Lecture8Ch161Feb18

Lecture8Ch161Feb18 - Summary of last lecture Equation of...

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Unformatted text preview: Summary of last lecture Equation of states general discussion EofState for an ideal gas Thermodynamic potentials and relation between them Dependence of TD potentials on number of particles and chemical potential - will be very important fairly soon when we do chemical reactions. Chemistry 161: Statistical Thermodynamics Lecture 8 Thermodynamic Relations, Stability Conditions and all that. Key Concepts and Lessons: a) Maxwell relations b) Physical meaning of TD potentials c) Stability conditions d) Cp > Cv always e) Throttling: a reporter on intermolecular interactions Reading: Reif, Ch5 Potential Finally we design a potential which is convenient when number of particles varies. In this case we have to switch from N to chemical potential as independent variable: = A N = A G = pV d = SdT pdV Nd When we discuss quantum gases we will see how useful this potential is! What is the meaning of all these TD potentials? Consider any process that takes place with a system at which some heat is exchanged. There can be additional processes inside a system at any rate entropy should obey the condition: dS dQ < T dt dt Where t means time. Equal sign is when the process is quasistatic . Now, using first law we write: dE dt pext dV dt dS T dt 0 now assume that a process occurs at constant T and V, (i.e. dT=dV=0) then dA d( E TS ) = dt dt Key conclusion: All processes that occur at constant T and V result in decrease of A! Equilibrium at given T and V corresponds to minimum of A!. What is the meaning of all these TD potentials?-cont Similarly if we consider processes that take place at constant T and p, the condition is dG dt 0 i.e. at constant T and p equilibrium corresponds to minimum of Gibbs potential G Stability Conditions how do we know that potentials are at Minimum? Consider now the isolated system partitioned into two subsystems with total numbers of particles, energy and volume conserved. Consider exchanges between two parts in entropy (heat), work and particles: E= 2 T =1 S p V + N But given the conservation of total entropy (equilibrium process!), volume and the number of particles we get: 0 E = T 1 T 2 S1 1 2 ( ) ( p1 p 2 V 1 + 1 2 N 1 1 2 1 2 ) ( ) Equilibrium is therefore achieved where E is minimum i.e.: T =T ; p = p ; = Stability Conditions - Cont Now we turn to the conditions of stability. They aim at assuring that E,A,G etc are at minimum (and not merely at an extremum!) Consider first E. the minimum requirement calls for the second variation to be positive (think about the same partitioning thought experiment!) ( 2 E ) S ,V , N 1 = 2 2 E S2 ( T ,N S (1) ) 2 1 + 2 2 E S2 ( S ( 2) ) 2 again we are at equilibrium therefore S 1 = 2 S 2 . Also we have (prove:10pt) E S2 = V ,n T S = V ,n T Cv We get therefore the condition: T Cv 0; or Cv > 0 Stability Conditions - Cont Now we can consider the condition that A is minimum and fluctuations with respect to volume partitioning (no T fluctuations because T is an intensive variable!!!!). ( 2 A ) T ,V , N 1 = 2 A V2 A V2 2 2 ( T ,N V (1) ) 2 1 + 2 A V2 2 ( V ( 2) ) 2 But = T ,N p V T ,N And we get the condition p V 0 or KT = T ,N 1 V V p 0 i.e. p decreases when V increases or isothermal compressibility >0 Maxwell Relations Recap all TD potentials: dE = TdS pdV + n i =1 i dN i i dN i i dN i i dN i E S H S A T G T = T; V E V H p A V G P = S p; dH = TdS + Vdp + dA = dG = SdT pdV + n i =1 n = T; P = V; S = V S; S; = T p; i =1 SdT + Vdp + n i =1 = P = V; T Now we notice that second derivative does not depend on order in which it is taken Maxwell relations (Cont) Now we note that second derivatives can be taken in any order. This implies several very useful relations: T V T p S V S p = S p S V S p T V T (from second derivatives of Energy) V = S (from second derivatives of Enthalpy) P = T (from second derivatives of Helmholtz free energy) V = T (from second derivatives of Gibbs free energy) P As promised: general relation between Cp and Cv The heat capacity at constant volume is given by: CV = dQ dT = T V dS dT V At constant pressure CP = dQ dT =T P dS dT dS dT P dQ = TdS = T dT + P dS dp dp = C p dT + T T dS dp dp T we can present dp = dQ = TdS = C p dT + T CV = C p + T dS dp dp dT dS dp dT + V dp dV dV and get: T T dp dT dT + V dp dV dV when dV=0 dQ = CV dT i.e. T T dp dT V general relation between Cp and Cv (cont) but how to deal with weird S p = T S p ? A Maxwell relation! T V T p T = V P where = 1 V V T is ''volume coefficient of expansion'' P What about dV = V T = V ? Here is the trick: V dT + P V p dp. We are interested in dV=0, i.e. T p T p T Cp V T V p P we define =- 1 V V p and get: T T = V and finally: 2 CV = > 0 because >0 for stability (see above!) A Practical Example: Throttling Throttling is used to cool gases and convert them to liquids. How does it work? Here gas is being `'squeezed'' through a porous partition. The walls are insulated so that no heat is exchanged with outside world. Now we note that gas passing from left to right actually makes some work (apparently at the expense of its own internal energy). Let us do some bookkeeping of this work. Imagine mass M of gas passing from left to right. Before crossing the plug it had volume V1 and was under pressure p1. After passing the plug it has volume V2 and pressure p2. What work W does it do? Well gas displaces V2 under pressure p2 on the right I.e. it does work p2V2. Is that all? No- because gas itself is being pushed out from left hand side and gas on the left does work on our mass M which is p1V1.. So the total work as mass M pushes through is W=p2V2-p1V1 Throttling, Cont The work done is at the expense of energy (lost or gained) and we have: dQ = dE + W = 0 E2 E1 + p2V2 p1V1 = 0 or finally: H 2 = E2 + p2V2 = H 1 = E1 + p1V1 Throttling is an isoenthalpic process! How does T change at throttling? These are prototypical constant enthalpy curves and their slopes show whether changing pressure is heating or cooling gas! Throttling, Cont Now we want to determine the effect of throttling on gas T. To this end we want to evaluate = T p we have the usual relation: H dH = TdS + Vdp = T C p dT + T S p T = H S p dp + T S T dT + Vdp = 0 i.e. P dp + V dp = 0 T T p S p S p dp + V T Cp V T = V P using Maxwell relation we get finally: = T = V Cp (T 1) Does T increase or decrease at throttling? For an ideal gas = 0 Prove:5pts Which means that T effect of throttling is directly related to intermolecular forces which act between particles! This opens an opportunity to explore intermolecular forces by analyzing throttling experiments. I will prove later that generally the equation of state for a non-ideal gas Presents an expansion in density of gas: expansion in density of gas n = N V 2 3 p = kT n + B2 n + B3n + ... where B2 , B3 ... are so-called vitial coffecients. 1 e 0 U (r ) kT E.g. we will show later that B2 = 2 r 2 dr 4v a where v is own volume of the core of the molecule T and a>0 comes from the attractive part of iteractions. For lpw densities n we can write p p V kT n + B2 n 2 and taking from the first term n NkT p 1+ B2 or V kT N kT + B2 which gives: p V T V = P p kT = 1 T CP N T CP B2 T B2 = N 2a CP T v T effect of throttling reports on balance between attraction and repulsion Recall B2 = 2v a at high T B2 > 0 i.e. repulsion dominates. At low T B2 < 0 i.e attraction dominates. T <0 at high T and >0 at low T. In other words when repulsion between molecules dominates throttling gas from low to high p results in its cooling. At low T the opposite is true: throttling gas from high to low pressure (expansion) results in gas cooling because average distance between molecules increases they do not attract each other so much their potential energy increases at the expense of the decrease in their kinetic energy - hence T becomes lower. T Next Lecture: Canonical Ensemble, connection to thermodynamics, partition function and its implications Reading: Reif Ch6. ...
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