Stuff 1st Law of Thermodynamics First Law Differential ---Tonight: Lecture 4 July 23 dU = ∂q PextdV − Form ---Assignment 2 has been posted. ∂U ∂U dU = dT + dV Total Differential ---Presentation Assignment posted. ∂T ∂V V,ni T,ni --Some more thermodynamics and then problem ∂U solving in class for Assignment #2. ∂q = C dT = v v ∂T V,n --Next week: Free Energy and In-class problem ∂U dU = C dT + dV solving session. v ∂V Total Differential T,ni Enthalpy Function: H(T,P) Heat capacity of gases depends on P and V we define: ∆H = ∆U + ∆PV Large changes dH =(∂q P dV )+P dV + VdP from large change ===> small change − Total Differential Heat q ∆U ∂qv Capacity dH = ∂q + VdP C = v = C = v ∆T ∆T v Constant ∂q ∂H ∂T V,n Volume C = p = p ∂T ∂T p P Heat ∂H ∂H qp ∆H ∂qp dH = dT + dP Cp = = Cp = Capacity ∂T P ∂P T ∆T ∆T ∂T P,n Constant Pressure ∂H dH = ∂qp = dT At dP = 0 ∂T P T2 T2 qv = CvdT qp = CpdT ∂H T1 T1 dH = CpdT + dP Recasting ∂P T,ni KMT and the “equipartition of energy” links internal If we look at the results of the equipartion theorem we find energy U and heat capacity, C of ideal gases. the internal energies for linear and non-linear molecules. 3 U IMPORTANT: This ∂qv U = RT (monoatomic) U = equation says that the 2 ∂T V internal energy of a 3 2 U = RT + RT +(3N 5)RT (linear molecule) gas only depends on 2 2 − 3 T and nothing else! U = nRT n moles of monoatomic gas 3 23 2 U = RT + RT +(3N 6)RT (non-linear molecule) 2 2 − We can take the derivative and link the heat capacity to 3 bulk properties C = R (monoatomic) Cv v 2 3 3 2 ∂q ∂( nRT ) 3 Cv = R + R +(3N 5)R (linear molecule) C = v = 2 = nR 2 2 − v ∂T ∂T 2 V 3 23 V Cv = R + R +(3N 6)R (non-linear molecule) 2 2 − Using the energies arising from equipartition of The 1st Law tells us that energy is conserved, energy, show that the heat capacity at constant and sets up a formal accounting system to track volume, Cv are the given values on the previous energy transformation as heat and work. slides. ∂q U = v ---With calculus 1st law ∂T V sets up experimental and theoretical parameters of 3 interest to measure. U = RT 2 (monoatomic) spontaneous not 3 2 spontaneous U = RT + RT +(3N 5)RT (linear molecule) 2 2 − ---But the 1st law does not give us predictive power 3 23 U = RT + RT +(3N 6)RT (non-linear molecule) on observation of 2 2 − naturally-occurring processes. Spontaneity refers to a process that appears to In the 1870’s, it was thought that !H determined the proceed “naturally” from an initial state to a final state “spontaneity of a chemical reaction”. This notion without outside intervention (does not mean “now”). proved wrong. T > 0˚C Examples of spontaneous reactions: 0 CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) !H = -890.4 kJ T < 0˚C T < 0C Freezing and melting of water 0 Hrnx does not H2O (l) H2O (s) !H = - 6.01 kJ ! determine whether Our goal in this Chapter T > 0C a reaction is 0 is to identify those H2O (s) H2O (l) !H = 6.01 kJ spontaneous! factors which determine whether a chemical H2O NH NO (s) NH +(aq) + NO - (aq) !H0 = 25 kJ reaction will proceed 4 3 4 3 spontaneously. The rusting of a nail What is the parameter or a bulk function There are many ways to express the 2nd Law that can predict whether A ==>B will occur of Thermodynamics “spontaneously” (irreversibly)? Kelvin Plank Statement: It is impossible for a system (engine) to undergo a cyclic process whose sole effects are flow of heat from a reservoir and the performance of an 2nd law of thermodynamics postulates that all equal amount of work by the system. processes that are spontaneous produce an increase in the entropy of the universe. Practical: Heat can not spontaneously pass from a cold body to a hot body. 1. Criteria for !Suniv = !Ssys + !Ssurr > 0 Spontaneous change! Deep: Work can be completely converted to heat, but heat can not be completely converted to work! (1st Law symmetry of work = heat is broken) !Suniv = !Ssys + !Ssurr < 0 2. No spontaneous change Spontaneous processes increase the total entropy of the !Suniv = !Ssys + !Ssurr = 0 3. Equilibrium condition universe. Entropy is too often described as positionally It appears that entropy focuses on the most statistically “disordered”. This is not correct, in the chemical favored position distribution entropy is more concerned with the fact that energy levels become more closely sense, but it is useful device to predict. spaced and more occupied in that most favored state. Gas expands spontaneously How many into larger volume ways can you have your 1.0 atm evacuated 0.5 atm room or a Quantum mechancis dictates closer energy deck of cards level spacing as V organized vs increases. disorganized? Same amount of energy is dispersed or Energy level spread among more Energy level energy levels. There is a thermodynamic state function called Entropy is a measure of the magnitude of energy entropy, S, that is a measure of the energy dispersal dispersal over available quantum states in a that occurs when a change of state occurs. chemical system. The more dispersed or spread out the energy is the higher the entropy. Macroscopic Microscopic E3 E6 E9 2. Molecular Description 1. Bulk defintion (it Increasing the E8 needs no molecules). volume of a E5 E7 system !S = k (ln Wf - ln Wi) E2 E4 E6 -qsys reversible decreases energy !S = level spacing E5 T increasing E3 E4 entropy by E2 E3 Change in populating more. E1 E2 Entropy E1 E1 L 2L 3L Boltzmann founded a thermodynamic state function A system can be described by bulk or macroscopic called entropy, S, that is a measure of the “dispersion or that we can observe and measure, and the spread of energy” that occurs when all spontaneous microscopic or molecular that we can’t see but can reactions occur. model statistically. You must see both! # of microstates Consider 4 labeled molecules A,B,C,D S = k ln W Microstate is particular distrubution Entropy that corresponds to some macrostate. Boltzman’s constant Multiplicity is the number of microstates 1.38 " 10#23 J/K. that give a specific macrostate. Boltzmann’s Tomb In Vienna, Austria !S = k ln Wf - k ln Wi Macrostate is the observed state of the When W > W then !S > 0 f i system that represents on a molecular Change in When Wf < Wi then !S < 0 level that microstate with the highest probability or number. Entropy 5-observable macrostates Microstate Macrostate Boltzman’s statistical interpretation of entropy Left Right Multiplicity Probability Bulk Property (dispersal or spread of energy) is connected to a Side Side macroscopic defintion of heat flow per unit temperature. A,B,C,D - 1 1/16 Macroscopic Microscopic A,B,C D A,B,D 1. Bulk defintion (it 2. Molecular Description C 4 4/16 A,C,D B needs no molecules). B,C,D A A,B C,D !S = k (ln Wf - ln Wi) A,C B,D -qsys reversible A,D B,C 6 6/16 !S = B,C A,D T B,D A,C C,D A,B Change in - A,B,C,D 1 1/16 Entropy 16 Increasing entropy or the dispersal or spread of Dissolution energy occurs in all spontaneous processes. It provides humans with an “arrow of time” as the There are many !S > 0 reverse situation never happens. chemical reactions that Dissolution Example: two block of different temperatures are lead to an brought together. increase in !S > 0 dispersal of 1 2 energy spread Mixing Thermal energy flows 100˚C 10˚C 100˚C 10˚C from the higher over a larger occupied energy number levels in the warmer microstates !S > 0 object into the entropy (!S > 0) unoccupied levels of Increasing T the cooler one until equal numbers of states are occupied. HOT COLD COMBINED Can We Determine Mathematically Criteria For KMT and the “equipartition of energy” links internal Spontaneous Change Using What We Know So energy U and heat capacity, C of ideal gases. Far? 3 U = nRT n moles of IMPORTANT: This 2 monoatomic gas equation says that the Usual approach is Carnot Cycle internal energy of a gas only depends on ∂q • Carnot analysis is long and the result unsatisfying U = v T and nothing else! ∂T KE gives this • Another approach is purely mathematical based on V the Euler criteria and gives the same result. We can take the derivative and link the heat capacity to • LET”S TRY IT bulk properties 3 ∂q ∂( nRT ) 3 C = v = 2 = nR v ∂T ∂T 2 V V Criteria For Spontaneous Change? 3 nRT ∂qrev = nR dT + dV Assume monoatomic ideal gas U = 3/2 nRT 2 V dU = ∂q + ∂w start with 1st law There is a math theorem that says we can transform an rev rev differential form inexact differential to an exact one by multiplying by and integrating factor. Can we find one? ∂qrev = dU + PextdV rearrange 3 from KMT 1 1 3 1 nRT ∂( nRT ) dS = qrev = nR dT + dV ∂U 2 3 T T 2 T V dU = dT = dT = nRdT ∂T ∂T 2 Show that this function is an exact differential using substituting Euler’s criteria.