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15.3 Kinetic theory of gases

Syllabus
9702–2028–2029
Topic
15.3
Level
A2

Learning objectives

15.3.1The basic assumptions of the kinetic theory of gases• state the basic assumptions of the kinetic theory of gases15.3.2How molecular movement causes the pressure exerted by a gas and derive and• explain how molecular movement causes the pressure exerted by a gas and derive and use the relationship pV = 1/3 Nm<c2>• Where <c2> is the mean-square speed (a simple model considering one-dimensional collisions and then extending to three dimensions using 1/3 <c2> = <cx 2> is sufficient)15.3.3The root-mean-square speed cr.m.s. is given by c<>2• understand that the root-mean-square speed cr.m.s. is given by c<>215.3.4PV = 1/3 Nm<c2> with pV = NkT to deduce that the average translational• compare pV = 1/3 Nm<c2> with pV = NkT to deduce that the average translational kinetic energy of a molecule is 3/2 kT, and recall and use this expression

The kinetic model treats an ideal gas as many randomly moving particles

Basic assumption Meaning in the ideal model
Molecules are in continuous random motion Every direction is equally likely in a large sample
Molecular volume is negligible compared with gas volume Molecules are treated as point particles for the model
No intermolecular forces act except during collisions Molecular potential energy is taken as zero and motion is uniform between collisions
Collisions are perfectly elastic Total kinetic energy is conserved in each collision
Collision duration is negligible Collisions are treated as instantaneous

These are modelling assumptions, not literal properties of every real gas. Ideal behaviour is a better approximation when molecules are far apart, so their own volume and intermolecular forces are negligible.

At very high pressure molecules are close together. Their volume or intermolecular forces may no longer be negligible, so the gas can depart from ideal behaviour.

“No intermolecular forces” means no forces between collisions. During a collision, forces act briefly, change molecular momentum and transfer momentum.

Derive gas pressure from molecular collisions with a wall

A molecule colliding elastically with a wall reverses its perpendicular momentum. The wall exerts a force on the molecule, so the molecule exerts an equal and opposite force on the wall. Many impacts across the wall area produce pressure.

Take a cube of side L and volume V=L³. One molecule has mass m and x-component of velocity cₓ toward a wall of area A=L².

momentumchangemagnitude=2mcxtimebetweensuccessivehitsonthesamewall=2L/cxaverageforcecontribution=(2mcx)/(2L/cx)=mcx2/Lmomentum change magnitude = 2mcₓ time between successive hits on the same wall = 2L/cₓ average force contribution = (2mcₓ)/(2L/cₓ) = mcₓ²/L

F=(m/L)Σcx2p=F/A=mΣcx2/L3pV=mΣcx2=Nm<cx2>F = (m/L)Σcₓ² p = F/A = mΣcₓ²/L³ pV = mΣcₓ² = Nm<cₓ²>

Randommotionisisotropic:<cx2>=<cγ2>=<cz2>and<c2>=<cx2>+<cγ2>+<cz2>Therefore<cx2>=(1/3)<c2>,sopV=(1/3)Nm<c2>.Random motion is isotropic: <cₓ²>=<cᵧ²>=<c_z²> and <c²>=<cₓ²>+<cᵧ²>+<c_z²> Therefore <cₓ²>=(1/3)<c²>, so pV=(1/3)Nm<c²>.

Symbol Meaning
N number of molecules
m mass of one molecule
<c²> mean of the squared molecular speeds

The factor 1/3 comes from three equivalent squared velocity components. It does not come from three walls. Also <c²> is the mean square speed, not <c>².

Root-mean-square speed is the square root of mean-square speed

crms=sqrt(<c2>)socrms2=<c2>c_rms = sqrt(<c²>) so c_rms²=<c²>

Step Operation on all molecular speeds
1 square each speed c
2 find the mean <c²>
3 take the square root

For speeds 2, 3 and 6 m s⁻¹, <c²>=(4+9+36)/3=16.3 m² s⁻², so c_rms=sqrt(16.3)=4.04 m s⁻¹. The arithmetic mean speed is 3.67 m s⁻¹, so the two means are not equal.

Foroneidealgasspecies,crms=sqrt(3kT/m),socrmsisproportionaltosqrt(T).AgraphofcrmsagainstthermodynamicTstartsattheoriginandriseswithdecreasinggradient.For one ideal gas species, c_rms=sqrt(3kT/m), so c_rms is proportional to sqrt(T). A graph of c_rms against thermodynamic T starts at the origin and rises with decreasing gradient.

c_rms is a statistical speed scale, not the speed of every molecule. Use thermodynamic temperature in kelvin; do not replace <c²> by <c>².

Average translational kinetic energy is three-halves kT

pV=(1/3)Nm<c2>andpV=NkT(1/3)Nm<c2>=NkT(1/3)m<c2>=kT(1/2)m<c2>=(3/2)kTTherefore<Ek>=(3/2)kT.pV=(1/3)Nm<c²> and pV=NkT (1/3)Nm<c²>=NkT (1/3)m<c²>=kT (1/2)m<c²>=(3/2)kT Therefore <E_k>=(3/2)kT.

The average translational kinetic energy per molecule depends only on thermodynamic temperature. At the same T, molecules of different ideal gases have the same average translational kinetic energy.

At T=400 K, <E_k>=(3/2)(1.38×10⁻²³)(400)=8.28×10⁻²¹ J per molecule.

Because(1/2)mcrms2=(3/2)kT,crms=sqrt(3kT/m).AtequalT,thelightermoleculehasthegreaterrmsspeedeventhoughaveragetranslationalkineticenergiesareequal.Because (1/2)m c_rms²=(3/2)kT, c_rms=sqrt(3kT/m). At equal T, the lighter molecule has the greater rms speed even though average translational kinetic energies are equal.

Do not omit the factor 1/2 from kinetic energy or use Celsius. This is an average per molecule: individual molecules have a distribution of speeds and energies.