1. Thermal Physics
Thermodynamics: Relates heat + work with empirical (observed, not derived)
properties of materials (e.g. ideal gas: PV = nRT).
2) Statistical Mechanics: Uses models (can be more complicated) of the molecules
to consider how their properties are distributed (as functions of temperature)
and then relates averages of these to measured properties (e.g. pressure).
a) Kinetic Theory: Calculates properties (e.g. pressure) of a gas in terms of a
very simple mechanical model.

2. The Ideal Gas
before
after
Kinetic Theory: Consider molecules (mass m0) inside a cylinder of length L and area A (V = AL) with a piston at one end which can move in x-direction.
When a molecule hits the piston, it has an elastic collision, so that px = -2m0vx .
If the collision lasts for a time tcollision, the force of the molecule on the piston is
Fon piston = (- the force of the piston on the molecule) = 2m0vx / tcollision .
The molecule will return to the piston after time ttravel = 2L/vx, so the time average force of that molecule on the piston is
<F(one molecule)on piston> = [2m0vx / tcollision ] [tcollision / ttravel] = m0vx2/L

3. <F(one molecule)on piston> = m0vx2/L
average over time and particles.
(Text’s notation: <vx2> = vx2)
<F(one molecule)on piston> = m0vx2/L
If there are N molecules in the cylinder, <Fon piston> = Nm0<vx2>/L
Pressure = <Fon piston> /A = Nm0<vx2>/(AL) = Nm0<vx2>/V
PV = Nm 0<vx2>
But <vx2> = <vy2> = <vz2> = 1/3 [<vx2> + <vy2> + <vz2>] = 1/3 <v2>
and average translational kinetic energy <Ktrans> = (m/2) <v2>
PV = (2/3) N <Ktrans>
but PV = NkBT
<Ktrans> = (3/2) kBT
vrms = <v2>1/2 = [(2/m) <Ktrans>]1/2 = [3kBT/m0]1/2
kB  R/NA, and NAm0  M (molecular weight)  vrms = [3RT/M]1/2
Avogadro’s number

4. This is an example of the Equipartition of Energy Theorem:
<Ktrans> = (m0/2) [<vx2> + <vy2> + <vz2>] = (3/2) kBT
This is an example of the Equipartition of Energy Theorem:
The average (i.e. thermal) value of the energy for each “classical”, “quadratic” degree of freedom = ½ kBT.
“Quadratic”: The energy associated with that degree of freedom is a quadratic function of a coordinate or a velocity.
i.e. Ktrans involves 3 quadratic degrees of freedom (vx, vy, and vz).
b) “Classical”: kBT >>  = the quantum of energy needed to go from the ground state to
the first excited state.
[In quantum mechanics, energy is not a continuous variable but is quantized, i.e. occurs
in steps. The classical limit means that there is enough thermal energy (kBT) to put the
molecules in lots of excited states. Note that (translation)  1/V2/3, so
(translation)  0 and K is “classical” for macroscopic volumes.]

5. vrms = [3RT/M]1/2

6. <Ktrans> = (3/2) kBT
Consider an ideal, monoatomic gas (e.g. He, Ne, Ar, …). If the atom is modeled as a “point particle”, i.e. with no internal structure,* then the only energy it can have is Ktrans:
Eint = 3/2 NKBT = 3/2 nRT
i.e. the energy needed to increase the temperature of 1 mole of monoatomic gas by T: Eint = 3/2 R T
If the volume is fixed, no work is done (W = - pdV), so this energy goes in as heat:
CV = (1/n) QV/T = 3/2 R
* This is a good approximation if <Ktrans> << 1 eV = 1.6 x J (a typical energy to excite an electron in the molecule)  T << (2/3) 1.6 x J / (1.38 x J/K)  8000 K.
molar specific heat
at constant volume
(for a monoatomic ideal gas)
natural unit for molar specific heat

7. <vrms> = [3RT/M]1/2
Statistical Mechanics Although vrms is a typical speed of the molecule, there is a wide distributions of speeds. This was first addressed by Boltzman, who suggested that the probability of a particle having energy E is given by:*
P(E) = n0(E) exp (-E/kBT)
**Here, n0 may also be a function of energy; it is defined such that P(E) dE = n0(E) exp(-E/kBT) dE = 1, where the integral goes from zero to infinity. [n0(E) is proportional to the “degeneracy of the energy”, the number of ways a particle can have energy E. For example, for a monoatomic gas where E = Ktrans, n0 is proportional to number of ways of arranging the velocity (v = vxi+vyj+vzk) such that ½ m0v2 = E.]
kB = Boltzman constant = R/NA = 1.38 x J/K
____________________________________________________________________
* The Boltzman relation is proven in statistical mechanics, for which T is the ideal gas temperature defined by T  PV/N for a non-interacting (i.e. dilute) gas.
** Note that this definition of n0 should replace the incorrect definition below Eqtn in the text, and we are using probabilities instead of number densities, nV.

8. P(E) = n0 exp (-E/kBT)
Before looking at the monoatomic gas problem, consider a simpler problem:
Example 21.4: Suppose the electrons in an atom can be in discrete energy levels. What is the relative probability of an electron being in an “excited state” with energy E2 = 1.50 eV above the ground state vs. being in the ground state (E1=0) if the temperature is 2500 K. [Assume that each molecule only has one ground state (E1=0) and one excited state at energy E2.]
Because each molecule has the same number of E2 and E1 states, n0(E2)=n0(E1) Then P(E2)/P(E1) = exp(-E2/kBT)/exp(-E1/kBT) = exp[-(E2-E1)/kBT]
P(E2)/P(E1) = exp{-1.5 eV (1.6 x J/eV)/[(1.38 x J/K) (2500 K)]}
= e-6.96 = 9.5 x 10-4
T (K)
(E2-E1)/kBT
P(E2)/P(E1)
30,000
0.58
5.6 x 10-1
10,000
1.74
1.8 x 10-1
3000
5.8
3.0 x 10-3
1000
17.4
2.8 x 10-8
300 58
6.6 x 10-26
100
174
3.0 x 10-76
The probability of being in the excited state becomes (exponentially) small at low temperature (kBT << E2-E1).
The relative probability  1 as T   (i.e. kBT >> E2-E1).

9. P(E) = n0(E) exp (-E/kBT) is the probability of having energy E.
Now let’s specialize to E = Ktrans = ½ m0v2, so specifying E is same as specifying v.
Then P(v)  Nv/N, where Nv is the number of particles that have
speeds between v and v+dv
Nv/N = P(v) = n0(v) exp(-mv2/2kBT).
Counting the number of ways of arranging vector velocity to get
same speed: n0(v) = A v2, where A = constant (indpt. of v,
but may be a function of T).
To determine A, set  dv [n0(v) exp(-m0v2/2kBT)] = A  dv [v2 exp(-m0v2/2kBT)] = 1
Let η  v (m0/2kBT)1/2  A = (m0/2kBT)3/2 / dη [η2 exp (-η2)]
= (m0/2kBT)3/2 /(1/2 / 4)
 P(v) = 4 (m0/2kBT)3/2 v2 exp(-m0v2/2kBT)
Nv = 4N (m0/2kBT)3/2 v2 exp(-m0v2/2kBT)

10. Nv = 4N (m0/2kBT)3/2 v2 exp(-m0v2/2kBT)
Nitrogen: N2
(m0 = 28 u)

11. Nv = 4N (m0/2kBT)3/2 v2 exp(-m0v2/2kBT)
Does vrms = (3RT/M)1/2 = (3kBT/m0)1/2 as desired?
[M/m0 = NA = R/kB]
vrms2  1/N dv (Nv v2) = 4 (m0/2kBT)3/2 dv v4 exp(-m0v2/2kBT)
Let η  v (m0/2kBT)1/2
vrms2 = 4 (m0/2kBT)3/2 (2kBT/m0)5/2  dη η4 exp(-η2)
vrms2 = (4/1/2) (2kBT/m0) (31/2/8) = (3kBT/m0) 

12. Nv = 4N (m0/2kBT)3/2 v2 exp(-m0v2/2kBT)
vrms = (3kBT/m0)1/2
Similarly, find average (mean) speed:
vavg  1/N dv (Nv v) = 4 (m0/2kBT)3/2 dv v3 exp(-m0v2/2kBT)
Let η  v (m0/2kBT)1/2
vavg = 4 (m0/2kBT)3/2 (2kBT/m0)2  dη η3 exp(-η2)
vavg = (4/1/2) (2kBT/m0)1/2 (1/2) = (8kBT/m0 )1/2
Most probable speed is the peak of the curve: dNv/dv = 0
2vmp exp[] = m0/kBT vmp3 []
vmp = (2kBT/m0)1/2

13. vmp = (2kBT/m0)1/2 = (2RT/M)1/2 vavg = (8kBT/m0 )1/2 = (8RT/M)1/2
vrms = (3kBT/m0)1/2 = (3RT/M)1/2
Calculate these speeds for nitrogen (N2) at T = 300 K:
M = 28 x 10-3 kg/mole
(RT/M)1/2 = [8.314 J/molK) (300 K) / (28 x 10-3 kg/mole)]1/2 = 298 m/s
vmp = 2 (RT/M)1/2 = 422 m/s
vavg = (8/) (RT/M)1/2 = 476 m/s
vrms = 3 (RT/M)1/2 = 517 m/s

14. vmp = (2kBT/m0)1/2 = (2RT/M)1/2 vavg = (8kBT/m0 )1/2 = (8RT/M)1/2
vrms = (3kBT/m0)1/2 = (3RT/M)1/2
Calculate these speeds for nitrogen (N2) at T = 300 K:
M = 28 x 10-3 kg/mole
(RT/M)1/2 = [8.314 J/molK) (300 K) / (28 x 10-3 kg/mole)]1/2 = 298 m/s
vmp = 2 (RT/M)1/2 = 422 m/s
vavg = (8/) (RT/M)1/2 = 476 m/s
vrms = 3 (RT/M)1/2 = 517 m/s
Calculate these speeds for hydrogen (H2) at 300 K:
Since M = 2 x10-3 kg/mole, the speeds should be (28/2)1/2 = 3.74 times larger than for N2.
vmp = m/s
vavg = m/s
vrms = m/s

15. Nitrogen (N2) at T = 300 K:
M = 28 x 10-3 kg/mole
vmp = 2 (RT/M)1/2 = 422 m/s
vavg = (8/) (RT/M)1/2 = 476 m/s
vrms = 3 (RT/M)1/2 = 517 m/s
Hydrogen (H2) at 300 K:
Since M = 2 x10-3 kg/mole
vmp = m/s
vavg = m/s
vrms = m/s
T = 300 K
Note that the escape velocity from earth’s gravity = 11.2 km/s. A much larger fraction of hydrogen molecules than nitrogen can escape at any time. Since the speed of a molecule fluctuates with time within the distribution, eventually “all” the hydrogen will leave the atmosphere (if it is not replenished).

16. Molar Specific Heat of Ideal Gases
Since Q depends on process, C  dQ/dT also depends on process.
Define a) molar specific heat at constant volume:
CV  (1/n) dQ/dT for constant V process.
b) molar specific heat at constant at constant pressure:
CP = (1/n) dQ/dT for constant P process.
Consider constant V process: W = 0 and Q = Eint .
Q = n CV dT (= nCV T if CV = constant)
Therefore Eint = nCV dT
Since Eint/n only depends on temperature end points (“Joule effect”),
Eint = nCV dT for any process.
Consider constant P process:.
Q = nCP dT and W = -PdV Since P = constant, dV = nR dT/P,
 dW = -  P (nR dT/P) = -nRdT
Q = Eint – W
nCP dT = nCV dT + nRdT  CP = CV + R (for any ideal gas)

17. CP = CV + R (any ideal gas)
 =1 + R/CV
e.g. for monoatomic ideal gas in 3-dimensions: CV = 3/2 R, CP = 5/2 R,  = 5/3
By equipartition theorem:
for monoatomic ideal gas in 2-dimensions, CV = 2/2 R, CP = 4/2 R,  = 2
for monoatomic ideal gas in 1-dimension, CV = 1/2 R, CP = 3/2 R,  = 3