Physics · Kinetic theory of gases

Pressure from kinetic theory

Q: How can I check a calculation involving Pressure from kinetic theory?

For $N=2.00\cdot10^{22}$, $m=6.64\cdot10^{-27}\,\mathrm{kg}$, $c_{\mathrm{rms}}=900\,\mathrm{m\,s^{-1}}$ and $V=0.0060\,\mathrm{m^3}$, $pV=35.856\,\mathrm J$ and $p=5976\,\mathrm{Pa}$. Check using $\rho=\frac{Nm}{V}\approx0.0221333\,\mathrm{kg\,m^{-3}}$ and $p=\frac13\rho c_{\mathrm{rms}}^2$. A second check uses total translational energy $E_{\mathrm{tr}}=\frac12Nmc_{\mathrm{rms}}^2=53.784\,\mathrm J$; it must give $pV=\frac23E_{\mathrm{tr}}$. Confirm that speed is thermal and the distribution isotropic. For a polyatomic gas, do not automatically equate this translational energy with complete internal energy. A pressure gauge reading also needs conversion to absolute pressure before applying the ideal gas law. Also ensure the stated volume is the actual accessible gas volume.

Q: How do I check units and scale for Pressure from kinetic theory?

$Nm\langle c^2\rangle$ has units $\mathrm{kg\,m^2\,s^{-2}}=\mathrm J$, so pV is an energy quantity. Dividing by V in m³ gives $\mathrm{J\,m^{-3}}=\mathrm{Pa}$. In the density form, kg/m³ multiplied by m²/s² also gives Pa. Actually square $c_{\mathrm{rms}}$; multiplying speed directly by mass would give momentum instead of the required quantity. An example with $N=10^{23}$, $m=4.65\cdot10^{-26}\,\mathrm{kg}$ and $c_{\mathrm{rms}}=500\,\mathrm{m\,s^{-1}}$ gives $pV=387.5\,\mathrm J$. At $V=0.010\,\mathrm{m^3}$, $p=38.75\,\mathrm{kPa}$. Doubling volume at unchanged particle count and thermal distribution halves p. Quadrupling $c_{\mathrm{rms}}$ at equal density instead multiplies pressure by sixteen.

Q: When should I combine Pressure from kinetic theory with another relation?

Combine momentum flux with $\frac12m\langle c^2\rangle=\frac32k_BT$ to obtain $pV=Nk_BT=nRT$. This turns the speed distribution into a thermal equation of state. In an ideal mixture at shared T, each species contributes $p_s=\frac{N_sk_BT}{V}$; partial pressures add. For 1.20·10²² and 0.80·10²² particles at 320 K in 0.0040 m³, contributions are 13.254 and 8.836 kPa, totaling 22.090 kPa. Their masses are unnecessary when shared T is known. For internal energy, $U=\frac32pV$ applies only to the appropriate monatomic model; for polyatomic gases it remains the translational sum.

Gas pressure can be understood as momentum transfer from random molecular motion to the container walls.

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Formula

Calculate

p V = \frac{1}{3} N m ⟨ v^{2} ⟩

p in Pa; V in m³; N molecule count; m in kg; ⟨v²⟩ in m²/s².

Variables & units

Variables & units
Symbol	Meaning	Unit
$p$	Absolute thermal gas pressure	Pa
$V$	Gas volume	m³
$N$	Number of molecules of the same species	dimensionless
$m$	Mass per molecule	kg
⟨v²⟩	Mean squared thermal speed relative to local mean gas flow	m²/s²

LaTeX

pV=\frac{1}{3}Nm\langle v^2\rangle

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A particle hits a stationary right wall. Normal component cx reverses sign while tangential component cy stays unchanged. The wall receives positive x momentum 2mcx. — Elastic wall collisions transfer normal momentum. The factor 1/3 arises only in the isotropic statistical average, not from one collision’s geometry. Velocities refer to the local gas rest frame.

Kinetic theory gas pressure calculator

Find the pressure of an ideal gas from its density and the RMS speed of its particles, or the other way round, also from particle number, particle mass and volume.

Calculation model

Find

Gas mass density ρ

Mass per volume of the gas, e.g. 1.2 kg/m³ for air at room temperature.

RMS speed of the particles cᵣ_ms

Root mean square particle speed without wind or flow, e.g. 500 m/s for air.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for p, substitute, evaluate

1. Rearrange for p
$p=\frac{\rho c_{\mathrm{rms}}^2}3$
2. Substitute values (in SI units)
$\frac{1.2\cdot 500\cdot 500}3$
3. Result
p = 100 kPa

Result: Gas pressure p (absolute)

p = 100 kPa

With ρ = 1.2 kg/m³ and cᵣ_ms = 500 m/s, p = 100 kPa.

Quick tips

Estimate for air: density 1.2 kg/m³ and c ≈ 500 m/s give p = 1.2 · 500² ÷ 3 = 100,000 Pa, roughly the air pressure of 1 bar.
Twice the particle speed, four times the pressure: p grows with c². Twice as many particles in the same volume only double the pressure.

Assumptions and limits

Classical nonrelativistic ideal gas with isotropic thermal motion and negligible interaction contributions. Speed is relative to local mean flow. The result is absolute thermal pressure, not gauge pressure, directed momentum flux or general pressure of an anisotropic distribution. Positive gas density and thermal RMS speed.

Derivation & background

Elastic wall reflection transfers twice the normal particle momentum. Mean momentum-flux balance gives $p_x=\frac{Nm\langle c_x^2\rangle}{V}$ ; isotropy supplies the factor 1/3. Upper notation v denotes thermal motion relative to local gas flow, and p is absolute pressure. Classical ideal equilibrium equipartition gives $pV=Nk_BT$ . Directed flow and anisotropic distributions need separate component balances.

Exam blueprint

Validity range

Classical ideal gas with isotropic thermal distribution and negligible interaction contributions. Wall derivation uses elastic reflection at a stationary wall. p is absolute pressure; c=v−u. Anisotropy requires separate normal pressure components.

Derivation steps

Combine momentum transfer per wall collision with collision rate; isotropy justifies the factor 1/3.

1Consider a stationary wall normal to x; L is the distance to its opposite wall.
2An elastic collision reverses normal velocity; the wall receives momentum $2m|c_x|$ .
3In the box model, return time is $\frac{2L}{|c_x|}$ ; mean force contribution is $\frac{mc_x^2}{L}$ . Zero normal velocity gives zero contribution.
4Sum particle contributions and divide by area A: $p_x=\frac{Nm\langle c_x^2\rangle}{AL}$ , with $AL=V$ .
5Isotropy gives $\langle c_x^2\rangle=\frac13\langle c^2\rangle$ , hence $pV=\frac13Nm\langle c^2\rangle$ .
6Using $m\langle c^2\rangle=3k_BT$ gives $pV=Nk_BT=nRT$ . Use velocity relative to mean gas motion; account for directed momentum flux separately.

Rearrangements

Pressure

p=\frac{Nm\langle v^2\rangle}{3V}

The speed denoted v here is thermal, relative to mean gas flow: $\vec c=\vec v_{\mathrm{lab}}-\vec u$ . Mean squared speed is $\langle c^2\rangle=c_{\mathrm{rms}}^2$ , not squared arithmetic mean speed.

Task variant

Calculate the quantity from the worked example for “Pressure from kinetic theory”.

For $N=1.0\cdot10^{23}$ , $m=4.65\cdot10^{-26}\,\mathrm{kg}$ and thermal $c_{\mathrm{rms}}=500\,\mathrm{m/s}$ , $pV=\frac13Nm c_{\mathrm{rms}}^2=387.5\,\mathrm J\approx388\,\mathrm J$ .
At $V=0.010\,\mathrm{m^3}$ , $p=38.75\,\mathrm{kPa}\approx38.8\,\mathrm{kPa}$ .

Common mistakes

Ignoring units or assumptions stated for the formula: p in Pa; V in m³; N molecule count; m in kg; ⟨v²⟩ in m²/s².

Convert units before substitution and verify the validity range; round only at the end.

Exam context

Interpret macroscopic pressure through microscopic particle motion.

Formula cluster

Kinetic theory of gases

Related relations: innere-energie-einatomiges-idealgas, mittlere-kinetische-energie-molekuel, rms-geschwindigkeit-gasmolekuel.

U = 3/2 nRT

Internal energy of a monatomic ideal gas

⟨Ekin⟩ = 3/2 kBT

Mean translational kinetic energy per gas molecule

vrms = √(3kBT/m)

Root-mean-square molecular speed

Worked example

For $N=1.0\cdot10^{23}$ , $m=4.65\cdot10^{-26}\,\mathrm{kg}$ and thermal $c_{\mathrm{rms}}=500\,\mathrm{m/s}$ , $pV=\frac13Nm c_{\mathrm{rms}}^2=387.5\,\mathrm J\approx388\,\mathrm J$ . At $V=0.010\,\mathrm{m^3}$ , $p=38.75\,\mathrm{kPa}\approx38.8\,\mathrm{kPa}$ .

Applications

Interpret macroscopic pressure through microscopic particle motion.

How do you calculate with Pressure from kinetic theory?

Here is how to work through a typical Pressure from kinetic theory (pV = 1/3 Nm⟨v²⟩) task step by step:

1
Task
Calculate the quantity from the worked example for “Pressure from kinetic theory”.
Solution path
1. For $N=1.0\cdot10^{23}$ , $m=4.65\cdot10^{-26}\,\mathrm{kg}$ and thermal $c_{\mathrm{rms}}=500\,\mathrm{m/s}$ , $pV=\frac13Nm c_{\mathrm{rms}}^2=387.5\,\mathrm J\approx388\,\mathrm J$ .
2. At $V=0.010\,\mathrm{m^3}$ , $p=38.75\,\mathrm{kPa}\approx38.8\,\mathrm{kPa}$ .

Understand and apply the equation

Gas pressure arises from momentum transfer to a wall. Squared speed therefore measures random motion, and the factor 1/3 distributes it among three spatial directions in isotropic equilibrium. The microscopic collision balance thus leads to the macroscopic ideal gas law.

Many individual impacts produce measurable pressure

One gas particle touches a wall only briefly. Many particles hit it at different times and transfer momentum. Their time-averaged momentum transfer produces a force; dividing by wall area gives pressure. Kinetic gas theory therefore translates microscopic motion into a macroscopic measurement.[1][3]

Two effects act together: more mass per volume increases momentum flux, and faster random motion increases it quadratically. Mean squared speed counts motion in both directions positively. Mean velocity can instead be zero even when gas pressure is substantial.[1]

We consider a dilute classical ideal gas with elastic wall collisions. A single thermal pressure requires equal second velocity moments in all directions. The exercises also show what remains without isotropy and why common flow must be treated separately.[1][4]

Momentum flux becomes pressure

\begin{aligned}&pV=\frac13Nm\langle c^2\rangle\\&p=\frac13\rho\langle c^2\rangle\\&\rho=\frac{Nm}{V}\end{aligned}

p is absolute thermal pressure; c is speed relative to local gas flow. N is particle count.[1]

Pressure has units $\mathrm{Pa}=\mathrm{N\,m^{-2}}$ ; pV has units joule. ρ denotes mass density, not number density N/V. The formula describes isotropic pressure, not an arbitrary directed beam hitting a wall. ⟨c²⟩ is the mean of squares, not the square of mean speed.

Thermal RMS speed is the square root of mean squared speed. Therefore use $\langle c^2\rangle=c_{\mathrm{rms}}^2$ ; arithmetic mean speed is a different statistical quantity. The density form combines this thermal square with mass per gas volume.[1]

One collision and its recurrence time

First consider a particle in a rectangular model box of length L along x and wall area A perpendicular to it. The wall is stationary. An elastic collision reverses cx while preserving tangential components. The wall receives momentum 2m|cx|. A return to the same wall requires a round trip.[3]

\begin{aligned}&\Delta t=\frac{2L}{|c_x|}\\&\overline F_x=\frac{2m|c_x|}{\Delta t}=\frac{mc_x^2}{L}\\&p_x=\frac{Nm\langle c_x^2\rangle}{AL}\end{aligned}

Summation and time averaging give the normal pressure component; $AL=V$ . The box model explains the momentum balance.[3]

For $c_x=0$ the particle does not hit that wall and contributes zero. Interparticle collisions in real gas alter individual paths. Equilibrium momentum-flux balance remains the appropriate statistical relationship while the classical ideal model applies.[1]

Exactly where the factor 1/3 comes from

\begin{aligned}\langle c_x^2\rangle&=\langle c_y^2\rangle=\langle c_z^2\rangle\\&=\frac13\langle c^2\rangle\\pV&=\frac13Nm\langle c^2\rangle\end{aligned}

Isotropy requires equal second moments, not equal instantaneous components for every particle.[1]

The assumed full isotropy also makes cross moments such as $\langle c_xc_y\rangle=0$ vanish. Equality of three diagonal moments alone would not establish equal normal pressure for arbitrary rotated wall orientations. Pressure is then equal for every wall orientation. For an anisotropic distribution, $p_x=\rho\langle c_x^2\rangle$ and the analogous components remain different. Replacing them without justification by one third of total mean squared speed would form only an average of diagonal components, not actual normal pressure on a particular wall.[3]

The bridge to the ideal gas law

\begin{aligned}\frac12m\langle c^2\rangle&=\frac32k_{\mathrm B}T\\pV&=Nk_{\mathrm B}T=nRT\end{aligned}

Equipartition connects mechanical pressure with thermal temperature.[2]

\begin{aligned}&pV=\frac23E_{\mathrm{tr,tot}}\\&c_{\mathrm{rms}}=\sqrt{\frac{3p}{\rho}}\\&N=\frac{3pV}{m\langle c^2\rangle}\end{aligned}

Etr,tot counts translational energy of all particles. For a polyatomic gas it need not be total internal energy.[1]

In an ideal mixture at shared T, species momentum contributions add: $p_s=\frac{N_sk_BT}{V}$ and $p=\sum_s p_s$ . Different masses produce different speeds but not different energy $\frac32k_BT$ per particle at equal temperature. More particles per volume give proportionally greater partial pressure.[1]

The ideal gas law describes the same equilibrium state relation using temperature instead of squared speed. The microscopic derivation explains the origin of pressure; it does not extend validity to dense real gases or directed beams.[1][2]

Three examples from particle model to measurement

Pressure from particle count and RMS speed

Given: $N=2.00\cdot10^{22}$ , $m=6.64\cdot10^{-27}\,\mathrm{kg}$ , $c_{\mathrm{rms}}=900\,\mathrm{m\,s^{-1}}$ , $V=0.0060\,\mathrm{m^3}$ . Classical isotropic ideal gas; speed relative to mean gas motion.
Find: Find pV, p and Etr,tot.

Approach: In the isotropic model use $pV=\frac13Nm c_{\mathrm{rms}}^2$ , with translational energy $E_{\mathrm{tr,tot}}=\frac12Nm c_{\mathrm{rms}}^2$ . Square RMS speed and calculate momentum flux.
$pV=\tfrac13(2.00\times10^{22})(6.64\times10^{-27})(900)^2=35.856\,\mathrm J$
Divide by volume.
$p=\frac{35.856}{0.0060}=5976\,\mathrm{Pa}$
Determine total translational energy separately.
$E_{\mathrm{tr,tot}}=\tfrac12Nm c_{\mathrm{rms}}^2=53.784\,\mathrm J$

Result: Absolute pressure is 5.976 kPa. pV is 35.856 J and total translational energy is 53.784 J.

Check: Density $\rho=\frac{Nm}{V}\approx0.0221333\,\mathrm{kg\,m^{-3}}$ gives $p=\frac{\rho c_{\mathrm{rms}}^2}{3}=\frac{\rho\cdot900^2}{3}\,\mathrm{m^2\,s^{-2}}=5976\,\mathrm{Pa}$ . Also pV is exactly two thirds of 53.784 J.

Two species sharing T

Given: $T=320\,\mathrm K$ , $V=0.0040\,\mathrm{m^3}$ . Particle counts: $N_A=1.20\cdot10^{22}$ , $N_B=0.80\cdot10^{22}$ . A and B denote species here, not the Avogadro constant. Ideal mixture at equilibrium.
Find: Find both partial pressures and total pressure.

Approach: At common T, each ideal species has $p_i=\frac{N_i k_{\mathrm B}T}{V}$ ; then $p=p_A+p_B$ . Each species contributes its own particle-count share.
$p_A=\frac{(1.20\times10^{22})k_{\mathrm B}(320)}{0.0040}=13254.2304\,\mathrm{Pa}$
Calculate the second contribution and add.
$\begin{aligned}p_B&=8836.1536\,\mathrm{Pa}\\{} p&=p_A+p_B=22090.384\,\mathrm{Pa}\end{aligned}$

Result: Partial pressures are about 13.254 and 8.836 kPa, totaling 22.090 kPa. Particle masses are unnecessary when shared T is known.

Check: $\frac{p_A}{p_B}=1.5$ matches the particle count ratio. Direct calculation with $N_{\mathrm{tot}}=2.00\cdot10^{22}$ gives the same total pressure.

Particle and density forms express the same momentum balance. The following comparison checks both methods with identical inputs.[1][2]

Particle count or density: the same pressure calculation

Given: An isotropic model gas contains $N=1.00\cdot10^{23}$ particles of mass $m=4.65\cdot10^{-26}\,\mathrm{kg}$ . $V=0.0100\,\mathrm{m^3}$ , thermal RMS speed $c_{\mathrm{rms}}=500\,\mathrm{m\,s^{-1}}$ .
Find: Find mass density and pressure; verify agreement between both pressure forms.

Approach: find total mass from particle count and particle mass, then divide by volume. Use the density to calculate isotropic thermal pressure.
$\begin{aligned}M_{\mathrm{gas}}&=Nm\\\rho&=\frac{Nm}{V}\\p&=\frac13\rho c_{\mathrm{rms}}^2\end{aligned}$
Substitute: square the speed, rather than summing signed velocities before averaging.
$\begin{aligned}M_{\mathrm{gas}}&=(1.00\cdot10^{23})(4.65\cdot10^{-26})\,\mathrm{kg}\\&=0.00465\,\mathrm{kg}\\\rho&=\frac{0.00465\,\mathrm{kg}}{0.0100\,\mathrm{m^3}}=0.465\,\mathrm{kg\,m^{-3}}\\p&=\frac13(0.465)(500^2)\,\mathrm{Pa}\\&=38750\,\mathrm{Pa}=38.75\,\mathrm{kPa}\end{aligned}$

Result: Density is 0.465 kg/m³ and pressure is 38.75 kPa.

Check: Check using the particle form: $pV=\frac13Nm c_{\mathrm{rms}}^2=387.5\,\mathrm J$ . Division by 0.0100 m³ gives the same pressure. Translational energy is $E_{\mathrm{tr}}=\frac32pV=581.25\,\mathrm J$ ; for polyatomic particles this is not automatically their whole internal energy.

State pressure type and reference frame

A gauge may display pressure relative to the surroundings. Before $pV=Nk_BT$ , form absolute pressure: $p_{\mathrm{abs}}=p_{\mathrm{gauge}}+p_{\mathrm{ambient}}$ . Zero pressure in this ideal gas equation does not mean merely a vessel balanced against its surroundings.[1]

In directed flow, decompose $\vec v=\vec u+\vec c$ . Thermal normal stress uses the second moment of c. The extra directed momentum flux $\rho u_x^2$ belongs to the flow balance and cannot indiscriminately be substituted as thermodynamic equilibrium pressure. Stagnation or shock processes need additional boundary and process conditions.[3]

Pressure grows with squared thermal speed

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At fixed mass density

\rho=0.80\,\mathrm{kg\,m^{-3}}

, the curve gives pressure for each thermal RMS speed. At 450 m/s it gives 54.0 kPa, as in Exercise 1. Doubling speed quadruples pressure if density truly stays fixed. Directed flow is not included.[1]

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Five exercises with solutions

Calculate · Foundation

An isotropic model gas has $\rho=0.80\,\mathrm{kg\,m^{-3}}$ and thermal RMS speed $c_{\mathrm{rms}}=450\,\mathrm{m\,s^{-1}}$ . Find pressure.

Show hint

Use density form $p=\frac13\rho c_{\mathrm{rms}}^2$ .

Learning goal: Calculate isotropic pressure from mass density and thermal RMS speed.

Show solution and calculation

Approach: multiply squared thermal RMS speed by density and divide by three.
Pressure is 54.0 kPa.

Check: units give $\mathrm{kg\,m^{-3}}\cdot\mathrm{m^2\,s^{-2}}=\mathrm{Pa}$ . This is thermal pressure; common flow is not part of the speed square.

\begin{aligned}p&=\frac13\rho c_{\mathrm{rms}}^2\\&=\frac{(0.80)(450^2)}3\,\mathrm{Pa}\\&=54000\,\mathrm{Pa}=54.0\,\mathrm{kPa}\end{aligned}

Interpret · Application

The same sealed rigid vessel is heated from 300 to 450 K. How do pressure and thermal RMS speed change?

Show hint

N and V remain unchanged, as does species.

Learning goal: Compare different temperature dependences of pressure and speed.

Show solution and calculation

Approach: at unchanged species and number density, pressure scales with temperature.
RMS speed scales only with its square root.

Check: squared speed increases by the same factor 1.5 as pressure; speed itself rises only by about 22.5 percent.

\begin{aligned}\frac{p_2}{p_1}&=\frac{T_2}{T_1}=\frac{450}{300}=1.50\\\frac{c_{\mathrm{rms},2}}{c_{\mathrm{rms},1}}&=\sqrt{\frac{T_2}{T_1}}=\sqrt{1.50}\\&\approx1.224745\end{aligned}

Transfer · Challenge

An anisotropic model distribution has $\rho=1\,\mathrm{kg\,m^{-3}}$ and $\langle c_x^2\rangle=90000\,\mathrm{m^2\,s^{-2}}$ , $\langle c_y^2\rangle=\langle c_z^2\rangle=40000\,\mathrm{m^2\,s^{-2}}$ . Find $p_x$ , $p_y$ and $p_z$ .

Show hint

Use component form; isotropy is explicitly absent.

Learning goal: Find directional pressure components instead of using an isotropic simplification.

Show solution and calculation

Approach: calculate the normal pressure component for each direction.
The isotropic simplification is unavailable here.
The average of all three components is about 56.67 kPa but represents none of the actual wall-normal pressures.

Check: stronger x motion gives the largest normal pressure. A single equilibrium temperature is not assumed for this distribution.

\begin{aligned}p_x&=\rho\langle c_x^2\rangle=90000\,\mathrm{Pa}=90\,\mathrm{kPa}\\p_y=p_z&=\rho\langle c_y^2\rangle=40\,\mathrm{kPa}\\\frac{p_x+p_y+p_z}{3}&=\frac{90+40+40}{3}\,\mathrm{kPa}\\&\approx56.67\,\mathrm{kPa}\end{aligned}

Interpret · Application

Two classical ideal gases have equal number density and temperature. A molecule in gas B has four times the mass of one in gas A. Both velocity distributions are isotropic in their respective gas rest frames. Compare pressure and thermal RMS speed. Does slower molecular motion alone imply lower pressure?

Show hint

At equal T, mean translational energy per particle agrees. Both mass density and mean squared speed change in the pressure formula.

Learning goal: Explain equal pressure despite different molecular thermal speeds.

Show solution and calculation

Approach: use equal thermal energy to obtain the mean-square speed ratio.
Gas B has one quarter of the mean square and half the RMS speed.
At equal number density its mass density is four times larger.
These factors cancel in the pressure formula, so pressures agree.

Check: the ideal expression $p=\frac NVk_BT$ likewise contains no molecular mass.

\begin{aligned}\frac{c_{\mathrm{rms},B}}{c_{\mathrm{rms},A}}&=\sqrt{\frac{m_A}{m_B}}=\frac12\\\frac{\rho_B}{\rho_A}&=4\\\frac{p_B}{p_A}&=\frac{\rho_B\langle c_B^2\rangle}{\rho_A\langle c_A^2\rangle}=4\cdot\frac14=1\end{aligned}

[1][2]

Error analysis · Challenge

An isotropic ideal gas has mass density 1.2 kg/m³ and thermal RMS speed 500 m/s. In the laboratory it additionally flows uniformly at 100 m/s. A solution inserts laboratory mean-square speed $(500\,\mathrm{m/s})^2+(100\,\mathrm{m/s})^2$ into $p=\frac13\rho\langle c^2\rangle$ and obtains 104 kPa. Find thermal pressure and explain the error. This asks neither stagnation pressure nor wall-directed bulk momentum flux.

Show hint

In this pressure equation, c is velocity relative to mean flow. A common drift must be subtracted first.

Learning goal: Exclude bulk motion from the molecular speed square used for thermal pressure.

Show solution and calculation

Approach: use the given thermal mean-square speed without the drift contribution.
Correct pressure is 100 kPa; the claimed 104 kPa contains 4 kPa from an invalid mixture of reference frames.

Check: changing to a comoving inertial frame does not change thermal pressure. Bulk momentum flux is an additional directed quantity and cannot simply be distributed isotropically by dividing by three.

\begin{aligned}\vec c&=\vec v-\vec u\\p&=\frac13\rho c_{\mathrm{rms}}^2=\frac13(1.2)(500^2)\,\mathrm{Pa}=100\,000\,\mathrm{Pa}\\\Delta p_{\mathrm{wrong}}&=\frac13\rho u^2=4000\,\mathrm{Pa}\end{aligned}

[1][3]

The model box does not describe every gas

The simple form assumes nonrelativistic classical particles, negligible interaction contributions and isotropic thermal motion. The wall model uses elastic reflection at a stationary wall. Dense real gases may have additional pressure contributions; quantum gases need different statistical relations.[4]

For a mixture without a known common equilibrium temperature, sum separate contributions using each species’ particle mass and second velocity moment. Example 3 fully develops the earlier additional numerical case: calculations using particle count and mass density must agree.[1]

What to carry into your next calculation

For isotropic random motion in the gas rest frame, $p=\frac13\rho c_{\mathrm{rms}}^2$ . The factor one third assigns the mean square equally to three directions. Particle mass, density and speed must describe the same gas state. Without isotropy use separate normal pressure components; common drift is not an extra thermal contribution.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant connections; preserved existing derivations, calculations, exercises, figures and model limits.
October 1, 2026Expanded explanations and learning transitions; added a further fully worked example and an original quantitative diagram; clarified assumptions and calculation steps.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Pressure from kinetic theory. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/kinetische-gastheorie-druck

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Frequently asked questions about Pressure from kinetic theory

What does Pressure from kinetic theory describe physically?+

The formula describes absolute thermal pressure as momentum transfer from random particle motion to a stationary wall: $pV=\frac13Nm\langle c^2\rangle$ . N counts particles, m is their mass and c is measured relative to local mean gas motion. An elastic collision reverses the normal velocity component; the wall receives corresponding momentum. The assumed isotropy makes the second moments equal in three spatial directions, yielding the factor one third. Directed flow therefore is not automatically thermodynamic pressure. With $\rho=\frac{Nm}{V}$ , the density form is $p=\frac13\rho c_{\mathrm{rms}}^2$ . Do not replace mean squared speed by squared arithmetic mean speed.

How can I check a calculation involving Pressure from kinetic theory?+

For $N=2.00\cdot10^{22}$ , $m=6.64\cdot10^{-27}\,\mathrm{kg}$ , $c_{\mathrm{rms}}=900\,\mathrm{m\,s^{-1}}$ and $V=0.0060\,\mathrm{m^3}$ , $pV=35.856\,\mathrm J$ and $p=5976\,\mathrm{Pa}$ . Check using $\rho=\frac{Nm}{V}\approx0.0221333\,\mathrm{kg\,m^{-3}}$ and $p=\frac13\rho c_{\mathrm{rms}}^2$ . A second check uses total translational energy $E_{\mathrm{tr}}=\frac12Nmc_{\mathrm{rms}}^2=53.784\,\mathrm J$ ; it must give $pV=\frac23E_{\mathrm{tr}}$ . Confirm that speed is thermal and the distribution isotropic. For a polyatomic gas, do not automatically equate this translational energy with complete internal energy. A pressure gauge reading also needs conversion to absolute pressure before applying the ideal gas law. Also ensure the stated volume is the actual accessible gas volume.

What assumptions and limits apply to Pressure from kinetic theory?+

The simple model assumes classical nonrelativistic particles, negligible interaction contributions and isotropic thermal motion. The illustrative wall derivation uses elastic reflection at a stationary wall. With an anisotropic distribution, $p_x=\rho\langle c_x^2\rangle$ remains direction dependent; the factor 1/3 is then unjustified for a specific wall. A directed beam or additional flow requires momentum-flux accounting in an appropriate reference frame. Dense real gases and quantum gases may need different pressure relations. In mixtures, sum species contributions separately unless shared equilibrium temperature is known. The stated p value is absolute pressure, not automatically gauge pressure relative to surroundings.

How do I check units and scale for Pressure from kinetic theory?+

$Nm\langle c^2\rangle$ has units $\mathrm{kg\,m^2\,s^{-2}}=\mathrm J$ , so pV is an energy quantity. Dividing by V in m³ gives $\mathrm{J\,m^{-3}}=\mathrm{Pa}$ . In the density form, kg/m³ multiplied by m²/s² also gives Pa. Actually square $c_{\mathrm{rms}}$ ; multiplying speed directly by mass would give momentum instead of the required quantity. An example with $N=10^{23}$ , $m=4.65\cdot10^{-26}\,\mathrm{kg}$ and $c_{\mathrm{rms}}=500\,\mathrm{m\,s^{-1}}$ gives $pV=387.5\,\mathrm J$ . At $V=0.010\,\mathrm{m^3}$ , $p=38.75\,\mathrm{kPa}$ . Doubling volume at unchanged particle count and thermal distribution halves p. Quadrupling $c_{\mathrm{rms}}$ at equal density instead multiplies pressure by sixteen.

When should I combine Pressure from kinetic theory with another relation?+

Combine momentum flux with $\frac12m\langle c^2\rangle=\frac32k_BT$ to obtain $pV=Nk_BT=nRT$ . This turns the speed distribution into a thermal equation of state. In an ideal mixture at shared T, each species contributes $p_s=\frac{N_sk_BT}{V}$ ; partial pressures add. For 1.20·10²² and 0.80·10²² particles at 320 K in 0.0040 m³, contributions are 13.254 and 8.836 kPa, totaling 22.090 kPa. Their masses are unnecessary when shared T is known. For internal energy, $U=\frac32pV$ applies only to the appropriate monatomic model; for polyatomic gases it remains the translational sum.

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[1]OpenStax, University Physics Vol. 2, §2.2: Pressure, Temperature, and RMS Speed

More formulas: Heat and thermodynamics

\eta_{Carnot} = 1 - \frac{T_{kalt}}{T_{warm}}

Carnot Efficiency

Q = m \cdot c \cdot \Delta T

Heat Capacity: Q = m·c·ΔT

\Delta U = Q - W

First Law of Thermodynamics

P = \sigma \cdot A \cdot T^4

Stefan-Boltzmann Law

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Pressure from kinetic theory

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Kinetic theory gas pressure calculator

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How do you calculate with Pressure from kinetic theory?

Many individual impacts produce measurable pressure

Momentum flux becomes pressure

One collision and its recurrence time

Exactly where the factor 1/3 comes from

The bridge to the ideal gas law

Three examples from particle model to measurement

Pressure from particle count and RMS speed

Two species sharing T

Particle count or density: the same pressure calculation

State pressure type and reference frame

Pressure grows with squared thermal speed

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An isotropic model gas has ρ=0.80 kg m−3\rho=0.80\,\mathrm{kg\,m^{-3}}ρ=0.80kgm−3 and thermal RMS speed crms=450 m s−1c_{\mathrm{rms}}=450\,\mathrm{m\,s^{-1}}crms​=450ms−1. Find pressure.

The same sealed rigid vessel is heated from 300 to 450 K. How do pressure and thermal RMS speed change?

The model box does not describe every gas

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More formulas: Heat and thermodynamics

Retain Pressure from kinetic theory for exams

An isotropic model gas has $\rho=0.80\,\mathrm{kg\,m^{-3}}$ and thermal RMS speed $c_{\mathrm{rms}}=450\,\mathrm{m\,s^{-1}}$ . Find pressure.