Thermal Dynamics

PHYS321 Thermodynamics and Statistical Physics

Thermal Physics
Deals with a collection of a large number of particles “More is different!” --- P.W. Anderson It is effectively impossible to follow the motion and trajectory of each particle two approaches in thermal physics • Thermodynamics (macroscopic) • Statistical mechanics (microscopic) “Four fundamental pillars of our physical theory: general relativity, quantum mechanics, the theory of elementary particles and statistical mechanics… No single, coherent and clear understanding as to just what the theory of statistical mechanics is or ought to be.” --- Physics and Chance, L. Sklar

Thermodynamics
From the Greek thermos meaning heat (Latin thermae: public baths) and dynamis meaning power Phenomenological theory of matter -- Draws its concepts directly from experiment Macroscopic theory (i.e. it doesn’t contain information about atoms, molecules, …) Does not depend upon microscopic detail (e.g. 1st & 2nd laws, heat flow from “hot” to “cold”, maximum efficiency of engines) Historically, thermodynamics developed out of the need to increase the efficiency of early steam engines

Thermodynamics – widely applied, very fundamental and reliable
“Thermodynamics has something to say about everything but does not tell us everything about anything.” --- Goldstein, The Refrigerator and the Universe “A theory is the more impressive the greater simplicity of its premises is, the more different things it relates, and the more extended is its applicability. Therefore the deep impression which classical thermodynamics made upon me; it is the only theory of universal content concerning which convinced that, within the framework of the applicability of its basic concepts, will never be overthrown.” ---Albert Einstein

Statistical Mechanics and Its Relation to Thermodynamics
Mechanics about microscopic physics: classical / quantum mechanics

theoretical Statistical Mechanics
(probabilistic)

Statistics Assumption: each possible microstate has equal probability

Infinite-large limit, i.e. thermodynamic limit

Statistical Mechanics provides an underlying explanation of thermodynamics

Thermodynamics
(deterministic)

phenomenological

Chapter 1 Revision on Thermodynamics

Thermodynamic Systems heat Q = 0 work W = 0

Isolated system:
Particle number N, energy E are fixed

Q≠0 W ≠0

Closed system:
N fixed E changeable

Q≠0 W ≠0

Open system:
N changeable E changeable

Macroscopic View
Thermodynamics describes macroscopic systems which consist of a large number of small particles (usually of the order of 1023) It is impossible and not useful to keep track of the exact state of motion of all the particles Instead, one describes the system by a few macroscopic quantities that we are interested in

Example: Ideal Gas
Macroscopically, described by volume (V) and pressure (P) Microscopically, it consists of a lot of particles

Each point represents a system

P

V

1.1 The Zeroth Law of Thermodynamics
Thermal Equilibrium and Temperature

Thermal Equilibrium
The idea of temperature is related to thermal equilibrium When two macroscopic systems are in contact, they can exchange different quantities Mechanical equilibrium: Exchange volume → same P Diffusive equilibrium: Exchange particles → same ρ Thermal equilibrium: Exchange energy → same “T” If no exchange is observed ⇔ equilibrium In particular, if no energy exchange ⇔ thermal equilibrium
Is this a equilibrium state?
0°C constant gradient 100°C

The time to reach equilibrium ⇔ relaxation time

Zeroth Law of Thermodynamics
If two systems are both in thermal equilibrium with a third system, then the two systems are in thermal equilibrium with each other.
--- Ralph H. Fowler, 1920s adiabatic wall (thermally insulating)
A C B

Remove the wall, no heat flow
A C B

In thermodynamics, it is an experimental fact Statistical mechanics provides an explanation

Temperature --- A measure of thermal energy
0’th law implies that we can assign a number to different systems at equilibrium ⇒ temperature 0’th law in terms of temperature: Two systems are in thermal equilibrium if and only if their temperatures are equal. Can T be defined in a non-equilibrium system? Usually hot = high temperature cold = low temperature

Temperature Scale and Units
Choose something that changes with temperature Pick two convenient temperatures and assign them arbitrary numbers, e.g. water boiling (100) and freezing (0) Mark off a number of equally spaced intervals in between the two end points and extrapolate above and below The ideal gas thermometer extrapolates to P = 0 at -273.15°C. We define it as the zero point of the temperature scale.

Absolute Temperature Scale
Also known as Kelvin scale Kelvin: one of the seven basic SI units. It is defined as 1/273.16 of the temperature of the triple point 三相 点 of water (0.01°C). The triple point of water is chosen because it is easy to reproduce practically [oC] ≣ [K] − 273.15 Melting temperature of water at 1 atmosphere: 273.152519± 0.000002 [K] or 0.002519(2)oC Most equations are correct only in the Kelvin scale.

Lord Kelvin ( William Thomson )

1.2 Equipartition Theorem
Degree of Freedom, Ideal Gas Model and Bead-Spring Model

The first model of a many-particle system: the Ideal Gas
The ideal gas model --works well for gases at low densities all the particles are identical the particles are tiny compared to their average separation (point masses); the particles do not interact with each other; the particles obey Newton’s laws of motion, their motion is random; collisions between the particles and the container walls are elastic.

The Ideal Gas Law

PV = Nk BT or PV = nRT

P – pressure V – volume T – the temperature in Kelvin n – number of moles of gas N = nNA total number of particles NA – the Avogadro’s number ~ 6.022045 ×1023 the number of particles in one mole R = NAkB = 8.31JMol-1K-1 ideal gas constant kB = 1.381 ×10–23J/K Boltzmann constant

very large

Kinetic Theory
Temperature

?
Kinetic Energy (K.E.)

Assumptions Smooth walls ⇒ symmetrical bounces (like a mirror) Molecules behave like elastic billiard balls bouncing around ⇒ vx has a constant magnitude, although periodically changing direction Low density ⇒ each molecule moves as if it is alone, i.e. non-interacting

Kinetic Theory of Ideal Gas
The relationship between energy and temperature (for monatomic ideal gas) L ∆t = 2 ∆px = 2mvx L vx 2 ∆px mvx Favg = = ∆t L
For N molecules, multiply by N
N 1 N 2 2 P = i =1 = mi vix = mvx ∑ A AL i =1 V i ∑F

N

where

2 2 vx = vx

1 = N

∑v i =1

N

2 ix

〈…〉 or …

represents average

Kinetic Theory of Ideal Gas
Relating microscopic and macroscopic descriptions
2 PV = Nmvx is a microscopic description

(derived from laws of mechanics + some assumptions)
PV = Nk BT is a macroscopic thermodynamic

description (“derived” from experiment)
2 We can identify k BT = mv x for an ideal gas; this is a

microscopic interpretation of T
1 1 2 k BT = mvx So 2 2

is the average Kinetic Energy of the

translational motion in the x-direction

Kinetic Theory of Ideal Gas
Similarly for the motions in the y and z directions
1 1 2 1 2 1 k BT = mv x = mv y = mv z2 2 2 2 2

The average K.E.
1 2 1 3 2 2 2 K .E. = N mv = N m(vx + v y + vz ) = Nk BT 2 2 2

Therefore, we interpret temperature as a measure of the average kinetic energy of the molecules The internal energy U of a monatomic ideal gas is independent of its volume, and depends only on T (∆U = 0 for an isothermal等温 process, T = const).

Animation of Monoatomic Gases

At equilibrium, same T, same K.E.

larger M → smaller v

Comparison with Experiment dU/dT(300K) (J/K·mole) Monatomic Helium Argon Neon Krypton Diatomic H2 N2 O2 CO Polyatomic H20 CO2 27.0 28.5 20.4 20.8 21.1 21 12.5 12.5 12.7 12.3

3 U = Nk BT 2
Testable prediction: if we put energy dU into 1 mol gas, and measure the resulting change dT, we expect to get

3 dU 3 = N A k B = R = 12.5 J/K ⋅ mole dT 2 2
Conclusion: diatomic and polyatomic gases can store thermal energy in forms other than the translational kinetic energy of the molecules.

Degree of Freedom自由度
The degrees of freedom of a system are a collection of independent variables required to characterize the system Monoatomic particle: ( ) three quadratic terms ⇒ the degree of freedom f = 3 vx, vy and vz fully describe the motion diatomic molecules have more than three degrees of freedom – they rotate (and vibrate at high temperature)
1 K .E. = M x 2 + y 2 + z 2 2

Diatomic Molecule
Diatomic molecule:
K .E. = 1 1 M x 2 + y 2 + z 2 + I xφx2 + I zφz2 2 2

(

)

(

)

Ix – the moment of inertia for rotations around the x-axis

f = 3 translational + 2 rotational = 5 = 2x3-1 degrees of freedom i.e. 5 “channels” to store energy Rotation along y axis leave the molecule unchanged ⇒ do not count
Question: f of 5 diatomic molecules = ?

Polyatomic Molecule
At low temperature
K .E . = 1 1 M x 2 + y 2 + z 2 + I xφx2 + I yφ y2 + I zφz2 2 2

(

)

(

)

3 translation + 3 rotation = 6 = 3x3-3 degrees of freedom

Vibrational Degrees of Freedom
The one-dimensional vibrational motion has
1 1 2 U = m x + k x2 2 2

1 kinetic + 1 potential = 2 quadratic vibrational degrees of freedom For a diatomic molecule, 3 transtional + 2 rotational + 2 vibrational = 7 degrees of freedom. It is usually very difficult to stretch molecules. We say that the stretching “mode” is not normally “excited” at room temperature, i.e ‘frozen out’.

Equipartition Theorem 能均分定理
At temperature T, the average thermal energy of each quadratic degree of freedom is ½ kBT f degrees of freedom, f “channels” to store energy ⇒ capacity of energy (heat capacity C) ∝ f U= f·kBT/2, C = f·kB/2 "equipartition" means "equal division“ The “temperature” in each degree T energy of freedom equilibrates to the same value after many collisions not true in quantum statistical mechanics (which is important at low temperature) f=3 f=5

“Frozen” degrees of freedom
U /kBT
7/2N Vibration 5/2N Rotation 3/2N Translation 10 100 1000

one mole of H2 U(x) E4 E3 E2 E1 x

kBT

T, K

Quantum ⇒ discrete E levels, ∆E finite If the difference between energy levels ∆E >> kT, then a typical collision cannot cause transitions to the higher (excited) states and thus cannot transfer energy to this degree of freedom: it is “frozen out”. For H2, T > 200 K to “activate” the rotational degree of freedom T > 3000 K to “activate” the vibrational degree of freedom

Bead-Spring Model of a Solid
Each atom has 6 degree of freedom 3 for kinetic, 3 for potential energy For small vibrations, the potential energy is like that for a harmonic oscillator, i.e. quadratic Thermal energy U thermal
⎛1 ⎞ = N ⋅ 6 ⋅ ⎜ k BT ⎟ ⎝2 ⎠
U(r)
1 2

k ( r − r0 )

2

r This is true only at ‘high’ temperature. r atomic separation At low temperature, some of the higher energy modes will be “frozen out” due to quantum effects. This example shows that the equipartition theorem is very powerful and useful, but it is only strictly correct in the realm of classical physics and harmonic potentials.
0

1.3 The First Law of Thermodynamics
Heat, Work and Internal Energy

Heat and Work
Heat & work: two different forms of energy transfer Heat: the form of energy which is transferred from one object to another due to a difference of temperature between the objects. Work: transfer of energy to or from a system by a change of the parameters other than temperature. E.g., compressing the piston of the gas container, stir a cup of water, run a current through a resistor…

First Law of Thermodynamics
For a closed system

∆U = Q + W
Q: Heat added to the system W: Work done on the system ∆U: the change of the internal energy U Sign convention: Q and W are positive if energy flows into the system. This is really just the Law of Conservation of Energy !

Internal Energy Energy of all molecules including Random motion of individual molecules
K.E. of translational motion K.E. of vibrational motion in molecules and in lattices (for solid) Chemical energy in bonds and interactions

DOES NOT INCLUDE Macroscopic motion of object Potential energy due to interactions with other objects, e.g. mgh

Total E = mc2 includes ALL energy

Mechanical Equivalent of Heat
Heat is not a substance, but a dynamical form of mechanical effect. SI unit: Joule 1 calorie (cal) = amount of heat or work needed to raise 1g of water by 1°C 1 cal = 4.186 J Calorie in food 1 Cal = 1000 cal

James Prescott Joule (1818 – 1889) English physicist

Joule's apparatus for measuring the mechanical equivalent of heat.

Works
There are different kinds of work (e.g. mechanical, electrical, magnetic…) Different Macrovariables (not required) first law fluid surface dielectricslab dU = dQ − pdV dU = dQ + γdA dU = dQ + Edρ variables (state functions) p, V , T heat capacity C p , CV Cγ , C A Cε , Cq CE , Cρ C H , Cm

γ (sureface tension ), A(area), T ε (emf ), q(charge), T
E (electric field), ρ (polarization ), T

electrochmicalcell dU = dQ + εdq

paramagnetic rod dU = dQ + µ 0 Hdm µ 0 H (magnetic field), m(magnetization ), T

Work
We focus on the simplest form of mechanical work.

Assumptions for ideal gas compression:
No “thermodynamic friction” – Quasistatic: the motion of the piston is so slow that the system always has time to reach equilibrium No mechanical friction: the piston does work on the gas only; no work is wasted in overcoming friction

Work
If ∆V is small, P ≈ constant.
W = F ∆x = PA∆x = P(−∆V ) ∆V = V final − Vinitial < 0 ⇒ W >0 dV P

W = - PdV applies to any shape of boundary If P changes with V, P = P(V)

W = − ∫ PdV
Vi

Vf

W is the area under the curve

Problem
An ideal gas changes from A → B → C. Fill in the signs of Q, W, and ∆U for each step.
P
A

Step A→B T=const
B C

Q

W

∆U

B→C C→A V

Can this process be non-quasistatic?

Work and Heat are Process-dependent
The system going from (Pi,Vi) to (Pf,Vf) via two different paths : U is a state function, process independent ∆U = 3 P V − 3 PV
2
f f

( Pi ,Vi )
Path 1 P Path 2

2

i i

Work is process-dependent W1 = pi (V f − Vi ) ≠ p f (V f − Vi ) = W2 Heat is process-dependent
Q1 = ∆U − W1 ≠ ∆U − W2 = Q2

( P ,V ) f f

V

the first law: dU = dW + dQ The bar through d means that the infinitesimal work or heat is an inexact differential. It is path-dependent.

State Function
U, P, T, and V are the state functions. Q and W are not (process quantities).

P

2

1 T

V

Analogy: in classical mechanics, if a force is not conservative (e.g., friction), the initial and final positions do not determine the work, the entire path must be specified.

Equation of state: relationship between different state functions f ( P, V , T ,...) = 0 at equilibrium In equilibrium, just a few macroscopic parameters are required to describe the state of a system.

Ideal Gas Compression: Two Kinds of Process
Isothermal: T = constant
The system is in contact with a heat bath (reservoir) of fixed temperature T; AND the process must be slow enough that heat can be transferred to or from the reservoir much faster than work is being done on the gas.

Adiabatic: No heat can enter or leave the gas (Q = 0) ; the gas cylinder is perfectly insulated; OR the process happens so quickly that heat does not have a chance to leave or enter

Isothermal process
Work

W = − ∫ PdV = − NkT ∫
Vi

Vf

Vf

Vi

dV V

Vi = − NkT (ln V f − ln Vi ) = NkT ln Vf

⎛1 ⎞ Internal energy (equal partition) U = N ⋅ f ⋅ ⎜ kT ⎟ ⎝2 ⎠ ⎛1 ⎞ ∆U = N ⋅ f ⋅ ⎜ k∆T ⎟ = 0 since ∆T = 0 (isothermal) ⎝2 ⎠
First law ∆U = Q + W

Vi Heat change Q = −W = − NkT ln < 0 for compression Vf

Adiabatic Processes
0 First law ∆U = Q + W
NkT ⎛1 ⎞ ⎛1 ⎞ dU = Nf ⎜ kdT ⎟ and dW = − PdV ⇒ Nf ⎜ kdT ⎟ = − PdV = − dV V ⎝2 ⎠ ⎝2 ⎠ Vf f dT dV f Tf ⇒ =− ⇒ ln = − ln ⇒ V f T f f / 2 = ViTi f / 2 2 T 2 Ti Vi V or VT f / 2 = constant

∆U = W > 0 ⇒ ∆T > 0
⎛ PV ⎞ ⇒ V⎜ ⎟ = constant ⎝ Nk ⎠ ⇒ PV γ = constant f /2

(PVγ = constant)

Isotherms:

γ=

f +2 f

is the adiabatic exponent

PV = nRTf PV = nRTi

Heat Capacity
Q C= ∆T

Heat capacity

Q

is the ratio of the amount of heat added to a system to its corresponding temperature rise

C c= m

Specific heat capacity or Specific heat

m

∆T

is the heat capacity per unit mass

C c' = n

Molar heat capacity

is the heat capacity per moles (n number of mole)

Heat Capacity
C is NOT a state function (since Q is not a state function) Path dependent: ( isothermic: C = ∞, adiabatic: C = 0 ) The two most common definitions (paths) are CV heat capacity at constant volume

C p heat capacity at constant pressure
C can be calculated from the first law

C=

∆U − W ∆T

Heat Capacity at Constant Volume, CV
V is constant, W=0 ⎛ ∂U ⎞ ⎛ ∆U ⎞ CV = ⎜ ⎟ ⎟ =⎜ ⎝ ∆T ⎠V ⎝ ∂T ⎠V
P Tf Ti V

This means at constant volume

For an ideal gas Nfk B ⎛1 ⎞ U = Nf ⎜ k BT ⎟ ⇒ CV = 2 ⎝2 ⎠ 3 3 e.g. for monatomic gas f = 3 ⇒ CV = Nk B = nR 2 2

Heat Capacity at Constant Pressure, CP
⎛ ∆U + P∆V ⎞ ⎛ ∆U − W ⎞ CP = ⎜ ⎟ ⎟ =⎜ ∆T ⎠p ⎝ ∆T ⎠ p ⎝ ⎛ ∂V ⎞ ⎛ ∂U ⎞ =⎜ ⎟ + P⎜ ⎟ ⎝ ∂T ⎠ p ⎝ ∂T ⎠ p
This portion of the added heat cause the internal energy to increase This portion of the added heat is used to do work of expansion

P Tf Ti V

For solid C P > CV except for some rare materials with negative thermal expansion For an ideal gas

⎛ ∂U ⎞ ⎛ ∂U ⎞ ≈⎜ ⎜ ⎟ ⎟ ∂T ⎠ p ⎝ ∂T ⎠V ⎝

f +2 f +2 Nk ⎛ ∂V ⎞ C p = CV + P⎜ = Nk = nR ⎟ = CV + P 2 2 P ⎝ ∂T ⎠ p

Dulong-Petit Law
For solids and liquids, is small, so CV ≈ C P For a crystal, f = 6 (K.E. and P.E. in x, y and z)
6 CV = N k = 3 Nk = 3nR 2 c ' = C / n = 3R = 24.9 Jmol−1K −1
3R

⎛ ∂V ⎞ ⎜ ⎟ ∂T ⎠ p ⎝

This Dulong-Petit law is true for all crystals, provided the temperature is high (compared to the Debye temperature – every material has a different unique Debye temperature). At low temperature, quantum effects become important!

Specific Heat of Water 4186 J / (kg °C) What is the relevance of water’s high specific heat?

∆T = Q / (c m)

The high specific heat of water makes ocean temperatures quite stable, creating a favorable environment for marine life.

Specific Heat of Water At Hong Kong, the temperature difference between January and July is 13oC Riyadh (Saudi Arabia) : Surface covered by sand c = 800 J / (kg°C) Temperature difference between January and July is 30oC

1.4 Latent Heat and Enthalpy

Latent Heat
During a 1st order phase transition (e.g. ice ↔ water, water ↔ stream), the system takes in or releases heat, but the temperature remains constant. During the transition, latent heat is used to break chemical bonds rather than increasing the kinetic energy of atoms. at the phase transition temperature, the heat capacity diverges C = Q = Q = ∞
∆T 0
Diverging C is commonly used to identify phase transitions

Continuous phase transition has no latent heat

Latent Heat
Q Latent heat is defined as L ≡ m

i.e. the heat needed to transform the system from one phase to another phase In order that L is well defined, the convention is that
P is constant (usually 1 atm) no other work is done besides the mechanical work of constant pressure expansion or compression

For example Ice melting: L = 80cal/g Water boiling L= 540cal/g Compare to Water (0°C → 100°C) 100 cal/g

Problem You put 1 kg of ice at 0oC together with 1 kg of steam at 100oC. What is the final temperature?
LF = 80 cal/g, Lv = 540 cal/g cwater = 1 cal/g oC
1) 2) 3) 4) 5) between 0°C and 50oC 50°C between 50°C and 100°C 100°C greater than 100°C

Problem Which will cause more severe burns to your skin: 100°C water or 100°C steam?
1) water 2) steam 3) both the same 4) it depends...

Enthalpy (焓)
Isobaric processes (P = const): dU = Q - P∆V = Q -∆(PV) ⇒ Q = ∆ U + ∆(PV) Define enthalpy,

H = U + PV

The enthalpy is a state function, because U, P, and V are state functions. Useful for considering energy changes under constant pressure conditions, usually 1 atm.

Enthalpy
Under constant P environment, the energy input will serve to increase the internal energy do work against (atmospheric) pressure
P P P

P volume expands (dV) P

dH = dU + PdV

P energy input (heat or work) P P P P external pressure (P) P system

P P PdV is associated with this part

The values of ∆H under different conditions have been measured and can be found in standard chemistry tables

Enthalpy
More generally, W = PdV + Wother Usually, Wother = 0,

dH = dU + PdV = dU − dW = dQ
Hence, the heat capacity at constant pressure can also be expressed as, i.e. enthalpy capacity.

⎛ ∂H ⎞ CP = ⎜ ⎟ ⎝ ∂T ⎠ P

First Order Phase Transition

One mole water H Cp

T

T

Freezing/melting and boiling are 1st order phase transition, not continuous transition

Enthalpy

If you create the system under the constant pressure environment, then Eother = ∆H = U + PV = H Enthalpy is the amount of energy you have to supply to create a system with internal energy U and volume V in a constant pressure environment

Enthalpy
If V = const: Q = ∆ U If P = const: Q = ∆ H In both cases, Q is path independent since U and H are state function Consequence: the energy released (absorbed) in chemical reactions at constant volume (pressure) depends only on the initial and final states of a system. The enthalpy of an ideal gas: f ⎛f ⎞ H = U + PV = Nk BT + Nk BT = ⎜ + 1⎟ Nk BT 2 ⎝2 ⎠ depends on T only

Example: Boiling of Water
Standard chemistry table says that the change in enthalpy when you boil one mole of water at 1 atm is 40,660 J Not all this energy goes into the internal energy of the steam, part of it is used to make room for the steam Under a constant pressure, the work required to make room for the steam is PV = RT = (8.31 J/K) ∙373 K = 3100 J The initial volume of the liquid water is negligible

Example: Boiling of Water
Hence the increase in internal energy when one mole of liquid water turns into steam is 40660J- 3100J = 37560 J About 8% of the energy supplied is used to push the surrounding air

Example: Hydrogen Combustion
Burning of hydrogen to form water: H2+1/2 O2 H2O Enthalpy of formation for one mole of water: –286 kJ When one mole of water is formed in the above reaction, amount of energy released is 286 kJ Not all this amount of energy comes from the change of internal energy of the reactant and the product A small amount of it, about 4 kJ, is due to the work done by the atmosphere when it collapses to fill the space left behind by the consumed gases