The ideal gas law, also called the general gas equation, is the equation of state of a hypothetical ideal gas. It is a good approximation of the behavior of many gases under many conditions, although it has several limitations. It was first stated by Benoît Paul Émile Clapeyron in 1834 as a combination of the empirical Boyle's law, Charles's law, Avogadro's law, and Gay-Lussac's law.[1] The ideal gas law is often written in an empirical form:
where
p
V
T
n
R
The state of an amount of gas 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 appropriate SI unit is the kelvin.[4]
The most frequently introduced forms are:where:
p
V
n
R
kB
NA
T
N
In SI units, p is measured in pascals, V is measured in cubic metres, n is measured in moles, and T in kelvins (the Kelvin scale is a shifted Celsius scale, where 0.00 K = −273.15 °C, the lowest possible temperature). R has for value 8.314 J/(mol·K) = 1.989 ≈ 2 cal/(mol·K), or 0.0821 L⋅atm/(mol⋅K).
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 the molar mass, M (in kilograms per mole):
n=
m | |
M |
.
pV=
m | |
M |
RT
p=
m | |
V |
RT | |
M |
p=\rho
R | |
M |
T
p=\rhoRspecificT
pv=RspecificT.
It is common, especially in engineering and meteorological applications, to represent the specific gas constant by the symbol R. In such cases, the universal gas constant is usually given a different symbol such as
\barR
R*
In statistical mechanics, the following molecular equation is derived from first principles
P=nkBT,
kB=
R | |
NA |
From this we notice that for a gas of mass, with an average particle mass of times the atomic mass constant,, (i.e., the mass is Da) the number of molecules will be given by
N=
m | |
\mumu |
,
P=
1 | |
V |
m | |
\mumu |
kBT=
kB | |
\mumu |
\rhoT.
In SI units, is measured in pascals, in cubic metres, in kelvins, and in SI units.
Combining the laws of Charles, Boyle and Gay-Lussac gives the combined gas law, which takes the same functional form as the ideal gas law says that the number of moles is unspecified, and the ratio of
PV
T
PV | |
T |
=k,
P
V
T
k
P1V1 | |
T1 |
=
P2V2 | |
T2 |
.
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, its potential energy is zero. Hence, all the energy possessed by the gas is the kinetic energy of the molecules, or atoms, of the gas.
E= | 3 |
2 |
nRT
This corresponds to the kinetic energy of n moles of a monoatomic gas having 3 degrees of freedom; x, y, z. The table here below gives this relationship for different amounts of a monoatomic gas.
Energy of a monoatomic gas | Mathematical expression | ||||
---|---|---|---|---|---|
Energy associated with one mole |
RT | ||||
Energy associated with one gram |
rT | ||||
Energy associated with one atom |
kBT |
The table below essentially simplifies the ideal gas equation for a particular processes, thus making this equation easier to solve using numerical methods.
A thermodynamic process is defined as a system that moves from state 1 to state 2, where the state number is denoted by 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, or H) is constant throughout the process.
For a given thermodynamics 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, or T) at state 2 can be calculated from the properties at state 1 using the equations listed.
Process | Constant | Known ratio or delta | p2 | V2 | T2 |
---|---|---|---|---|---|
Isobaric process | Pressure | V2/V1 | p2 = p1 | V2 = V1(V2/V1) | T2 = T1(V2/V1) |
T2/T1 | p2 = p1 | V2 = V1(T2/T1) | T2 = T1(T2/T1) | ||
Isochoric process (Isovolumetric process) (Isometric process) | Volume | p2/p1 | p2 = p1(p2/p1) | V2 = V1 | T2 = T1(p2/p1) |
T2/T1 | p2 = p1(T2/T1) | V2 = V1 | T2 = T1(T2/T1) | ||
Isothermal process | Temperature | p2/p1 | p2 = p1(p2/p1) | V2 = V1(p1/p2) | T2 = T1 |
V2/V1 | p2 = p1(V1/V2) | V2 = V1(V2/V1) | T2 = T1 | ||
Isentropic process (Reversible adiabatic process) | p2/p1 | p2 = p1(p2/p1) | V2 = V1(p2/p1)(−1/γ) | T2 = T1(p2/p1)(γ − 1)/γ | |
V2/V1 | p2 = p1(V2/V1)−γ | V2 = V1(V2/V1) | T2 = T1(V2/V1)(1 − γ) | ||
T2/T1 | p2 = p1(T2/T1)γ/(γ − 1) | V2 = V1(T2/T1)1/(1 − γ) | T2 = T1(T2/T1) | ||
Polytropic process | P Vn | p2/p1 | p2 = p1(p2/p1) | V2 = V1(p2/p1)(−1/n) | T2 = T1(p2/p1)(n − 1)/n |
V2/V1 | p2 = p1(V2/V1)−n | V2 = V1(V2/V1) | T2 = T1(V2/V1)(1 − n) | ||
T2/T1 | p2 = p1(T2/T1)n/(n − 1) | V2 = V1(T2/T1)1/(1 − n) | T2 = T1(T2/T1) | ||
Isenthalpic process (Irreversible adiabatic process) | p2 − p1 | p2 = p1 + (p2 − p1) | T2 = T1 + μJT(p2 − p1) | ||
T2 − T1 | p2 = p1 + (T2 − T1)/μJT | T2 = T1 + (T2 − T1) |
a. In an isentropic process, system entropy (S) is constant. Under these conditions, p1V1γ = p2V2γ, where γ is defined as the heat capacity ratio, which is constant for a calorifically perfect gas. The value used for γ is typically 1.4 for diatomic gases like nitrogen (N2) and oxygen (O2), (and air, which is 99% diatomic). Also γ is typically 1.6 for mono atomic gases like the noble gases helium (He), and argon (Ar). In internal combustion engines γ varies between 1.35 and 1.15, depending on constitution gases and temperature.
b. In an isenthalpic process, system enthalpy (H) is constant. In the case of free expansion for an ideal gas, there are no molecular interactions, and the temperature remains constant. For real gasses, the molecules do interact via attraction or repulsion depending on temperature and pressure, and heating or cooling does occur. This is known as the Joule–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 both molecular size and intermolecular attractions, it is most accurate for monatomic gases at high temperatures and low pressures. The neglect of 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 increasing thermal kinetic energy, i.e., with increasing temperatures. More detailed equations of state, such as the van 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:
where P stands for pressure, V for volume, N for number of particles in the gas and T for temperature; where
C1,C2,C3,C4,C5,C6
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 keeping N and T 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 the gas 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
P1,V1,N1,T1
Say, starting to change only pressure and volume, according to Boyle's law, then:After this process, the gas has parameters
P2,V2,N1,T1
Using then equation to change the number of particles in the gas and the temperature,After this process, the gas has parameters
P2,V2,N2,T2
Using then equation to change the pressure and the number of particles,After this process, the gas has parameters
P3,V2,N3,T2
Using then Charles's law (equation 2) to change the volume and temperature of the gas,After this process, the gas has parameters
P3,V3,N3,T3
Using simple algebra on equations,, and yields the result: or where
kB
Another equivalent result, using the fact that
nR=NkB
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, and 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, and you would be able to get all six equations because combining and will yield, then and will yield, then and will yield, as well as would the combination of and as is explained in the following visual relation:
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:
then only volume and temperature:
then as we can choose any value for
V3
V1=V3
combining equations and yields
P1 | |
T1 |
=
P2 | |
T2 |
See main article: Kinetic theory of gases.
The ideal gas law can also be derived from first principles using the kinetic 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
PV=
1 | |
3 |
2. | |
Nmv | |
rms |
Consider a container in the
xyz
x
y
z
v
P0
S
x
t
vtS
x
NvtS/V
vtS
x
N'
S
t
NvtS/(6V)
When a molecule bounces off the wall of the container, it changes its momentum
p1
p2=-p1
|p2-p1|=2mv
S
t
|\Deltap|=2mvN'=NtSmv2/(3V)
F=|\Deltap|/t
P0=F/S
P | Nm | ||||
|
v2 | |
V |
.
v
P0
P
v2
2. | |
v | |
rms |
P= | 1 | Nm |
3 |
| |||||||
V |
Using the Maxwell–Boltzmann distribution, the fraction of molecules that have a speed in the range
v
v+dv
f(v)dv
f(v)=4\pi\left(
m | |
2\pikT |
| ||||
\right) |
v2
| ||||
e |
k
2 | |
v | |
rms |
=
infty | |
\int | |
0 |
v2f(v)dv=4\pi\left(
m | |
2\pikT |
| ||||
\right) |
infty | |
\int | |
0 |
v4
| ||||
e |
dv.
Using the integration formula
infty | |
\int | |
0 |
x2n
| ||||
e |
dx=\sqrt{\pi}
(2n)! | \left( | |
n! |
a | |
2 |
\right)2n+1, n\inN,a\inR+,
2 | |
v | |
rms |
=4\pi\left(
m | |
2\pikT |
| ||||
\right) |
\sqrt{\pi}
4! | \left( | |
2! |
| ||||
PV=
1 | Nm\left( | |
3 |
3kT | |
m |
\right)=NkT.
See main article: Statistical mechanics.
Let q = (qx, qy, qz) and p = (px, py, pz) denote the position vector and momentum vector of a particle of an ideal gas, respectively. Let F denote the net force on that particle. Then (two times) the time-averaged kinetic energy of the particle is:
\begin{align} \langleq ⋅ F\rangle &=\left\langleqx
dpx | |
dt |
\right\rangle+ \left\langleqy
dpy | |
dt |
\right\rangle+ \left\langleqz
dpz | |
dt |
\right\rangle\\ &=-\left\langleqx
\partialH | |
\partialqx |
\right\rangle- \left\langleqy
\partialH | |
\partialqy |
\right\rangle- \left\langleqz
\partialH | |
\partialqz |
\right\rangle=-3kBT, \end{align}
3NkBT=-\left\langle
N | |
\sum | |
k=1 |
qk ⋅ Fk\right\rangle.
By Newton'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 pressure P of the gas. Hence
N | |
-\left\langle\sum | |
k=1 |
qk ⋅ Fk\right\rangle=P\ointsurfaceq ⋅ dS,
\nabla ⋅ q=
\partialqx | + | |
\partialqx |
\partialqy | + | |
\partialqy |
\partialqz | |
\partialqz |
=3,
P\ointsurfaceq ⋅ dS =P\intvolume\left(\nabla ⋅ q\right)dV =3PV,
Putting these equalities together yields
3NkBT=-\left\langle
N | |
\sum | |
k=1 |
qk ⋅ Fk\right\rangle=3PV,
PV=NkBT=nRT,
For a d-dimensional system, the ideal gas pressure is:[8]
P(d)=
NkBT | |
Ld |
,
Ld