* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download August 30, 2016 Lecture 1: Thermodynamics vs. Statistical Mechanics
Superconductivity wikipedia , lookup
Marcus theory wikipedia , lookup
Degenerate matter wikipedia , lookup
Eigenstate thermalization hypothesis wikipedia , lookup
Chemical potential wikipedia , lookup
Spinodal decomposition wikipedia , lookup
Heat transfer physics wikipedia , lookup
Temperature wikipedia , lookup
State of matter wikipedia , lookup
Statistical mechanics wikipedia , lookup
Glass transition wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Equilibrium chemistry wikipedia , lookup
Maximum entropy thermodynamics wikipedia , lookup
Vaporβliquid equilibrium wikipedia , lookup
Transition state theory wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Gibbs paradox wikipedia , lookup
Equation of state wikipedia , lookup
Non-equilibrium thermodynamics wikipedia , lookup
Van der Waals equation wikipedia , lookup
History of thermodynamics wikipedia , lookup
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 ππππ ππ2 ππ defined as the inflection point of an isotherm or ππππ = ππ2 π£π£ = 0. (d) Show that in terms of the reduced quantities, π£π£β² β‘ π£π£βπ£π£ππ , ππβ² β‘ ππβππππ , ππβ² β‘ ππβππππ , the van der 3 1 8 Waals equation can be written as οΏ½ππβ² + π£π£β²2 οΏ½ οΏ½π£π£ β² β 3οΏ½ = 3 ππβ². 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 ο mechanical ο 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) ππππ = 2π¦π¦π¦π¦π¦π¦ + 3π₯π₯π₯π₯π₯π₯, (b) ππππ = π¦π¦π¦π¦π¦π¦ + (2π₯π₯ β π¦π¦ 2 )ππππ, (c) ππππ = (π₯π₯ 4 + π¦π¦ 4 )ππππ β π₯π₯π¦π¦ 3 ππππ. 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. πΏπΏπΏπΏ = ππππππ 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 Ξ΄W and Ξ΄Q an exact differential, dV and dS. (We will use Ξ΄ from now on for inexact differentiation) 6. Entropy change defined by Ξ΄Q/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) ππππ = 2π¦π¦π¦π¦π¦π¦ + 3π₯π₯π₯π₯π₯π₯, (b) ππππ = π¦π¦π¦π¦π¦π¦ + (2π₯π₯ β π¦π¦ 2 )ππππ, (c) ππππ = (π₯π₯ 4 + π¦π¦ 4 )ππππ β π₯π₯π¦π¦ 3 ππππ. 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. ππππ = ππππππ β ππππππ 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 Eulerβs 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 ππ(ππ, ππ, ππππ ) is a homogeneous function of the first order, then ππ = ππππ β ππππ + βππ ππππ ππππ . 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) ππ 2 1. Given Van der Waals equation as οΏ½ππ + οΏ½ππ οΏ½ πποΏ½ (ππ β ππππ) = ππππππ, show that the (a) Internal ππ 1 1 energy ππ(ππ, ππ) = ππ0 (ππ0 , ππ0 ) + β«ππ πΆπΆπ£π£ (ππ)ππππ β ππ 2 ππ οΏ½ππ β ππ οΏ½ and (b) Entropy ππ(ππ, ππ) = ππ 0 ππβππππ ππ0 (ππ0 , ππ0 ) + πΆπΆππ ππππ ππ + ππππππππ ππ βππππ 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 π·π·ππ ππππ and πΉπΉπΉπΉ 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) ππ(ππ, ππ) = ππ0 (ππ0 , ππ0 ) β ππ 5β2 ππ0 ππππππππ οΏ½οΏ½ππ οΏ½ 5ππ ππ 0 5 οΏ½ ππ οΏ½οΏ½ + οΏ½2 β π π 0 οΏ½ ππ(ππ β ππ0 ). (b) Using Eulerβs equation, show that ππ0 (ππ0 , ππ0 ) = οΏ½ 2 β ππ0 οΏ½ ππ0 . From (a) and (b) it is easy to see that the chemical potential of an ideal gas is ππ ππ 5β2 ππ0 ππ(ππ, ππ) = ππππ οΏ½ππππ0 β ππππ οΏ½οΏ½ππ οΏ½ 0 0 οΏ½ ππ οΏ½οΏ½οΏ½. 3. Using the exact differentials of E, H, G and F, derive the Maxwell relations: ππππ ππππ ππππ ππππ ππππ ππππ ππππ ππππ (a) οΏ½ππππ οΏ½ = οΏ½πππποΏ½ (b) οΏ½ππππ οΏ½ = β οΏ½πππποΏ½ (c) οΏ½πππποΏ½ = β οΏ½ππππ οΏ½ (d) οΏ½πππποΏ½ = οΏ½ ππππ οΏ½ ππ ππ ππ ππ ππ ππ ππ ππ 4. Using Maxwell relations and S(T,V) and S(T,P), derive the two TdS equations: ππππ ππππ (a) ππππππ = πΆπΆππ ππππ + ππ οΏ½ππππ οΏ½ ππππ (b) ππππππ = πΆπΆππ ππππ β ππ οΏ½ππππ οΏ½ ππππ ππ π π π π ππ ππ 5. The Dieterici equation of state is ππ = π£π£βππ ππ βππβπ π π π π π . Show that (a) ππππ = 4ππ 2 ππ2 , π£π£ππ = 2ππ, and ππ ππππ = 4π π π π . (b) Find ππππ π£π£ππ π π ππππ and compare with experimental values of real gases.