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Thermodynamic Potentials Why are thermodynamic potentials useful Consider U=U(T,V) Complete knowledge of equilibrium properties of a simple thermodynamic.

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Presentation on theme: "Thermodynamic Potentials Why are thermodynamic potentials useful Consider U=U(T,V) Complete knowledge of equilibrium properties of a simple thermodynamic."— Presentation transcript:

1 Thermodynamic Potentials Why are thermodynamic potentials useful Consider U=U(T,V) Complete knowledge of equilibrium properties of a simple thermodynamic System requires in addition P=P(T,V) equation of state U=U(T,V) and P=P(T,V)complete knowledge of equilibrium properties However U(T,V) is not a thermodynamic potential We are going to show: U=U(S,V) complete knowledge of equilibrium properties U(S,V): thermodynamic potential

2 The thermodynamic potential U=U(S,V) Consider first law in differential notation inexact differentials expressed by exact differentials 2 nd law Legendre Transformations dU: differential of the function U=U(S,V) natural coordinates Note: exact refers here to the coordinate differentials dS and dV. T dS and PdV are inexact as we showed previously.

3 Legendre transformation Special type of coordinate transformation Example: coordinates Partial derivatives of U(S,V) ( vector field components ) Legendre transformation:One ( or more ) of the natural coordinates becomes a vector field component while the associated coefficient becomes new coordinate. Back to our example becomes a coordinate becomes a coefficient in front of dP Click for graphic example

4 easy check: Product rule =:H (enthalpy) H=H(S,P)is a thermodynamic potentialEnthalpy

5 Geometrical interpretation of the Legendre transformation - 1-dimensional example Note: natural variable of f is Y’ Y and X have to be expressed as Y=Y(Y’) and x=x(Y’) mapping between the graph of the function and the family of tangents of the graph Legendre transformation:

6 Legendre transformation from (S,V) to (T,V): Helmholtz free energy (T,P): Gibbs free energy

7 equilibrium thermodynamics and potentials complete knowledge of equilibrium properties Consider Helmholtz free energy F=F(T,V) Differential reads: and Entropy Equation of state Response functions from 2 nd derivatives and etc. thermodynamics potential

8 Maxwell relations differential of the function F=F(T,V) dF is an exact differential In general: relations which follow from the exactness of the differentials of thermodynamic potentials are called Maxwell relations

9 Properties of an ideal gas derived from the Helmholtz free energy Helmholtz free energy F=U-TS Reminder: Equation of state:

10 Equation of state derived from F Note: U derived from F S(T,V)obtained from U(T,V)obtained from U=F+TS Heat capacity at constant volume Isothermal bulk modulus etc.

11 System  Heat Reservoir R Systems in Contact with Reservoirs Entropy statement of 2 nd law: entropy always increased in an adiabatically isolated system What can we say about evolution of systems which are not adiabatically isolated T=const. adiabatic wall  changes from initial state with to final state with remain constant Consider system at constant temperature and pressure

12 From Entropy change of  : Entropy change Aim: Find the total entropy change and apply 2 nd law of the reservoir: Heat Q R that, e.g., leaves the reservoir flows into the system  Q = -Q R With 1 st law: Heat reservoir: T=const.

13 Entropy statement of 2 nd law:for an adiabatically isolated system Gibbs free energy never increases in a process at fixed pressure in a system in contact with a heat reservoir. Gibbs free energy will decrease if it can, since in doing so it causes the total entropy to increase. (T=const, P=const.) System with V=const. in contact with a heat reservoir Special case, very important for problems in solid state physics (T=const, V=const.) Q = -Q R

14 Summary: Thermodynamic potentials for PVT systems T=const,P=constT=const,V=constIsobaric process 1 st law:Properties Maxwell relations Vector field components dG=-SdT+VdPdF=-SdT-PdVdH=TdS+VdPdU=TdS-PdVdifferential Gibbs free energy G(T,P) G=U –TS+PV Helmholtz free energy F(T,V) F=U -TS Enthalpy H(S,P) H=U+PV Internal energy U(S,V)Potential

15 Open Systems and Chemical Potentials Open systemParticle exchange with the surrounding allowed Heat Reservoir R T=const. Thermodynamic potentials depend on variable particle number N Example: U=U(S,V,N) Particle reservoir

16 U( S, V, N) = 2 U(S,V,N) 22 2 In general: SVN holds  and in particular for =1 (homogeneous function of first order)

17 keep N constant as in closed systems Chemical potential

18 Intuitive meaning of the chemical potential μ First law: with mechanical work PdVwork μdN required to change # of particles by dN + How do the other potentials change when particle exchange is allowed Helmholtz free energy F=U-TS

19 Gibbs free energy G=U -TS+PV Properties of μ With and both extensiveintensive ( independent of N ) F

20 Equilibrium Conditions Adiabatically isolating rigid wall System1: T 1,P 1,  1 System2: T 2,P 2,  2 From differentials of entropy changes

21 Total entropy change 2 nd law In equilibrium With conservation of -total internal energy -total volume -total # of particles

22  small changes dU 1, dV 1, dN 1 000 Equilibrium conditions T 1 = T 2, P 1 = P 2,  1 =  2 Remark: T 1 = T 2, P 1 = P 2 and  1 =  2 no new information for system in a single phase but Important information if system separated into several phases ( see next chapter )


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