structure of matter lecture notes - week 2 1

4 Thermodynamics of Gases

4.1 Quasistatic work on PV diagram

Figure 7: Expansion of a gas

The work done BY gas is area under curve during the expansionof the gas:

∆W = −∫ V1

V0

P dV < 0

Figure 8: Compression of a gas

The work done ON gas is area under curve during the compres-

sion of the gas: ∆W = −∫ V1

V0

P dV > 0

structure of matter lecture notes - week 2 2

4.2 W & Q are process dependent

The work performed and therefore the area under a PV curve de-pends on path. In fig. 9 there can be any number of paths from initialstate A to final state B. However, if the process is known, for exam-ple an isothermal process where T remains constant; for an ideal gasP0 V0 = P1 V1

Figure 9: Two equilibrium states Ato B

Figure 10: process 1; isothermal

Figure 11: process 2; isobaric com-pression followed by isochoricheating

Process 1: isothermal → PV = NkBT → P ∝ (1/V)

W1 = −∫ B

AP dV = −

∫ B

A(NkBT/V)dV (from A→ B)

= −(NkBT)∫ V1

V0

(1/V)dV (V goes from V0 → V1)

= −(NkBT) ln(V1/V0) > 0 for V1 < V0 (work done ONsystem)

Process 2: isobaric compression followed by isochoric heating.A→ C : const P→ T falls

WA→C = −∫

P dV (A→ C)

= −P0

∫dV (V goes from V0 → V1)

= −P0(V1 −V0) (again > 0 since V1 < V0)

C→ B : const. V, T rises, WC→B = 0 since dV = 0so W2 = WA→C→B = P0(V0 −V1) = (NkBT)(1−V1/V0) < W1

Since ∆U = U f in −Uinit → depends only on initial and final states∴ NOT process dependent.

BUT W is process dependent and ∴ so is Q [= ∆U −W]

4.3 Heat Capacity

Quantity of heat, ∆Q, required to raise the temperature of a materialof mass m, from T1 to T2 is found to be approximated by

∆Q = mc∆TFor an infinitesimal temperature change→ dQ = NmCm dT

Nm - number of molesCm - molar heat capacity

c = specific heat (lower case) = heat capacity per kg = Cm/MM is the mass per mole (molar mass or molecular weight).dQ is process dependent ∴ so is C.For ideal gas: U = nd

12 NkBT, where nd is the number of degrees of

freedomdifferentiating dU = 1

2 ndNkB dT

structure of matter lecture notes - week 2 3

1st Law:

dQ = dU − dW = 12 ndNkB dT + P dV (5)

for constant V (i.e. dV = 0)→ dQ = 12 ndNkB dT

∴ CV = 12 ndNkB Heat capacity at constant volume

and we can write U = CV T (for an ideal gas)

For 1 mole of gas CVm =nd2

R.

Figure 12: Variation of CV for H2as a function of T. As T increases,higher rotational and vibrationalstates (degrees of freedom) becomeaccessible.

Now differentiating PV = NkBT (applying product rule)d(PV) = V dP + P dV = NkB dT → P dV = NkB dT −V dP

using eqn. 5, dQ = 12 ndNkB dT + NkB dT −V dP

for constant P (i.e. dP = 0)→ dQ = (CV + NkB)dT→ CP = CV + NkB Heat capacity at constant pressureFor 1 mole of ideal gas, CPm = CVm + RAs T rises for constant P → V rises→ gas does work→ need

extra Q to compensate→ CP > CV

Also note, CP = 12 ndNkB + NkB = 1

2 ndNkB(1 + (2/nd)) = γCv

where γ = CP/CV = (1 + (2/nd)) = (nd + 2)/nd

γ (the adiabatic constant) is a dimensionless ratio and depends onthe state of the gas. For normal air, γ ≈ 1.4. This is because airis mainly comprised of diatomic molecules (78%N2 and 21%O2)which have 5 degrees of freedom (3 translational and 2 rotational),the further two vibrational degrees of freedom are only relevant at There are two degrees of freedom

associated with vibrational motion sincethere are both kinetic and potentialcomponents of the vibrational energy.

high temperatures (see fig. 12). Thus, at RTP for a diatomic moleculeγ = 7

5 = 1.4.

4.4 Adiabatic process

Adiabatic process dQ = 0, i.e. naturally reversible.1st Law becomes: dU = −P dV

structure of matter lecture notes - week 2 4

For ideal gas: U = nd12 NkBT = nd

12 PV

so dU = (nd/2)(P dV + V dP) = −P dV→ (nd + 2)P dV = −ndV dP(nd + 2)

nd

∫(1/V)dV = −

∫(1/P)dP

γ ln V = − ln P + constantln PVγ = constant→ PVγ = constant (adiabatic relation)Note that the final pressure (for the same volume change) is higher

for adiabatic changes than for isotherms, hence, adiabats are steeperthan isotherms see fig. 13).

adiabat

Figure 13: PV diagram showingadiabat and isotherm.

structure of matter lecture notes - week 2 5

5 The Boltzmann Law

The atmosphere of a planet is the gaseous envelope that surroundsit. At a microscopic level, it consists of molecules in motion and largedifferences can exist between the atmospheres of different planetsin both chemical composition and physical structure. How does thedensity of the Earth’s atmosphere change with altitude? How doesthe distribution of speeds of molecules in a gas change as the tem-perature of the gas increases? What can we say about the probabilityof observing a certain type of energy associated with one particularatom or oscillator? Addressing such questions will lead to the Boltz-mann distribution, which uses a statistical approach and is applicableto a wide variety of physical, chemical, and biological phenomena.

5.1 Isothermal atmosphere

Referring to the Earth’s atmosphere, a complete understandingwould require knowledge of its’ composition, structure, densitydistribution and motion. That said, much headway can be madeby starting with some basic assumptions; the first of which is that theatmosphere is comprised of layers which are isothermal. Consider aslab of air in mechanical equilibrium (ie macroscopic variables notchanging) at a certain height z in the atmosphere (fig 14). The airpressure at any height in the atmosphere is due to the force per unitarea exerted by the weight of all of the air lying above that height.Consequently, atmospheric pressure decreases with increasing height abovethe ground.

Area A

dz

slab of air

Earth z (height)

Figure 14: Change in pressure due toslab of width dz

Net force acting on slab = PA− (P + dP)A− ρA dz g1: pressure from below, P2: pressure from above is P+ dP at z+ dz [should find that dP < 0]3: weight of slab is (ρ = density) (Adz = volume)(g assumed con-

stant)but, in mechanical equilibrium, net force = 0

→ PA− PA− AdP− ρA dz g = 0

structure of matter lecture notes - week 2 6

dPdz

= −ρg (6)

(Hydrostatic equation) NB. P falls with height and providesan example of non-uniform equilibrium[this is consistent with observation; forexample the air density is so low on topof Mount Everest that most climberscarry oxygen tanks].

We can define a z dependent density as

ρ(z) = no. density [particles/m3]×mass of particles = n(z)m

Assume an isothermal ideal gas: P(z) = n(z)kBT and differentiatingwrt z yields

dPdz

= kBTdndz

(7)

Equating dPdz in Eqns 6 and 7 and substituting for ρ yields

kBTdndz

= −mng

dndz

= − mgkBT

n

z

P

1 atm

0 λ

Figure 15: Plot of pressure with height

with the following exponential solution:

n = n0 e−z/λ (8)

where n0 = n(z = 0) [ie. at ground level], λ = kBT/mg = scalelength [i.e. the vertical distance over which the pressure of the atmo-sphere changes by a factor of e (fig. 15)]. For the Earth’s atmosphereλ ≈ 8.5km−1

Similarly we can write ∴

P = nkBT = P0 e−z/λ (9)

Estimating the atmospheric weightTo estimate the pressure due to the particles in the atmosphere,

consider a column of air of area L2, that is infinitely high (see fig-ure 16):

z=∞

z=0

↑

L

L

Figure 16: Column of air - squarelength L

The number of particles that lie between z and z+ dz = n0 exp(−z/λ)L2dz

∴ N = total number of particles =∫ ∞

0n0 exp(−z/λ)L2dz

= −n0L2λ[e−∞ − e0]

= n0L2λ

At the Earth’s surface P ≈ 105Pa and T ≈ 293K so

n0 =P

kBT=

105

kB293= 2.5× 1025m−3

structure of matter lecture notes - week 2 7

So the total number of particles contained in the rectangular col-umn above each m2 is n0L2λ

≈ 2.5× 1025 × 8.5× 103 = 2.1× 1029

The weight of these particles is 1.66× 10−27 × 29× 2.1× 1029 =

104kg = 10 tonnes. The force due to 104kg is balanced by the atmo-spheric pressure of ≈ 105N.m−2

structure of matter lecture notes - week 2 8

5.2 Particle probabilities

Assume there are N particles in a box.

z=0

z=L

y=0

y=L

x=0 x=L

x+!xxFigure 17: Number of particles withx-displacement between x and x + δx

The probability of a given molecule existing in a region of the box= number of particles in region of the box / total number of particles(N)

With a uniform distribution of particles, the particle density in abox with sides of length L is n = N/L3

In a thin slab of thickness δx shown if fig. 17 the number of parti-cles between x and x + δx = nδxL2 = (N/L3)δxL2 = Nδx/L

so probability of given particles lying between x and x + δx;p(x) = δx/L,

similarly, the probability of given particles lying between y andy + δy; p(y) = δy/L

z=0

z=L

y=0

y=L

x=0 x=L

x+!xx

y+!y

y

Figure 18: Number of particles withdisplacement between x and x + δxAND between y and y + δy

Now number of particles between x and x + δx AND between yand y + δy = δxδyL× n = δxδyL(N/L3) as depicted in fig. 18

Thus the no. of particles = Nδxδy/L2

∴ the probability p(x, y) =(

Nδxδy/L2) /N = (δx/L)(δy/L) =

p(x)× p(y)→ If two events are independent, probability of both = prob(1)

× prob(2)

5.3 Boltzmann Law

Let us now return to the isothermal atmosphere:The probability of particles lying between z & z + dz; p(z) =̄ (no. of

particles between z & z + dz)/N

=n0e(−z/λ)L2dz

n0L2λ

substituting for λ we obtain

=

(mgkBT

)e(−mgz

kBT

)dz

hence the probability of particles lying between z & z + dz is pro-portional to exp(−mgz/kBT)dz

However the gravitational potential energy is given by E(z) = mgz,so p(z) ∝ p(E) ∝ exp(−E(z)/kBT)

IN FACT generally when in thermal equilibrium;

P(E) ∝ exp(−E/kBT)

(Boltzmann’s Law)

structure of matter lecture notes - week 2 9

NB. E can be any form of energy; for example the energy of a single(ideal) gas molecule in the gravitational field near the Earth’s surface(excluding rest energy, nuclear energy, and electronic energy) is givenby:

ETotal = Etrans + Evib + Erot + mgzwhere Evib and Erot are the vibrational and rotational energies

relative to the centre of mass. If T is the same everywhere in the gasthen the probability that a particular molecule will have a certainamount of energy isP(ETotal) ∝ exp−ETotal/kBT = e−(Etrans+Evib+Erot+mgz)/kBT

It is useful to group the terms of the Boltzmann factor accordingly→ P(ETotal) ∝ [e−Etrans/kBT ][e−Evib/kBT ][e−Erot/kBT ][e−mgz/kBT ]

The first bracket is associated with the distribution of velocities,the second with the distribution of vibrational energy, the third withdistribution of rotational energy and the fourth with the distributionof gravitational p.e (as discussed previously). Clearly temperatureplays an important role in all forms of energy distribution, and weshall continue to explore each of the other three terms in bracketsseparately in sections 6 & 7.

An important application of the Boltzmann distribution relatesto the temperature dependance of chemical reaction rates, kr. It canbe expressed mathematically in the form of the Arrhenius equationwhereby:

kr = Ae(−Ea/kBT)

Arrhenius equationwhere the pre-exponential term A is known as the frequency factor

and has the same units as kr, and Ea is the activation energy - theminimum kinetic energy that reactants must have in order to formproducts.This lies at the heart of a branch of physical chemistry known aschemical kinetics and the theory goes as follows: when two moleculescollide there is a chance that they will react to form new products(or not). At close proximity, particles will experience a repulsivepotential that reaches a maximum and provided the particles havesufficient energy to overcome this ‘potential barrier’ - a reactionwill proceed. Physical chemists use this formula (or variations ofit) to predict rates of various chemical reactions at temperatures. Itcan also be shown that A in this instance, can be interpreted as thefraction of collisions that have enough kinetic energy to lead to reactions[Although slightly beyond of the scope of this course, further detailscan be found in Ch. 20 & 21 of Physical Chemistry by Atkins & Paula,9th Edition (Oxford)].

structure of matter lecture notes - week 2 10

6 Distribution Functions

6.1 Collisions

For an ideal gas we ignored collisions, however we know thatatoms have a finite size of diameter d. Let’s consider a particle travel-ling with an average velocity v̄.

Figure 19: Collision cross-section

Figure 20: Mean free path

The effective cross-sectional area of particles during a collision isA = πd2, so the volume swept out by particle with mean velocity v̄ intime t is V = v̄t πd2

∴ Ncoll = no. of particles in this volume = n v̄t πd2,and mean distance between collisions = mean free path λ =v̄t

nπd2v̄t=

1nπd2

Correcting for relative motion of particles (vrel =√

2v̄) finallygives,

λ =1√

2nπd2(10)

6.2 Velocity Component Distribution

Boltzmann’s Law states that the probability that a particle has en-ergy between E and E+ dE is given by p(E) = g(E)dE = A exp(−E/kBT)dE,where A is the normalisation constant. NB. neither g(E) or f (vx) are pure

probabilities (which are just numbers),g(E) is the probability per unit intervalof energy, and f (vx) is the probabilityper unit of velocity OR the distributionfunction.

We can apply Boltzmann’s Law to an ideal gas since energy asso-ciated with vx is 1

2 mv2x; therefore the probability of particle having vx

between vx & vx + dvx is,

structure of matter lecture notes - week 2 11

f (vx)dvx = A exp(− 1

2mv2

x/kBT)dvx (11)

∴ f (vx) = A exp(−αv2x) where α = m/2kBT

Any particle velocity must take some value vx from +∞ to −∞, so

Figure 21: 1D Boltzmann distributionfor two temperatures

∫ +∞

−∞f (vx)dvx = 1 (as distribution function must be normalised)

A = (m/2πkBT)12 (12)

It follows that there must be the same distribution function for vy

& vz, so the probability of a particle lying between vx & vx + dvx andvy & vy + dvy and vz & vz + dvz is

f (vx, vy, vz)dvxdvydvz = A exp(−αv2x)dvx × A exp(−αv2

y)dvy × A exp(−αv2z)dvz

= A3 exp(−α(v2x + v2

y + v2z))dvxdvydvz

= A3 exp(−αv2)d3v (5.2.3)

6.3 Velocity Space

vx, vy and vz are co-ordinates in virtual ‘velocity space’→ plotparticles ‘position’, v =‘distance’ from origin, so v2 = v2

x + v2y + v2

z

Figure 22: Phase space examples withsame |v|, v1 = vx − 2vy + 3vz orv2 = −2vx + 3vy − vz

Figure 23: particles with speed betweenv & v + dv, occupy a spherical shell ofthickness dv and radius v, equivalentlymade of many cubes with dvx , dvy anddvz

Volume of small box is Vbox = dvxdvydvz

Volume of shell (radius v), Vshell = 4πv2 dv∴ Number of boxes in shell = Vshell/Vbox = 4πv2dv/(dvxdvydvz)

Assuming particles are isotropically distributed in velocity space;prob. of particles in shell; f (v) dv = prob. of particles in one box

× num. boxes in shell= A3 exp(−αv2)dvxdvydvz · 4πv2dv/(dvxdvydvz) = A34πv2 exp(−αv2)dv

structure of matter lecture notes - week 2 12

6.4 Maxwell-Boltzmann speed distribution

From the last section we established that the probability of particlehaving speed between v & v + dv is given by

f (v) dv = A34πv2 exp(−αv2) dv

f (v) = A34πv2 exp(−αv2) (13)

(Maxwell-Boltzmann Speed Distribution Function)A plot of the M-B distribution for different relative temperatures isshown below (fig. 24).

v

f(v)

2T

T

4T

8T

Figure 24: Maxwell-Boltzmann speeddistribution for different temperaturesT

By considering the form the equation 13, we can conclude the follow-ing:

• The decaying exponential term implies that the fraction of moleculeswith very hight speeds will be very low

• The α term multiplying v2 is large when particle mass is large→exponential factor goes most rapidly to zero when m is large i.e.heavy molecules are unlikely to be found with high speeds.

• For high T, α is small→ exponential factor decreases slowly as vincreases, i.e. a greater fraction of particles are expected to havehigher speeds at high T than at low T.

• Particles with very low speeds v are unfavoured due to the v2

factor (before the e), as this factor goes to zero as v goes to zero.This ensures that only a small fraction of particles have very lowspeeds.

structure of matter lecture notes - week 2 13

• The remaining terms ensure normalisation such that when all thefractions over the entire range of speeds from zero to infinity areadded, it equals 1.

• Most probable speed occurs at the turning point and gives vmp =

(2kBT/m)12

6.5 Average Speed

To determine the average speed, < v >, of molecules in a gas, wemust weight the speed by the number of molecules that have thatspeed, and then divide by the total by the number of molecules.

→ < v > = (v1 + v2 + v3 + v4 + ...) /N

= (∑ Nvv) /N (Nv = number of particles with speed v)

= (∑ vN f (v)dv) /N

→ < v > =∫ ∞

0v f (v) dv

=∫ ∞

0vA3 4πv2 exp(−αv2) dv

= A34π∫ ∞

0v3 exp(−αv2) dv

using a standard integral –∫ ∞

0v3 exp(−αv2) dv =

12α2

∴ < v > = (8kBT/πm)12 ≈ 1.13 vmp

structure of matter lecture notes - week 2 14

7 Equipartition of Energy

7.1 Translation kinetic energy

For a gas confined in a container, quantum mechanics predicts thatthe kinetic energy, 1/2m < v2 > is quantised. However, if the con-tainer is of ordinary (macroscopic) size then the energy quanta areextremely small compared to the average kinetic energy, even at lowtemperatures.To calculate the mean translational energy of particles in a gas,

〈 12 mv2〉 = 1

2 m (1/N) ∑v Nvv2 (Nv = number of particles with speed v)

= 12 m (1/N) ∑v v2N f (v)dv

=12

m∫

v2 f (v)dv =12

m〈v2〉

→ 〈v2〉 =∫ ∞

0v2 f (v) dv

=∫

v2 A3 4πv2 exp(−αv2) dv

= A34π∫

v4 exp(−αv2) dv

using a standard integral –∫

v4 exp(−αv2) dv =38

( π

α5

)1/2

so the mean squared velocity 〈v2〉 = (3kBT/m)

the root mean squared velocity (rms) 〈v2〉1/2 = (3kBT/m)1/2 ≈1.22 vmp

and the mean translational energy 12 m〈v2〉 = 3

2 kBT

7.2 Translation degrees of freedom

To find the contribution from each translational degree of free-dom, we need to take averages involving vx i.e. using f (vx) =

(m/2πkBT)12 exp(−mv2

x/2kBT)

〈vx〉 =∫ ∞

−∞vx f (vx) dvx = 0 Problem sheet 2

〈v2x〉 =

∫ ∞

−∞v2

x f (vx) dvx = kBT/m (14)

Problem sheet 3

Similarly 〈v2y〉 = 〈v2

z〉 = kBT/m – root mean square!k.e associated with vx = 1

2 m〈v2x〉 = 1

2 kBTand also k.e associated with vy = k.e associated with vz = 1

2 kBT

structure of matter lecture notes - week 2 15

but total k.e. 〈Etrans〉 = 12 m〈v2〉 = 1

2 m〈v2x + v2

y + v2z〉

= 12 m〈v2

x〉+ 12 m〈v2

y〉+ 12 m〈v2

z〉= 3× 1

2 kBT as above.

Hence each degree of freedom contributes exactly 13 of the total

energy. Note also that 〈v2〉 6= (〈v〉)2

7.3 Theorem of Equipartition of Energy

For a system in thermal equilibrium, each degree of freedom contributes12 kBT per molecule to the internal energy of the system.

7.4 Diatomic molecules : rotational degrees of freedom

Consider the rotational-energy portion of the M-B distribution forwhich the Boltzmann factor is

e−Erot/kBT

For a monotomic gas there is no rotational term, but for diatomicgases (which account for most of the Earth’s atmosphere) there is akinetic energy associated with the rotation of the ‘rigid rotor’ consist-ing of the two nuclei.

Hence each degree of freedom contributes exactly 13 of the total energy.

Note also that �v2� �= (�v�)2 (cf 6.5.2, PS 2)

Theorem of Equipartition of Energy: For a system in thermal equilibrium, each degree of freedom con-tributes 1

2kBT per molecule to the internal energy of the system.

7.3 Diatomic molecules : rotational degrees of freedom

Figure 1: (molecule) CofM → 3 translationaldegrees of freedom

Figure 2: (rot axis) mol. can rotate about 2axes (assuming point particles)

Each rotational degree of freedom has energy 12Iω2. Can have any ω from ∞ to +∞

Boltzmann law: probability of having ω1 between ω1 & ω1 + dω1 = C1 exp(−Iω21/2kBT ) dω1

(distribution function for ω1)

Identical to f(vx) except vx ↔ ω1 and m ↔ I . Substitute into previous results

�ω2� =

�ω2f(ω) dω = kBT/I (cf 7.2.1)

∴ �ke�rotation about each axis = 12kBT

7.4 Diatomic molecules: vibrational degrees of freedom

Figure 3: Atoms vibrate about equilibrium separation (d0) with displacement d = d0 + δ

2 degrees of freedom for vibrations (1 for ke and 1 for pe)

p.e. = 12kδ2 [k = force/displacement = spring constant]

2

vx

vz

vy

motion of centre of massm

2 degrees of rotation

2r

m/2

I=mr2 - moment of intertia of molecule

ω1

ω2

168.64d

xSpring Constant k

vvib/2vvib/2

m/2m/2

Figure 25: (molecule) CofM→ 3 trans-lational degrees of freedom

I=mr2 - moment of intertia of molecule

!"#$%&$$'"()"&(*+*,(-

2r

m/2 ω1

ω2

Figure 26: (rot axis) mol. can rotateabout 2 axes (assuming point particles)

Each rotational degree of freedom has energy Erot =12 Iω2 and can

have any ω from −∞ to +∞We now apply Boltzmann law→ probability of having ω1 between

ω1 & ω1 + dω1 = C1 exp(−Iω21/2kBT) dω1

(distribution function for ω1).This is identical to f (vx) except vx ↔ ω1 and m ↔ I. By substitutinginto previous results:

〈ω2〉 =∫

ω2 f (ω) dω = kBT/I (cf. < v2x > eqn. 14)

We can conclude therefore that at high temperatures the rotationalmotion of a diatomic molecule has an average energy that is approx-imately 2( 1

2 kBT) = kBT distributed of over the two rotational degreesof freedom

∴ 〈ke〉rotation about each axis =12 kBT

7.5 Diatomic molecules: vibrational degrees of freedom

We have treated the distribution of position, velocity, and rotation.Next we examine the distribution of vibrational energy with theBoltzmann factor e−(Evib/kBT)

168.64d

xSpring Constant k

vvib/2 vvib/2

m/2 m/2

Figure 27: Atoms vibrate about equilib-rium separation (d0) with displacementd = d0 + δ

For a monotomic gas (such as helium) there is no vibrational en-ergy term. However, for diatomic molecules such as N2 or HCl thereare 2 degrees of freedom for vibrations (1 for ke and 1 for pe)

p.e. = 12 kδ2 [k = force/displacement = spring constant]

structure of matter lecture notes - week 2 16

k.e. (vx part) = 12 mv2

x + 12 (m/4)v2

vib (PS2) – (m = total mass ofmolecules)

1: 1st term is part of translational ke2: v = dδ/dt i.e. 2nd term is ke of vibration.Evib = 1

2 kδ2 + 12 (m/4)v2

vib (p.e. + k.e)Boltzmann’s Law: probability of molecule having δ (displacement

from equilibrium separation) between δ & δ+ dδ = C2 exp(−kδ2/2kBT) dδ

Subs vx → δ and m→ k with eqn. 14. → 〈δ2〉 = kBT/k〈pe〉vib = 〈 1

2 kδ2〉 = 12 kBT

Boltzmann’s Law: probability of molecule having n between vvib &vvib + dvvib is C3 exp(−(m/4)v2

vib/2kBT) dvvib

Substitute vx → vvib , m → (m/4) into eqn. 14 → 〈v2vib〉 =

kBT/(m/4)→ 〈k.e〉vib = 1

2 (m/4) kBT/ (m/4) = 12 kBT

∴ 〈Evib〉 = kBTWe can conclude therefore that at high temperatures the vibra-

tional motion of a diatomic molecule (treated as a 1D oscillator) hasan average energy that is approximately 2( 1

2 kBT) = kBT distributed ofover the two vibrational degrees of freedom (1 k.e. and 1 p.e.)

7.6 Breakdown of equipartition

Most diatomic gases have nd = 5 NOT nd = 7. Rotation and vibrationare quantised,

At low T, all molecules are in the lowest rotation & vibration states→ molecule has nd = 3

Higher T: rotation important → nd = 5v. high T : vib. Important → nd = 7∴ vibrational and rotational degrees of freedom (“frozen out”) at

low tempsCv = heat capacity at constant vol. = (nd/2)NkB – (see section 3.3)We can estimate the temperature at which the degrees of freedom

become “unfrozen" using Boltzmann’s Law. From QM, the minimum

rotational energy for H2 is Ermin =h2

4π2 I, where the moment of inertia

I = 4.7× 10−48 kgm2.→ Ermin = 2.4× 10−21 J ≈ 0.01 eV.∴ the first excited state becomes significantly populated when

kBT > Ermin , i.e. for T > 170 K.Similarly, the minimum vibrational energy is given by the QM SHM(2nd yr QM) Evmin = hν where ν is a characteristic vibration fre-quency. For H2 , ν = 2.6× 1014

→ Evmin = 1.72× 10−19 J ≈ 1.1 eV.This becomes populated for T > 10, 000 K. In practise this is close toits dissociation temperature, and so at common temperatures, for H2,

structure of matter lecture notes - week 2 17

nd = 5.

structure of matter lecture notes - week 2 18

Appendix: Useful Standard Integrals

∫ ∞

0e−αx2

dx =12

(π

α

)1/2

∫ ∞

−∞e−αx2

dx =(π

α

)1/2

∫ ∞

0xe−αx2

dx =1

2α∫ ∞

−∞xe−αx2

dx = 0

∫ ∞

0x2e−αx2

dx =14

( π

α3

)1/2

∫ ∞

−∞x2e−αx2

dx =12

( π

α3

)1/2

∫ ∞

0x3e−αx2

dx =1

2α2∫ ∞

−∞x3e−αx2

dx = 0

∫ ∞

0x4e−αx2

dx =38

( π

α5

)1/2

∫ ∞

−∞x4e−αx2

dx =34

( π

α5

)1/2