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