Download August 30, 2016 Lecture 1: Thermodynamics vs. Statistical Mechanics

August 30, 2016
Lecture 1: Thermodynamics vs. Statistical Mechanics
Statistical mechanics is using a handful of quantities to describe a large number of particles. Its
macroscopic counterpart is thermodynamics. We will review the essential concepts and
mathematics of thermodynamics in the first two weeks.
References
Carter, Ch. 1-2
Griner, Neise, Stocker, Ch.1
Pathria, Ch. 1
Key words
Thermodynamics
State variables
chemical potential
Phase transition
Temperature
Ideal gas law
Points of discussion
1. Why thermodynamics is different from mechanics, electrodynamics and quantum
mechanics?
2. What can statistical mechanics explain but not in thermodynamics?
3. How do we understand the ideal gas law?
Short presentation topics
1. How to understand van der Waals equation of state?
Homework (Due 9/13/16)
1. (a) Draw a schematic diagram of isotherms (a PV diagram) of a van der Waals gas.
(b) Find a few values of a and b for different materials and discuss them.
(c) Get the critical values of vc, Tc and Pc in terms of parameter a and b. The critical point is
𝜕𝜕𝜕𝜕
𝜕𝜕2 𝑃𝑃
defined as the inflection point of an isotherm or 𝜕𝜕𝜕𝜕 = 𝜕𝜕2 𝑣𝑣 = 0.
(d) Show that in terms of the reduced quantities, 𝑣𝑣′ ≡ 𝑣𝑣⁄𝑣𝑣𝑐𝑐 , 𝑃𝑃′ ≡ 𝑃𝑃⁄𝑃𝑃𝑐𝑐 , 𝑇𝑇′ ≡ 𝑇𝑇⁄𝑇𝑇𝑐𝑐 , the van der
3
1
8
Waals equation can be written as �𝑃𝑃′ + 𝑣𝑣′2 � �𝑣𝑣 ′ − 3� = 3 𝑇𝑇′.
September 1, 2016
Lecture 2: Basic Thermodynamics
Equilibrium thermodynamics only describes the relations among system variables in equilibrium
states. These states can be reached via reversible or irreversible processes. Nonetheless the
changes of state variables are independent of the processes.
References
Carter, Ch. 1-2
Griner, Neise, Stocker, Ch.1-2
Pathria, Ch. 1
Key words
Van der Waals equation of state
Reversible and irreversible process
Open and closed system
Adiabatic and isothermal processes
Extensive and intensive variables
State quantities
Integrating factor
Work
Points of discussion
1. Van der Waals equation can describe 1st order phase transition.
2. Ideal gas laws hold quite well for real gases until very low temperatures.
3. Ideal gas law does not allow liquefaction of gases.
4. exact and inexact differtial
5. A loop integral on exact differentials will be 0 while non-zero for inexact differentials.
6. Thermodynamics laws
7. Thermodynamic equilibrium = thermal  mechanical  chemical equilibrium
8. Thermometers: gas, thermocouple, electric resistance
Short presentation topics
2. Calculate the number of molecules per cm3 at 25ºC for 650 torr (Logan pressure) and 1x10-10
torr (common ultrahigh vacuum system in a lab).
3. Is it always possible to find an integrating factor to make an inexact differential of two
variables into an exact differential?
Homework (Due 9/13/16)
2. Show that the following dz is not an exact differential. Find the integrating factor µ and check
that µdz is exact.
(a) 𝑑𝑑𝑑𝑑 = 2𝑦𝑦𝑦𝑦𝑦𝑦 + 3𝑥𝑥𝑥𝑥𝑥𝑥, (b) 𝑑𝑑𝑑𝑑 = 𝑦𝑦𝑦𝑦𝑦𝑦 + (2𝑥𝑥 − 𝑦𝑦 2 )𝑑𝑑𝑑𝑑, (c) 𝑑𝑑𝑑𝑑 = (𝑥𝑥 4 + 𝑦𝑦 4 )𝑑𝑑𝑑𝑑 − 𝑥𝑥𝑦𝑦 3 𝑑𝑑𝑑𝑑.
September 6, 2016
Lecture 3: 1st and 2nd Law of Thermodynamics
First law of thermodynamics equates change of internal energy of a system with work and heat in
or out of the system by either a reversible or an irreversible process. The second law, however,
inhibits certain energy conservation processes to happen.
References
Carter, Ch. 1-2
Griner, Neise, Stocker, Ch.1-2
Pathria, Ch. 1
Key words
Temperature
Entropy
Heat engine
efficiency
Carnot theorem
Kelvin-Planck statement of the 2nd law
Clausius statement of the 2nd law
Points of discussion
1. First law relates the internal energy (state quantity) with heat and work which are not state
quantities.
2. 𝛿𝛿𝛿𝛿 = 𝑇𝑇𝑇𝑇𝑇𝑇 is only true for reversible processes.
3. Carnot engine can be run irreversibly but only the reversible one will achieve the highest
efficiency.
4. If one can construct another engine running between 2 reservoirs with a higher efficiency than
the Carnot engine, one can hook them up to make a composite system that will violet either
the Planck-Kelvin or Clausius statement of the 2nd Law.
5. An intensive quantity 1/P and 1/T can be an integrating factor to make an inexact differential
δW and δQ an exact differential, dV and dS. (We will use δ from now on for inexact
differentiation)
6. Entropy change defined by δQ/T is only for reversible processes. In the case of free gas
expansion in an adiabatic system, there is no heat change but entropy change is not 0.
Homework (Due 9/13/16)
3. Show that the following dz is not an exact differential. Find the integrating factor µ and check
that µdz is exact.
(a) 𝑑𝑑𝑑𝑑 = 2𝑦𝑦𝑦𝑦𝑦𝑦 + 3𝑥𝑥𝑥𝑥𝑥𝑥, (b) 𝑑𝑑𝑑𝑑 = 𝑦𝑦𝑦𝑦𝑦𝑦 + (2𝑥𝑥 − 𝑦𝑦 2 )𝑑𝑑𝑑𝑑, (c) 𝑑𝑑𝑑𝑑 = (𝑥𝑥 4 + 𝑦𝑦 4 )𝑑𝑑𝑑𝑑 − 𝑥𝑥𝑦𝑦 3 𝑑𝑑𝑑𝑑.
4. Show that the coefficient of performance of a Carnot refrigerator operating between Tc and Th
is Tc/(Th-Tc).
5. An electric current of 1A flows for 10s in a resistor of resistance 25 ohm. The resistor is
submerged in a large volume of water, the temperature of which is 250K. What is the change
in the entropy of the resistor? Of the water?
September 8, 2016
Lecture 4: Reversible Processes and Legendre Transformation
Entropy increases in all spontaneous (irreversible) processes until the maximum is reached for
the equilibrium state while the energies (or potentials) are striving for minima. In addition to
internal energy, there are other energy forms which can be related to internal energy by Legendre
transformations.
References
Carter, Ch. 6, 8
Griner, Neise, Stocker, Ch.1-2
Pathria, Ch. 1
Key words
Reversible and irreversible process
available energy
Legendre transformation
Enthalpy
Helmholtz free energy
Gibbs free energy
Short presentation topics
4. What is the entropy change for a free expansion gas from V1 to V2?
Points of discussion
1. For an isolated system irreversible processes increase entropy while reversible processes
conserve entropy.
2. Thermal equilibrium is reached when entropy reaches its maximum.
3. Maximum work is obtained when a process takes place reversibly.
4. The 1st law applies to both reversible and irreversible processes. 𝑑𝑑𝑑𝑑 = 𝑇𝑇𝑇𝑇𝑇𝑇 − 𝑃𝑃𝑃𝑃𝑃𝑃 relates the
state variables between two neighboring equilibrium states.
5. The free expansion of ideal gases and the stirring in an isolated system serve as examples of
how the irreversible processes satisfies the 1st law.
6. Legendre transformation creates thermodynamic potentials by pairing the intensive variables
(T, P) and extensive variables (S, V) into various canonically conjugate pairs: (P,V) and (T,
S). The independent variables can be one from the two thermal variables (T, S) and one from
the two mechanical variables (P, V).
7. The independent variables for U is (S,V); enthalpy H(S,P), Helmholtz free energy A(T,V) and
Gibbs free energy G(T,P).
September 13, 2016
Lecture 5: Helmholtz and Gibbs Free Energy
Direction of chemical reactions depends on enthalpy and entropy change. For constant
temperature and pressure conditions, it is the Gibbs free energy that dictates the direction of
chemical reactions. Gibbs free energy is a sum of all chemical potentials.
References
Carter, Ch. 8
Griner, Neise, Stocker, Ch.3-4
Pathria, Ch. 1
Key words
Spontaneity of chemical reactions
Euler’s equation
Gibbs-Duhem relation
Phase diagram
Clausius-Clapeyron equation
Points of discussion
1. The geometric meaning of Legendre transformation is to replace using coordinates of points to
describe a curve by lines tangent to the curve where the lines are defined by slopes and
intercepts.
2. At constant temperature and temperature, the Helmholtz free energy is at minimum at
equilibrium.
3. At constant temperature and pressure, the Gibbs free energy is at minimum at equilibrium.
4. Assuming 𝑈𝑈(𝑆𝑆, 𝑉𝑉, 𝑁𝑁𝑖𝑖 ) is a homogeneous function of the first order, then 𝑈𝑈 = 𝑇𝑇𝑇𝑇 − 𝑃𝑃𝑃𝑃 +
∑𝑖𝑖 𝜇𝜇𝑖𝑖 𝑁𝑁𝑖𝑖 . Note that this is different from the first law of thermodynamics.
6. Gibbs-Duhem relates intensive variables.
7. We can calculate change of state functions of an irreversible process by a reversible process.
8. Chemists set many of the formation Helmholtz and Gibbs free energy to be 0 at 298K, 760
torr. Then entropy is, however, not necessarily 0 even for molecules at these conditions.
9. Particles move from higher chemical potential to lower one.
Homework (Due 9/27/16)
𝑁𝑁 2
1. Given Van der Waals equation as �𝑝𝑝 + �𝑉𝑉 � 𝑎𝑎� (𝑉𝑉 − 𝑁𝑁𝑁𝑁) = 𝑁𝑁𝑁𝑁𝑁𝑁, show that the (a) Internal
𝑇𝑇
1
1
energy 𝑈𝑈(𝑉𝑉, 𝑇𝑇) = 𝑈𝑈0 (𝑉𝑉0 , 𝑇𝑇0 ) + ∫𝑇𝑇 𝐶𝐶𝑣𝑣 (𝑇𝑇)𝑑𝑑𝑑𝑑 − 𝑁𝑁 2 𝑎𝑎 �𝑉𝑉 − 𝑉𝑉 � and (b) Entropy 𝑆𝑆(𝑉𝑉, 𝑇𝑇) =
𝑇𝑇
0
𝑉𝑉−𝑁𝑁𝑁𝑁
𝑆𝑆0 (𝑉𝑉0 , 𝑇𝑇0 ) + 𝐶𝐶𝑉𝑉 𝑙𝑙𝑙𝑙 𝑇𝑇 + 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑉𝑉 −𝑁𝑁𝑁𝑁
0
0
0
September 15, 2016
Lecture 6: Phase Diagrams and Maxwell Construction
Van der Waals equation is the first attempt to include attractive forces between gas molecules
into the equation of state of gas. Not surprisingly at some temperature and pressure, the forces
will condense the vapor into liquid; i.e., phase transition. The ratio of 𝑷𝑷𝒄𝒄 𝒗𝒗𝒄𝒄 and 𝑹𝑹𝑹𝑹 turns out to
be a universal number for gases.
References
Griner, Neise, Stocker, Ch.3
Key words
Maxwell construction
Liquid saturation and vapor saturation curves
critical point drying
Gibbs phase rule
Points of discussion
1. Calculated vapor pressure of a liquid with simplified assumptions.
2. When in equilibrium, the temperature, pressure and chemical potential of each species in the
system are equal.
3. The three intensive variables T, P, µ are functions of extensive variables (S, V, N).
4. The degrees of freedom described in the Gibbs phase rule are the number of intensive
variables to define a system. The extensive variables are r+2 where r is the number of species
in the system. The extensive variable number is independent of the number of phases. For
example, for a water in a bottle, the extensive variables are S, V, N. If at room temperature,
there are only 2 phases, so the only intensive variable to define the system is T. The pressure
is fixed by temperature.
5. Gibbs phase rule describes the (P, T) degree of freedom but not V. Thus a triple point in a P-T
diagram can be a triple line when V is included.
6. 4He liquid can go through a phase transition to become superfluid when all He atoms have the
same wavefunction and behave in a coherent manner. This is a macroscopic manifestation of
the quantum phenomenon.
7. The breakdown of Van der Waals equation by having positive dP/dV implies phase transition.
8. The liquid and vapor saturation curve bound the liquid-vapor mixture region in the PV phase
diagram.
9. During the vapor to liquid phase transition at a constant temperature, the pressure remains
constant while the volume reduces. This is accomplished by decreasing the number of
molecules in gas phase and increasing the number of molecules in liquid phase.
10. The constant pressure during isothermal phase transition is determined solely by temperature
not by volume.
11. Maxwell construction can determine the constant pressure line during the phase transition at
a particular temperature.
12. Beyond critical point temperature, isothermal processes cannot induce phase transitions
between liquid and vapor. The two fluid phases cannot be distinguished.
13. One can use an isothermal compression in the liquid-vapor mixture region and isothermal
expansion in the vapor region to achieve critical point drying which is commonly used for
preparing biological samples for high vacuum electron microscopy and the fabrication of freestanding microscopic membranes.
Homework (Due 9/27/16)
2. Using the ideal gas entropy expression of Prob. 2, show that the (a) 𝜇𝜇(𝑝𝑝, 𝑇𝑇) = 𝜇𝜇0 (𝑝𝑝0 , 𝑇𝑇0 ) −
𝑇𝑇 5⁄2 𝑝𝑝0
𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 ��𝑇𝑇 �
5𝑘𝑘
𝑆𝑆
0
5
� 𝑝𝑝 �� + �2 − 𝑠𝑠0 � 𝑘𝑘(𝑇𝑇 − 𝑇𝑇0 ). (b) Using Euler’s equation, show that 𝜇𝜇0 (𝑝𝑝0 , 𝑇𝑇0 ) =
� 2 − 𝑁𝑁0 � 𝑇𝑇0 . From (a) and (b) it is easy to see that the chemical potential of an ideal gas is
𝜇𝜇
𝑇𝑇 5⁄2 𝑝𝑝0
𝜇𝜇(𝑝𝑝, 𝑇𝑇) = 𝑘𝑘𝑘𝑘 �𝑘𝑘𝑘𝑘0 − 𝑙𝑙𝑙𝑙 ��𝑇𝑇 �
0
0
� 𝑝𝑝 ���.
3. Using the exact differentials of E, H, G and F, derive the Maxwell relations:
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
(a) �𝜕𝜕𝜕𝜕 � = �𝜕𝜕𝜕𝜕� (b) �𝜕𝜕𝜕𝜕 � = − �𝜕𝜕𝜕𝜕� (c) �𝜕𝜕𝜕𝜕� = − �𝜕𝜕𝜕𝜕 � (d) �𝜕𝜕𝜕𝜕� = � 𝜕𝜕𝜕𝜕 �
𝑉𝑉
𝑇𝑇
𝑉𝑉
𝑆𝑆
𝑇𝑇
𝑃𝑃
𝑆𝑆
𝑃𝑃
4. Using Maxwell relations and S(T,V) and S(T,P), derive the two TdS equations:
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
(a) 𝑇𝑇𝑇𝑇𝑇𝑇 = 𝐶𝐶𝑉𝑉 𝑑𝑑𝑑𝑑 + 𝑇𝑇 �𝜕𝜕𝜕𝜕 � 𝑑𝑑𝑑𝑑 (b) 𝑇𝑇𝑇𝑇𝑇𝑇 = 𝐶𝐶𝑃𝑃 𝑑𝑑𝑑𝑑 − 𝑇𝑇 �𝜕𝜕𝜕𝜕 � 𝑑𝑑𝑑𝑑
𝑉𝑉
𝑅𝑅𝑅𝑅
𝑃𝑃
𝑎𝑎
5. The Dieterici equation of state is 𝑃𝑃 = 𝑣𝑣−𝑏𝑏 𝑒𝑒 −𝑎𝑎⁄𝑅𝑅𝑅𝑅𝑅𝑅 . Show that (a) 𝑃𝑃𝑐𝑐 = 4𝑒𝑒 2 𝑏𝑏2 , 𝑣𝑣𝑐𝑐 = 2𝑏𝑏, and
𝑎𝑎
𝑇𝑇𝑐𝑐 = 4𝑅𝑅𝑅𝑅 . (b) Find
𝑃𝑃𝑐𝑐 𝑣𝑣𝑐𝑐
𝑅𝑅𝑇𝑇𝑐𝑐
and compare with experimental values of real gases.