Ideal gas law

The most frequently introduced forms are:

$${\displaystyle pV=nRT=nk_{\text{B}}N_{\text{A}}T=Nk_{\text{B}}T}$$

where:

• ${\displaystyle p}$ is the absolute pressure of the gas,
• ${\displaystyle V}$ is the volume of the gas,
• ${\displaystyle n}$ is the amount of substance of gas (also known as number of moles),
• ${\displaystyle R}$ is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant,
• ${\displaystyle k_{\text{B}}}$ is the Boltzmann constant,
• ${\displaystyle N_{A}}$ is the Avogadro constant,
• ${\displaystyle T}$ is the absolute temperature of the gas,
• ${\displaystyle N}$ is the number of particles (usually atoms or molecules) of the gas.

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):

${\displaystyle n={\frac {m}{M}}.}$

By replacing n with m/M and subsequently introducing density ρ = m/V, we get:

${\displaystyle pV={\frac {m}{M}}RT}$

${\displaystyle p={\frac {m}{V}}{\frac {RT}{M}}}$

${\displaystyle p=\rho {\frac {R}{M}}T}$

Defining the specific gas constant R_specific as the ratio R/M,

${\displaystyle p=\rho R_{\text{specific}}T}$

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 the specific volume v, the reciprocal of density, as

${\displaystyle pv=R_{\text{specific}}T.}$

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 ${\displaystyle {\bar {R}}}$ or ${\displaystyle R^{*}}$ 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]

In statistical mechanics, the following molecular equation is derived from first principles

${\displaystyle P=nk_{\text{B}}T,}$

where P is the absolute pressure of the gas, n is the number density of the molecules (given by the ratio n = N/V, in contrast to the previous formulation in which n is the number of moles), T is the absolute temperature, and k_B is the Boltzmann constant relating temperature and energy, given by:

${\displaystyle k_{\text{B}}={\frac {R}{N_{\text{A}}}}}$

where N_A is the Avogadro constant.

From this we notice that for a gas of mass m, with an average particle mass of μ times the atomic mass constant, m_u, (i.e., the mass is μ Da) the number of molecules will be given by

${\displaystyle N={\frac {m}{\mu m_{\text{u}}}},}$

and since ρ = m/V = nμm_u, we find that the ideal gas law can be rewritten as

${\displaystyle P={\frac {1}{V}}{\frac {m}{\mu m_{\text{u}}}}k_{\text{B}}T={\frac {k_{\text{B}}}{\mu m_{\text{u}}}}\rho T.}$

In SI units, P is measured in pascals, V in cubic metres, T in kelvins, and k_B = 1.38×10⁻²³ J⋅K⁻¹ 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 ${\displaystyle PV}$ to ${\displaystyle T}$ is simply taken as a constant:[6]

${\displaystyle {\frac {PV}{T}}=k,}$

where ${\displaystyle P}$ is the pressure of the gas, ${\displaystyle V}$ is the volume of the gas, ${\displaystyle T}$ is the absolute temperature of the gas, and ${\displaystyle k}$ is a constant. When comparing the same substance under two different sets of conditions, the law can be written as

${\displaystyle {\frac {P_{1}V_{1}}{T_{1}}}={\frac {P_{2}V_{2}}{T_{2}}}.}$

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:

• Boyle's law (Equation 1)
  $${\displaystyle PV=C_{1}\quad {\text{or}}\quad P_{1}V_{1}=P_{2}V_{2}}$$

  
• Charles's law (Equation 2)
  $${\displaystyle {\frac {V}{T}}=C_{2}\quad {\text{or}}\quad {\frac {V_{1}}{T_{1}}}={\frac {V_{2}}{T_{2}}}}$$

  
• Avogadro's law (Equation 3)
  $${\displaystyle {\frac {V}{N}}=C_{3}\quad {\text{or}}\quad {\frac {V_{1}}{N_{1}}}={\frac {V_{2}}{N_{2}}}}$$

  
• Gay-Lussac's law (Equation 4)
  $${\displaystyle {\frac {P}{T}}=C_{4}\quad {\text{or}}\quad {\frac {P_{1}}{T_{1}}}={\frac {P_{2}}{T_{2}}}}$$

  
• Equation 5
  $${\displaystyle NT=C_{5}\quad {\text{or}}\quad N_{1}T_{1}=N_{2}T_{2}}$$

  
• Equation 6
  $${\displaystyle {\frac {P}{N}}=C_{6}\quad {\text{or}}\quad {\frac {P_{1}}{N_{1}}}={\frac {P_{2}}{N_{2}}}}$$

  

Relationships between Boyle's, Charles's, Gay-Lussac's, Avogadro's, combined and ideal gas laws, with the Boltzmann constant k = R/N_A = n R/N (in each law, properties circled are variable and properties not circled are held constant)

where P stands for pressure, V for volume, N for number of particles in the gas and T for temperature; where ${\displaystyle C_{1},C_{2},C_{3},C_{4},C_{5},C_{6}}$ 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 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 ${\displaystyle P_{1},V_{1},N_{1},T_{1}}$

Say, starting to change only pressure and volume, according to Boyle's law (Equation 1), then:

$${\displaystyle P_{1}V_{1}=P_{2}V_{2}}$$

(7)

After this process, the gas has parameters ${\displaystyle P_{2},V_{2},N_{1},T_{1}}$

Using then equation (5) to change the number of particles in the gas and the temperature,

$${\displaystyle N_{1}T_{1}=N_{2}T_{2}}$$

(8)

After this process, the gas has parameters ${\displaystyle P_{2},V_{2},N_{2},T_{2}}$

Using then equation (6) to change the pressure and the number of particles,

$${\displaystyle {\frac {P_{2}}{N_{2}}}={\frac {P_{3}}{N_{3}}}}$$

(9)

After this process, the gas has parameters ${\displaystyle P_{3},V_{2},N_{3},T_{2}}$

Using then Charles's law (equation 2) to change the volume and temperature of the gas,

$${\displaystyle {\frac {V_{2}}{T_{2}}}={\frac {V_{3}}{T_{3}}}}$$

(10)

After this process, the gas has parameters ${\displaystyle P_{3},V_{3},N_{3},T_{3}}$

Using simple algebra on equations (7), (8), (9) and (10) yields the result:

$${\displaystyle {\frac {P_{1}V_{1}}{N_{1}T_{1}}}={\frac {P_{3}V_{3}}{N_{3}T_{3}}}}$$

or

$${\displaystyle {\frac {PV}{NT}}=k_{\text{B}},}$$

where ${\displaystyle k_{\text{B}}}$ stands for the Boltzmann constant.

Another equivalent result, using the fact that ${\displaystyle nR=Nk_{\text{B}}}$, where n is the number of moles in the gas and R is the universal gas constant, is:

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

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:

$${\displaystyle P_{1}V_{1}=P_{2}V_{2}}$$

(1')

then only volume and temperature:

$${\displaystyle {\frac {V_{2}}{T_{1}}}={\frac {V_{3}}{T_{2}}}}$$

(2')

then as we can choose any value for ${\displaystyle V_{3}}$, if we set ${\displaystyle V_{1}=V_{3}}$, equation (2') becomes:

$${\displaystyle {\frac {V_{2}}{T_{1}}}={\frac {V_{1}}{T_{2}}}}$$

(3')

combining equations (1') and (3') yields ${\displaystyle {\frac {P_{1}}{T_{1}}}={\frac {P_{2}}{T_{2}}}}$, which is equation (4), of which we had no prior knowledge until this derivation.

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

${\displaystyle PV={\frac {1}{3}}Nmv_{\text{rms}}^{2}.}$

Consider a container in the ${\displaystyle xyz}$ Cartesian coordinate system. For simplicity, we assume that a third of the molecules moves parallel to the ${\displaystyle x}$-axis, a third moves parallel to the ${\displaystyle y}$-axis and a third moves parallel to the ${\displaystyle z}$-axis. Next we temporarily assume that all molecules with the same velocity ${\displaystyle v}$. We choose an area ${\displaystyle S}$ on a wall of the container, perpendicular to the ${\displaystyle x}$-axis. When time ${\displaystyle t}$ elapses, all molecules in the volume ${\displaystyle vtS}$ moving in the positive direction of the ${\displaystyle x}$-axis will hit the area. There are ${\displaystyle NvtS/V}$ molecules in a part of volume ${\displaystyle vtS}$ of the container, but only one sixth (i.e. a half of a third) of them moves in the positive direction of the ${\displaystyle x}$-axis. Therefore the number of molecules ${\displaystyle N'}$ that will hit the area ${\displaystyle S}$ when the time ${\displaystyle t}$ elapses is ${\displaystyle NvtS/(6V)}$.

When a molecule bounces off the wall of the container, it changes its momentum ${\displaystyle \mathbf {p} _{1}}$ to ${\displaystyle \mathbf {p} _{2}=-\mathbf {p} _{1}}$. Hence the magnitude of change of the momentum of one molecule is ${\displaystyle |\mathbf {p} _{2}-\mathbf {p} _{1}|=2mv}$. The magnitude of the change of momentum of all molecules that bounce off the area ${\displaystyle S}$ when time ${\displaystyle t}$ elapses is then ${\displaystyle |\Delta \mathbf {p} |=2mvN'=NtSmv^{2}/(3V)}$. From ${\displaystyle F=|\Delta \mathbf {p} |/t}$ and ${\displaystyle P=F/S}$ we get

${\displaystyle P={\frac {1}{3}}Nm{\frac {v^{2}}{V}}.}$

We supposed that all molecules move with the same velocity ${\displaystyle v}$, but in fact they move with different velocities, so we replace ${\displaystyle v^{2}}$ in the equation by the arithmetic mean of all squares of all velocities of the molecules, i.e. by ${\displaystyle v_{\text{rms}}^{2}.}$ Therefore

${\displaystyle P={\frac {1}{3}}Nm{\frac {v_{\text{rms}}^{2}}{V}}}$

which gives the desired formula.

Using the Maxwell–Boltzmann distribution, the fraction of molecules that have a speed in the range ${\displaystyle v}$ to ${\displaystyle v+dv}$ is ${\displaystyle f(v)\,dv}$, where

${\displaystyle f(v)=4\pi \left({\frac {m}{2\pi kT}}\right)^{\!{\frac {3}{2}}}v^{2}e^{-{\frac {mv^{2}}{2kT}}}}$

and ${\displaystyle k}$ denotes the Boltzmann constant. The root-mean-square speed can be calculated by

${\displaystyle v_{\text{rms}}^{2}=\int _{0}^{\infty }v^{2}f(v)\,dv=4\pi \left({\frac {m}{2\pi kT}}\right)^{\frac {3}{2}}\int _{0}^{\infty }v^{4}e^{-{\frac {mv^{2}}{2kT}}}\,dv.}$

Using the integration formula

${\displaystyle \int _{0}^{\infty }x^{2n}e^{-{\frac {x^{2}}{a^{2}}}}\,dx={\sqrt {\pi }}\,{\frac {(2n)!}{n!}}\left({\frac {a}{2}}\right)^{2n+1},}$

it follows that

${\displaystyle v_{\text{rms}}^{2}=4\pi \left({\frac {m}{2\pi kT}}\right)^{\!{\frac {3}{2}}}{\sqrt {\pi }}\,{\frac {4!}{2!}}\left({\frac {\sqrt {\frac {2kT}{m}}}{2}}\right)^{\!5}={\frac {3kT}{m}},}$

from which we get the ideal gas law:

${\displaystyle PV={\frac {1}{3}}Nm\left({\frac {3kT}{m}}\right)=NkT.}$

Let q = (q_x, q_y, q_z) and p = (p_x, p_y, p_z) 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 the time-averaged kinetic energy of the particle is:

${\displaystyle {\begin{aligned}\langle \mathbf {q} \cdot \mathbf {F} \rangle &=\left\langle q_{x}{\frac {dp_{x}}{dt}}\right\rangle +\left\langle q_{y}{\frac {dp_{y}}{dt}}\right\rangle +\left\langle q_{z}{\frac {dp_{z}}{dt}}\right\rangle \\&=-\left\langle q_{x}{\frac {\partial H}{\partial q_{x}}}\right\rangle -\left\langle q_{y}{\frac {\partial H}{\partial q_{y}}}\right\rangle -\left\langle q_{z}{\frac {\partial H}{\partial q_{z}}}\right\rangle =-3k_{\text{B}}T,\end{aligned}}}$

where the first equality is Newton's second law, and the second line uses Hamilton's equations and the equipartition theorem. Summing over a system of N particles yields

${\displaystyle 3Nk_{B}T=-\left\langle \sum _{k=1}^{N}\mathbf {q} _{k}\cdot \mathbf {F} _{k}\right\rangle .}$

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

${\displaystyle -\left\langle \sum _{k=1}^{N}\mathbf {q} _{k}\cdot \mathbf {F} _{k}\right\rangle =P\oint _{\text{surface}}\mathbf {q} \cdot d\mathbf {S} ,}$

where dS is the infinitesimal area element along the walls of the container. Since the divergence of the position vector q is

${\displaystyle \nabla \cdot \mathbf {q} ={\frac {\partial q_{x}}{\partial q_{x}}}+{\frac {\partial q_{y}}{\partial q_{y}}}+{\frac {\partial q_{z}}{\partial q_{z}}}=3,}$

the divergence theorem implies that

${\displaystyle P\oint _{\text{surface}}\mathbf {q} \cdot d\mathbf {S} =P\int _{\text{volume}}\left(\nabla \cdot \mathbf {q} \right)dV=3PV,}$

where dV is an infinitesimal volume within the container and V is the total volume of the container.

Putting these equalities together yields

${\displaystyle 3Nk_{\text{B}}T=-\left\langle \sum _{k=1}^{N}\mathbf {q} _{k}\cdot \mathbf {F} _{k}\right\rangle =3PV,}$

which immediately implies the ideal gas law for N particles:

${\displaystyle PV=Nk_{B}T=nRT,}$

where n = N/N_A is the number of moles of gas and R = N_Ak_B is the gas constant.