Specific heat of gases

Specific heat of monatomic gas I

Add heat keeping the volume constant. Energy can only go into kinetic energy of atoms:

Conclusion:

TnRTnC

UQ

V

23

RCV 23

Specific heat of a monatomic gas II

Add heat keeping the pressure constant. Energy can go into kinetic energy of atoms, or into work when the gas expands:

Some mathematical gymnastics:

VpTnCTnC

WUQ

Vp

constant when pTp

nRVT

pnR

V

Specific heat of a monatomic gas III

Substitute:

Conclusion:

Valid for all ideal gases

TnRTnCTnC

VpTnCTnC

Vp

Vp

RCC Vp

Equipartition of energy

According to classical mechanics, molecules in thermal equilibrium have an average energy of associated with each independent degree of freedom of their motion provided that the expression for energy is quadratic.

monatomic gas: x, y, z motion: since the terms are quadratic

kT21

kTkTU23

213

2212

212

21 ,, zyx mvmvmv

Degrees of freedom for molecules

We take m to be the mass of the molecule, then as for monatomic gases

However, there is now internal energy due to rotation and vibration:

kTmv 232

21

kTU f2

Diatomic molecule: rotation

Think of molecule as dumbbell:

Centre-of-mass (CM) moves with K.E.

Rotation about CM with K.E.

All quadratic: f = 6

2212

212

21

zyx MvMvMv

2332

12222

12112

1 III

Diatomic molecule: vibration

Atoms attract each other via Lennard-Jones potential

Atoms stay close to equilibrium

Can be approximated by a harmonic oscillator potential 2

021 )( rrkV

0.10 0.12 0.14 0.16 0.18 0.20

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Lennard-Jones harmonic oscillator

V (

eV)

r (nm)

Harmonic oscillator

Average P.E. = average K.E.

P.E. is quadratic so associated with it

K.E. is the same so also

Total: kT or f = 2.

202

1 )( rrkV kT21

kT21

Diatomic molecules: conclusion?

According to classical physics there should be 8 degrees of freedom for each diatomic molecule…

From Cp and CV measurements for H2: below 70 K: f = 3 at room temperature: f = 5 above 5000 K: f = 7

Question

If e.g. vibration of H2 does not contribute to the specific heat, then that is because

a) The vibration is independent of temperature

b) The vibration doesn’t change in collisions

c) The vibration is like a very stiff spring

Quantum mechanical effects

Translation: 3 degrees

Rotation: 2 degrees

Vibration: at room temperature often also “frozen out” at high temperatures K.E. + P.E. = 2 degrees

Rotation about this axis is “frozen out”

Rotation about these axes imparts energy

Rotational states

Quantum mechanics: rotational and vibrational energies not continuous but quantised

Rotation: spacing ~1/I small steps, near-continuous about two axes big steps about axis through atoms If step >> kT : not excited, no rotation Even at T=104 K no rotation about third axis

Vibrational states

Vibration: harmonic oscillator Spacing ~1/m step >> kT for light molecules (O2, N2)

but excitations possible for heavy molecules such as I2

Potential energy: on average equal to kinetic energy

vibration contributes kT when it does

Solids

Model for solid:

Elemental (“monatomic”) solids have K.E. and P.E. to give C = 3R (Rule of Dulong and Petit)

RT23

RT23

Freezing out - again

At lower temperatures, some vibrational levels cannot be reached with kT energy. At very low temperatures C T3.

For metals the electrons (which can move freely through the metal) contribute and C = aT + bT3 (Einstein-Debye model)

Specific molar heat

Dulong & Petit: elemental solids C 25 J mol-1 K-1

Question 2

A room measures 4.002.402.40 m3. Assume the “air molecules” all have a velocity of 360 ms-1 and a mass of 510-26 kg. The density of air is about 1 kg m-3. Calculate the pressure in this room.

Isothermal expansion of an ideal gas

Isothermal expansion: pressure drops as volume increases since pV = nRT = constant

The internal energy only depends on the temperature so it doesn’t change

Two equations hold at the same time:

pV = nRT and T2 = T1

Question

An identical volume of the same gas expands adiabatically to the same volume. The pressure drops

a) more than in the isothermal process

b) by the same amount

c) less than in the isothermal process

Adiabatic expansion of an ideal gas I

Q = 0 so U + W = 0

Remember: U = nCV T

Use ideal gas law:

Substitute:

VV

nRTVpW

0 VV

nRTTnCV

Adiabatic expansion of an ideal gas II

Divide by nCV T:

Prepare for integration:

Define so that

0 0 VV

CR

TT

VV

nRTTnC

VV

0dd

becomes 0 V

VCR

TT

VV

CR

TT

VV

V

p

C

C 1

V

Vp

V C

CC

CR

Adiabatic expansion of an ideal gas III

Integrate:

Play around with it:

Use

constantln)1(ln 0d

)1(d VT

VV

TT

constant (another)

constantln1

1

TV

TV

nRTpVpVnRpV

T whileconstant :

Question

Consider isothermal and adiabatic expansion of a Van der Waals gas. Do T2 = T1 and pV= C hold for this gas?

a) yes; yes

b) yes; no

c) no; yes

d) no; no

Question

A gas is compressed adiabatically. The temperature

a) rises because work is done on the gas

b) rises depending on how much heat is added

c) drops because work is done by the gas

d) drops depending on how much heat is added

PS225 – Thermal Physics topicsThe atomic hypothesisHeat and heat transferKinetic theoryThe Boltzmann factorThe First Law of ThermodynamicsSpecific HeatEntropyHeat enginesPhase transitions