Thermodynamic Processes: The Limits of Possible Thermodynamics put severe restrictions on what processes involving a change of the thermodynamic state of a system (U,V,N,…) are possible. In a quasi-static process system moves from one equilibrium state to another via a series of other equilibrium states . All quasi-static processes fall into two main classes: reversible and irreversible processes . Processes with decreasing total entropy of a thermodynamic system and its environment are prohibited by Postulate II Notes Graphic representation of a single thermodynamic system Phase space of extensive coordinates The fundamental relation S(1) =S(U (1) , X (1) ) of a thermodynamic system defines a hypersurface in the coordinate space S(1) S(1) U(1) U(1) X(1) X(1) S(1) – entropy of system 1 (1) (1) (1) (1) (1) U – energy of system 1 X = V , N 1 , …N m – coordinates of system 1 Notes Graphic representation of a composite thermodynamic system Phase space of extensive coordinates The fundamental relation of a composite thermodynamic system S = S (1) (U (1 ), X (1) ) + S (2) (U-U(1) ,X (2) ) (system 1 and system 2). defines a hyper-surface in the coordinate space of the composite system S(1+2) S(1+2) U (1,2) X = V, N 1, …N m – coordinates U of subsystems (1 and 2) X(1,2) (1,2) S – entropy of a composite system X U – energy of a composite system Notes Irreversible and reversible processes If we change constraints on some of the composite system coordinates, (e.g. Xi and Xj) the composite system relaxes to a new equilibrium state with the maximum possible entropy consistent with the new constraints Reversible process, (1+2) Irreversible process, S(1+2) S entropy increases Entropy is constant (most processes are (idealized situation, (1,2) like that) B X (1,2) there is no real Xi i drive for the system as Postulate II Is not at work) B (1,2) X (1,2) Xj A j A Notes Heat and Entropy If we have flow of heat between two subsystems, then: δQ1 = −δQ2 δQ1 = T1dS 1 δQ2 = T2dS 2 If temperatures of the two subsystems are equal to each other: T1 = T2 dS 1 = −dS 2 dS = dS 1 + dS 2 = 0 then the process is reversible. Heat flow between systems with different temperatures is always irreversible . Notes Heat capacity and heat flow Let us consider a situation of T1>T 2 before we remove adiabatic wall between the two systems. After new equilibrium is reached, the final temperature of the two systems is Tf. Let us also make a simplifying assumption that heat capacity (CV = dU/dT ) at constant volume for these systems does not depend on temperature: ∆U1 = C1(Tf −T1 ) ∆U 2 = C2 (Tf −T2 ) Notes Equilibrium temperature ∆U1 = C1(Tf −T1 ) ∆U 2 = C2 (Tf −T2 ) Total energy of the composite system must be conserved: ∆U = ∆U1 + ∆U 2 = (C1 + C2 )Tf − (C1T1 + C2T2 ) = 0 (C1T1 + C2T2 ) Tf = ()C1 + C1 Notes Entropy Change δQ C dT T 1 1 1 ln ln ln f dS 1 = = ∆S1 = C1()()Tf − ()T1 = C1 T1 T1 T1 For the composite system: T T f f ∆S = ∆S1 + ∆S2 = C1 ln + C2 ln T1 T2 + + C1T1 C2T2 C1T1 C2T2 ∆S = C1 ln + C2 ln > 0 T1()C1 + C2 T2 ()C1 + C2 Notes Heat flow is irreversible + + C1T1 C2T2 C1T1 C2T2 ∆S = C1 ln + C2 ln > 0 T1()C1 + C2 T2 ()C1 + C2 Taking for simplicity, C1=C 2 : + + + 2 T1 T2 T1 T2 (T1 T2 ) ∆S = C ln + C ln = C ln > 0 2T1 2T2 4T2T1 Irreversible process T1 > T2 Notes Utilizing Entropy to do Work Let us start thinking about how to extract work from a thermodynamic system. We have a system that goes form one equilibrium state to another: A→B. As it undergoes this transformation, it does work -δW and releases heat -δQ. We would like to know, what is the maximum work that this system can do in the process A→B. In particular, we would like to compare reversible and irreversible processes . Notes Heat Engine Model We model the environment of the system as two objects: - reversible work source (sink) RWS - reversible heat source (sink) RHS Heat engine model ∆U = U B –UA -δQ RHS -δW System A →B RWS Heat sink Work sink TRHS T Rigid walls, so no work can Adiabatic walls, so no heat can be be done on the heat sink ! transferred to the work sink! ( dS RWS =0 ) Notes Energy-efficient use of Entropy Problem: to minimize energy E spent to lift a weight using a stretched rubber band. Lift by hand, no help Lift by hand at constant Lift by warming the from the rubber band T, rubber band helps rubber band E = W '= ? EΓ=const = Q = ? E0 = W = Mgh T =const Spontaneous heat flow h h h min (EΓ=const , ET =const ) = ? Notes.