. The table here below gives this relationship for different amounts of a monoatomic gas.
The table below essentially simplifies the ideal gas equation for a particular process, making the equation easier to solve using numerical methods.
Athermodynamic process is defined as a system that moves from state 1 to state 2, where the state number is denoted by a subscript. As shown in the first column of the table, basic thermodynamic processes are defined such that one of the gas properties (P,V,T,S, orH) is constant throughout the process.
For a given thermodynamic process, in order to specify the extent of a particular process, one of the properties ratios (which are listed under the column labeled "known ratio") must be specified (either directly or indirectly). Also, the property for which the ratio is known must be distinct from the property held constant in the previous column (otherwise the ratio would be unity, and not enough information would be available to simplify the gas law equation).
In the final three columns, the properties (p,V, orT) at state 2 can be calculated from the properties at state 1 using the equations listed.
^a. In an isentropic process, systementropy (S) is constant. Under these conditions,p1V1γ =p2V2γ, whereγ is defined as theheat capacity ratio, which is constant for a calorificallyperfect gas. The value used forγ is typically 1.4 for diatomic gases likenitrogen (N2) andoxygen (O2), (and air, which is 99% diatomic). Alsoγ is typically 1.6 for mono atomic gases like thenoble gaseshelium (He), andargon (Ar). In internal combustion enginesγ varies between 1.35 and 1.15, depending on constitution gases and temperature.
^b. In an isenthalpic process, systementhalpy (H) is constant. In the case offree expansion for an ideal gas, there are no molecular interactions, and the temperature remains constant. For real gases, the molecules do interact via attraction or repulsion depending on temperature and pressure, and heating or cooling does occur. This is known as theJoule–Thomson effect. For reference, the Joule–Thomson coefficient μJT for air at room temperature and sea level is 0.22 °C/bar.[7]
The equation of state given here (PV =nRT) applies only to an ideal gas, or as an approximation to a real gas that behaves sufficiently like an ideal gas. There are in fact many different forms of the equation of state. Since the ideal gas law neglects bothmolecular size and intermolecular attractions, it is most accurate formonatomic gases at high temperatures and low pressures. The molecular size becomes less important for lower densities, i.e. for larger volumes at lower pressures, because the average distance between adjacent molecules becomes much larger than the molecular size. The relative importance of intermolecular attractions diminishes with increasingthermal kinetic energy, i.e., with increasing temperatures. More detailedequations of state, such as thevan der Waals equation, account for deviations from ideality caused by molecular size and intermolecular forces.
The empirical laws that led to the derivation of the ideal gas law were discovered with experiments that changed only 2 state variables of the gas and kept every other one constant.
All the possible gas laws that could have been discovered with this kind of setup are:
whereP stands forpressure,V forvolume,N for number of particles in the gas andT fortemperature; where are constants in this context because of each equation requiring only the parameters explicitly noted in them changing.
To derive the ideal gas law one does not need to know all 6 formulas, one can just know 3 and with those derive the rest or just one more to be able to get the ideal gas law, which needs 4.
Since each formula only holds when only the state variables involved in said formula change while the others (which are a property of the gas but are not explicitly noted in said formula) remain constant, we cannot simply use algebra and directly combine them all. This is why: Boyle did his experiments while keepingN andT constant and this must be taken into account (in this same way, every experiment kept some parameter as constant and this must be taken into account for the derivation).
Keeping this in mind, to carry the derivation on correctly, one must imagine thegas being altered by one process at a time (as it was done in the experiments). The derivation using 4 formulas can look like this:
at first the gas has parameters
Say, starting to change only pressure and volume, according toBoyle's law (Equation 1), then:
7
After this process, the gas has parameters
Using then equation (5) to change the number of particles in the gas and the temperature,
8
After this process, the gas has parameters
Using then equation (6) to change the pressure and the number of particles,
9
After this process, the gas has parameters
Using thenCharles's law (equation 2) to change the volume and temperature of the gas,
10
After this process, the gas has parameters
Using simple algebra on equations (7), (8), (9) and (10) yields the result: or where stands for theBoltzmann constant.
Another equivalent result, using the fact that, wheren is the number ofmoles in the gas andR is theuniversal gas constant, is: which is known as the ideal gas law.
If three of the six equations are known, it may be possible to derive the remaining three using the same method. However, because each formula has two variables, this is possible only for certain groups of three. For example, if you were to have equations (1), (2) and (4) you would not be able to get any more because combining any two of them will only give you the third. However, if you had equations (1), (2) and (3) you would be able to get all six equations because combining (1) and (2) will yield (4), then (1) and (3) will yield (6), then (4) and (6) will yield (5), as well as would the combination of (2) and (3) as is explained in the following visual relation:
Relationship between the six gas laws
where the numbers represent the gas laws numbered above.
If you were to use the same method used above on 2 of the 3 laws on the vertices of one triangle that has a "O" inside it, you would get the third.
For example:
Change only pressure and volume first:
1'
then only volume and temperature:
2'
then as we can choose any value for, if we set, equation (2') becomes:
3'
Combining equations (1') and (3') yields, which is equation (4), of which we had no prior knowledge until this derivation.
The ideal gas law can also be derived fromfirst principles using thekinetic theory of gases, in which several simplifying assumptions are made, chief among which are that the molecules, or atoms, of the gas are point masses, possessing mass but no significant volume, and undergo only elastic collisions with each other and the sides of the container in which both linear momentum and kinetic energy are conserved.
First we show that the fundamental assumptions of the kinetic theory of gases imply that
Consider a container in the Cartesian coordinate system. For simplicity, we assume that a third of the molecules moves parallel to the-axis, a third moves parallel to the-axis and a third moves parallel to the-axis. If all molecules move with the same velocity, denote the corresponding pressure by. We choose an area on a wall of the container, perpendicular to the-axis. When time elapses, all molecules in the volume moving in the positive direction of the-axis will hit the area. There are molecules in a part of volume of the container, but only one sixth (i.e. a half of a third) of them moves in the positive direction of the-axis. Therefore, the number of molecules that will hit the area when the time elapses is.
When a molecule bounces off the wall of the container, it changes its momentum to. Hence the magnitude of change of the momentum of one molecule is. The magnitude of the change of momentum of all molecules that bounce off the area when time elapses is then. From and we get
We considered a situation where all molecules move with the same velocity. Now we consider a situation where they can move with different velocities, so we apply an "averaging transformation" to the above equation, effectively replacing by a new pressure and by the arithmetic mean of all squares of all velocities of the molecules, i.e. by Therefore
Letq = (qx,qy,qz) andp = (px,py,pz) denote the position vector and momentum vector of a particle of an ideal gas, respectively. LetF denote the net force on that particle. Then (two times) the time-averaged kinetic energy of the particle is:
ByNewton's third law and the ideal gas assumption, the net force of the system is the force applied by the walls of the container, and this force is given by the pressureP of the gas. Hence
where dS is the infinitesimal area element along the walls of the container. Since thedivergence of the position vectorq is
An additional derivation is possible using theparticle in a box model ofquantum mechanics.[8] In a rectangular box of dimensions, the possible quantized energy levels are given as
where, and are the quantum numbers for translational motion in the three base directions,, and are the kinetic energies associated with translational motion in these directions. The force () acting upon the wall perpendicular to direction is calculated as the derivative of the particle energy with respect to a change in side length
The overall force is calculated as the sum of the contributions form independent particles as
where is the extent of thed-dimensional domain in which the gas exists. Thequantity dimension of the pressure-like quantity changes with the space dimensionalityd: it corresponds to a force per length (ford=1), force per area (d=2), or force per volume (d=3).
^Khotimah, Siti Nurul; Viridi, Sparisoma (2011-06-07). "Partition function of 1-, 2-, and 3-D monatomic ideal gas: A simple and comprehensive review".arXiv:1106.1273.{{cite arXiv}}: CS1 maint: missing class (link) A bot will complete this citation soon.Click here to jump the queue
