What is meant by ideal gas equation?

The temperature, T

The temperature has to be in kelvin. Don't forget to add 273 if you are given a temperature in degrees Celsius.

Using the ideal gas equation

Calculations using the ideal gas equation are included in my calculations book (see the link at the very bottom of the page), and I can't repeat them here. There are, however, a couple of calculations that I haven't done in the book which give a reasonable idea of how the ideal gas equation works.

The molar volume at stp

If you have done simple calculations from equations, you have probably used the molar volume of a gas.

1 mole of any gas occupies 22.4 dm3 at stp (standard temperature and pressure, taken as 0°C and 1 atmosphere pressure). You may also have used a value of 24.0 dm3 at room temperature and pressure (taken as about 20°C and 1 atmosphere).

These figures are actually only true for an ideal gas, and we'll have a look at where they come from.

We can use the ideal gas equation to calculate the volume of 1 mole of an ideal gas at 0°C and 1 atmosphere pressure.

First, we have to get the units right.

0°C is 273 K. T = 273 K

1 atmosphere = 101325 Pa. p = 101325 Pa

We know that n = 1, because we are trying to calculate the volume of 1 mole of gas.

And, finally, R = 8.31441 J K-1 mol-1.

Slotting all of this into the ideal gas equation and then rearranging it gives:

And finally, because we are interested in the volume in cubic decimetres, you have to remember to multiply this by 1000 to convert from cubic metres into cubic decimetres.

The molar volume of an ideal gas is therefore 22.4 dm3 at stp.

And, of course, you could redo this calculation to find the volume of 1 mole of an ideal gas at room temperature and pressure - or any other temperature and pressure.

Finding the relative formula mass of a gas from its density

This is about as tricky as it gets using the ideal gas equation.

The density of ethane is 1.264 g dm-3 at 20°C and 1 atmosphere. Calculate the relative formula mass of ethane.

The density value means that 1 dm3 of ethane weighs 1.264 g.

Again, before we do anything else, get the awkward units sorted out.

A pressure of 1 atmosphere is 101325 Pa.

The volume of 1 dm3 has to be converted to cubic metres, by dividing by 1000. We have a volume of 0.001 m3.

The temperature is 293 K.

Now put all the numbers into the form of the ideal gas equation which lets you work with masses, and rearrange it to work out the mass of 1 mole.

The mass of 1 mole of anything is simply the relative formula mass in grams.

So the relative formula mass of ethane is 30.4, to 3 sig figs.

Now, if you add up the relative formula mass of ethane, C2H6 using accurate values of relative atomic masses, you get an answer of 30.07 to 4 significant figures. Which is different from our answer - so what's wrong?

There are two possibilities.

• The density value I have used may not be correct. I did the sum again using a slightly different value quoted at a different temperature from another source. This time I got an answer of 30.3. So the density values may not be entirely accurate, but they are both giving much the same sort of answer.

• Ethane isn't an ideal gas. Well, of course it isn't an ideal gas - there's no such thing! However, assuming that the density values are close to correct, the error is within 1% of what you would expect. So although ethane isn't exactly behaving like an ideal gas, it isn't far off.

If you need to know about real gases, now is a good time to read about them.

© Jim Clark 2010 (last modified July 2017)

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. 

The terms are:

p = pressure, in pascals (Pa).

V = volume, in m3.

n = number of moles.

R = the gas constant, 8.31 J K-1 mol-1 (you will be given this value).

T = temperature, in kelvin (K).

In an exam question, you will be normally be given 4 of the terms and asked to work out the 5th. The equation can be rearranged to work out each of the different terms. For example, to calculate the number of moles, n:

pV = nRT is rearranged to n = RT/pV.

The hardest part of the question is often using the correct SI units, as given above. Often you will have to convert a term from the incorrect to the correct units before using it in the equation. 

Pressure may be given in atmospheres or kPa, for which the conversions are:

1 atm = 101,325 Pa. 

100 kPa = 100,000 Pa,

Temperature may be given in degrees celsius. To convert this to kelvin, you simply add 273. 

Volume may be given in cm3 or dm3, for which the conversion to m3 is:

1 m3 = 1000 dm3 = 1,000,000 cm3

Once you have the terms in the correct SI units, you can simply plug them into your rearranged equation.

Equation of the state of a hypothetical ideal gas

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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:

p V = n R T {\displaystyle pV=nRT}

p {\displaystyle p}

n {\displaystyle n}

R {\displaystyle R}

Equation

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]

Common forms

The most frequently introduced forms are:

p V = n R T = n k B N A T = N k B T {\displaystyle pV=nRT=nk_{\text{B}}N_{\text{A}}T=Nk_{\text{B}}T}

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

Molar form

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

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

p V = m M R T {\displaystyle pV={\frac {m}{M}}RT}

p = m V R T M {\displaystyle p={\frac {m}{V}}{\frac {RT}{M}}}

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

Defining the specific gas constant Rspecific(r) as the ratio R/M,

p = ρ R specific T {\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

p v = R specific T . {\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 R ¯ {\displaystyle {\bar {R}}}

R ∗ {\displaystyle R^{*}}

Statistical mechanics

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

P = n k B T , {\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 kB is the Boltzmann constant relating temperature and energy, given by:

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

where NA 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, mu, (i.e., the mass is μ u) the number of molecules will be given by

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

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

P = 1 V m μ m u k B T = k B μ m u ρ T . {\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 kB = 1.38×10−23 J⋅K−1 in SI units.

Combined gas law

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 P V {\displaystyle PV}

P V T = k , {\displaystyle {\frac {PV}{T}}=k,}

where P {\displaystyle P}

k {\displaystyle k}

P 1 V 1 T 1 = P 2 V 2 T 2 . {\displaystyle {\frac {P_{1}V_{1}}{T_{1}}}={\frac {P_{2}V_{2}}{T_{2}}}.}

Energy associated with a gas

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 n R T {\displaystyle E={\frac {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	E = 3 2 R T {\displaystyle E={\frac {3}{2}}RT}
Energy associated with one gram	E = 3 2 r T {\displaystyle E={\frac {3}{2}}rT}
Energy associated with one atom	E = 3 2 k B T {\displaystyle E={\frac {3}{2}}k_{B}T}

Applications to thermodynamic processes

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.

^ 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]

Deviations from ideal behavior of real gases

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.

Derivations

Empirical

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)

  P V = C 1 or P 1 V 1 = P 2 V 2 {\displaystyle PV=C_{1}\quad {\text{or}}\quad P_{1}V_{1}=P_{2}V_{2}}

  

• Charles's law (Equation 2)

  V T = C 2 or V 1 T 1 = V 2 T 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)

  V N = C 3 or V 1 N 1 = V 2 N 2 {\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)

  P T = C 4 or P 1 T 1 = P 2 T 2 {\displaystyle {\frac {P}{T}}=C_{4}\quad {\text{or}}\quad {\frac {P_{1}}{T_{1}}}={\frac {P_{2}}{T_{2}}}}

  

• Equation 5

  N T = C 5 or N 1 T 1 = N 2 T 2 {\displaystyle NT=C_{5}\quad {\text{or}}\quad N_{1}T_{1}=N_{2}T_{2}}

  

• Equation 6

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

  

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

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 P 1 , V 1 , N 1 , T 1 {\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:

P 1 V 1 = P 2 V 2 {\displaystyle P_{1}V_{1}=P_{2}V_{2}}

(7)

After this process, the gas has parameters P 2 , V 2 , N 1 , T 1 {\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,

N 1 T 1 = N 2 T 2 {\displaystyle N_{1}T_{1}=N_{2}T_{2}}

(8)

After this process, the gas has parameters P 2 , V 2 , N 2 , T 2 {\displaystyle P_{2},V_{2},N_{2},T_{2}}

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

P 2 N 2 = P 3 N 3 {\displaystyle {\frac {P_{2}}{N_{2}}}={\frac {P_{3}}{N_{3}}}}

(9)

After this process, the gas has parameters P 3 , V 2 , N 3 , T 2 {\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,

V 2 T 2 = V 3 T 3 {\displaystyle {\frac {V_{2}}{T_{2}}}={\frac {V_{3}}{T_{3}}}}

(10)

After this process, the gas has parameters P 3 , V 3 , N 3 , T 3 {\displaystyle P_{3},V_{3},N_{3},T_{3}}

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

P 1 V 1 N 1 T 1 = P 3 V 3 N 3 T 3 {\displaystyle {\frac {P_{1}V_{1}}{N_{1}T_{1}}}={\frac {P_{3}V_{3}}{N_{3}T_{3}}}}

P V N T = k B , {\displaystyle {\frac {PV}{NT}}=k_{\text{B}},}

Another equivalent result, using the fact that n R = N k B {\displaystyle nR=Nk_{\text{B}}}

P V = n R T , {\displaystyle PV=nRT,}

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:

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:

V 2 T 1 = V 3 T 2 {\displaystyle {\frac {V_{2}}{T_{1}}}={\frac {V_{3}}{T_{2}}}}

(2')

then as we can choose any value for V 3 {\displaystyle V_{3}}

V 1 = V 3 {\displaystyle V_{1}=V_{3}}

V 2 T 1 = V 1 T 2 {\displaystyle {\frac {V_{2}}{T_{1}}}={\frac {V_{1}}{T_{2}}}}

(3')

combining equations (1') and (3') yields P 1 T 1 = P 2 T 2 {\displaystyle {\frac {P_{1}}{T_{1}}}={\frac {P_{2}}{T_{2}}}}

Theoretical

Kinetic theory

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.

The fundamental assumptions of the kinetic theory of gases imply that

P V = 1 3 N m v rms 2 . {\displaystyle PV={\frac {1}{3}}Nmv_{\text{rms}}^{2}.}

Using the Maxwell–Boltzmann distribution, the fraction of molecules that have a speed in the range v {\displaystyle v}

v + d v {\displaystyle v+dv}

f ( v ) d v {\displaystyle f(v)\,dv}

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

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

v rms 2 = ∫ 0 ∞ v 2 f ( v ) d v = 4 π ( m 2 π k T ) 3 2 ∫ 0 ∞ v 4 e − m v 2 2 k T d v . {\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

∫ 0 ∞ x 2 n e − x 2 a 2 d x = π ( 2 n ) ! n ! ( a 2 ) 2 n + 1 , {\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

v rms 2 = 4 π ( m 2 π k T ) 3 2 π 4 ! 2 ! ( 2 k T m 2 ) 5 = 3 k T m , {\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:

P V = 1 3 N m ( 3 k T m ) = N k T . {\displaystyle PV={\frac {1}{3}}Nm\left({\frac {3kT}{m}}\right)=NkT.}

Statistical mechanics

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 the time-averaged kinetic energy of the particle is:

⟨ q ⋅ F ⟩ = ⟨ q x d p x d t ⟩ + ⟨ q y d p y d t ⟩ + ⟨ q z d p z d t ⟩ = − ⟨ q x ∂ H ∂ q x ⟩ − ⟨ q y ∂ H ∂ q y ⟩ − ⟨ q z ∂ H ∂ q z ⟩ = − 3 k B T , {\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

3 N k B T = − ⟨ ∑ k = 1 N q k ⋅ F k ⟩ . {\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

− ⟨ ∑ k = 1 N q k ⋅ F k ⟩ = P ∮ surface q ⋅ d S , {\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

∇ ⋅ q = ∂ q x ∂ q x + ∂ q y ∂ q y + ∂ q z ∂ q z = 3 , {\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

P ∮ surface q ⋅ d S = P ∫ volume ( ∇ ⋅ q ) d V = 3 P V , {\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

3 N k B T = − ⟨ ∑ k = 1 N q k ⋅ F k ⟩ = 3 P V , {\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:

P V = N k B T = n R T , {\displaystyle PV=Nk_{B}T=nRT,}

where n = N/NA is the number of moles of gas and R = NAkB is the gas constant.

Other dimensions

For a d-dimensional system, the ideal gas pressure is:[8]

P ( d ) = N k B T L d , {\displaystyle P^{(d)}={\frac {Nk_{B}T}{L^{d}}},}

where L d {\displaystyle L^{d}}

See also

• 
  Physics portal

• Boltzmann constant – Physical constant relating particle kinetic energy with temperature
• Configuration integral – Function in thermodynamics and statistical physics
• Dynamic pressure – Kinetic energy per unit volume of a fluid
• Gas laws
• Internal energy – Energy contained within a system
• Van der Waals equation – Gas equation of state which accounts for non-ideal gas behavior

References

1. ^ Clapeyron, E. (1835). "Mémoire sur la puissance motrice de la chaleur". Journal de l'École Polytechnique (in French). XIV: 153–90. Facsimile at the Bibliothèque nationale de France (pp. 153–90).
2. ^ Krönig, A. (1856). "Grundzüge einer Theorie der Gase". Annalen der Physik und Chemie (in German). 99 (10): 315–22. Bibcode:1856AnP...175..315K. doi:10.1002/andp.18561751008. Facsimile at the Bibliothèque nationale de France (pp. 315–22).
3. ^ Clausius, R. (1857). "Ueber die Art der Bewegung, welche wir Wärme nennen". Annalen der Physik und Chemie (in German). 176 (3): 353–79. Bibcode:1857AnP...176..353C. doi:10.1002/andp.18571760302. Facsimile at the Bibliothèque nationale de France (pp. 353–79).
4. ^ "Equation of State". Archived from the original on 2014-08-23. Retrieved 2010-08-29.
5. ^ Moran; Shapiro (2000). Fundamentals of Engineering Thermodynamics (4th ed.). Wiley. ISBN 0-471-31713-6.
6. ^ Raymond, Kenneth W. (2010). General, organic, and biological chemistry : an integrated approach (3rd ed.). John Wiley & Sons. p. 186. ISBN 9780470504765. Retrieved 29 January 2019.
7. ^ J. R. Roebuck (1926). "The Joule-Thomson Effect in Air". Proceedings of the National Academy of Sciences of the United States of America. 12 (1): 55–58. Bibcode:1926PNAS...12...55R. doi:10.1073/pnas.12.1.55. PMC 1084398. PMID 16576959.
8. ^ Khotimah, Siti Nurul; Viridi, Sparisoma (2011-06-07). "Partition function of 1-, 2-, and 3-D monatomic ideal gas: A simple and comprehensive review". Jurnal Pengajaran Fisika Sekolah Menengah. 2 (2): 15–18. arXiv:1106.1273. Bibcode:2011arXiv1106.1273N.

Further reading

• Davis; Masten (2002). Principles of Environmental Engineering and Science. New York: McGraw-Hill. ISBN 0-07-235053-9.

External links

• "Website giving credit to Benoît Paul Émile Clapeyron, (1799–1864) in 1834". Archived from the original on July 5, 2007.
• Configuration integral (statistical mechanics) where an alternative statistical mechanics derivation of the ideal-gas law, using the relationship between the Helmholtz free energy and the partition function, but without using the equipartition theorem, is provided. Vu-Quoc, L., Configuration integral (statistical mechanics), 2008. this wiki site is down; see this article in the web archive on 2012 April 28.
• Gas equations in detail

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