Lecture 3 Clausius Inequality
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1 Lecture 3 Clausius Inequality Rudolf Julius Emanuel Clausius 2 January August 1888 Defined Entropy Greek, en+tropein content transformative or transformation content The energy of the universe is constant. The entropy of the universe tends to a maximum 1865 CY T. Pradeep
2 Assume reversible and irreversible paths between two states. Reversible path produces more work. du is the same for both the paths. du = dq + dw = dq rev + dw rev dq rev dq = dw - dw rev 0 dq rev /T dq/t ds dq/t Clausius inequality System is isolated. ds 0 Clausius inequality
3 Equilibrium Entropy Process
4 Spontaneous processes entropy increases. Entropy is Time s Arrow Arthur Stanley Eddington ( )
5 How do we derive conditions for equilibrium and spontaneity? For an isolated system S 0, > sign for a spontaneous process and = for equilibrium. In the case of open or closed system, there are two ways 1. Evaluate S for systems and surroundings. S total = S system + S surroundings S 0
6 2. Other way is to define entropy change of the system alone. dstotal = dssystem + dssurroundings ds - dq/t 0 Clausius inequality Consider constant volume: ds - du/t 0 TdS du (constant V and so no work due to expansion) At constant U or at constant S, the expression is: 1. ds U,V 0 2. du S,V 0 Criterion of spontaneity 1. is the common statement of second law. 2. Entropy is unchanged, for sponteneity, entropy of the surroundings must increase for which U of the system as to decrease.
7 At constant pressure, TdS dh 1. ds H,P 0 2. dh S,P 0 Interpretations are the same. The inequalities mean, du TdS 0 dh TdS 0 da = du TdS dg = dh TdS (da) T, V 0 (dg) T, P 0 We define, A = U TS Helmoltz energy G = H TS Gibbs energy
8 Hermann von Helmholtz Born: 31 Aug 1821 in Potsdam, Germany Died: 8 Sept 1894 in Berlin, Germany
9 What is A? du = dq + dw TdS dq First law du TdS + dw dw du-tds = da Most negative value of W is W max and that is equal to da. Under constant T and V can the system do work? A is not defined only for this condition!!
10 G = H TS H = U + PV dh = dq + dw + d(pv) = U + PV TS dg = dh TdS SdT = dq + dw + d(pv) TdS SdT At constant T, dg = dq + dw + d(pv) TdS When the change is reversible, dw = dw rev, dq = dq rev = TdS dg = TdS + dw rev + d(pv) TdS = dw rev + d(pv) dw rev = -PdV + dw additional System can do work other than PdV also dg = dw rev + d(pv) = [-PdV + dw additional ]+ PdV + VdP dg = dw additional + VdP At constant P and T, dg = dw additional Work function Free energy Here work done by the system is taken as expansion work, -PdV Carnot limitation Decrease in free energy, G, at constant temperature and pressure corresponds to the maximum work other than the P V work that the system is capable of doing under reversible conditions.
11 Conditions of equilibrium (ds) U, q 0 (TdS) U, V 0 (da) T, V 0 (dg) T, P 0 Josiah Willard Gibbs February 11, 1839 April 28, 1903
12 G is a function of P and T G = f(p, T) dg = ( G/ P) dp + ( G/ T) dt 1 T P G = H TS = U + PV TS dg = du + PdV + VdP TdS SdT du = TdS PdV dg = VdP SdT 2 Comparing 1 and 2 ( G/ P) = V T ( G/ T) = S P One component system
13 Variation of G with T Gas G Solid Liquid T ( G/ T) P = -S
14 Variation of G with P Gas G Liquid Solid P ( G/ P) T = V
15 S and V are always positive quantities. G should increase with P at constant temperature and decrease with temperature at constant pressure. For a finite change in free energy at constant temperature, P2 P1 dg = P2 P1VdP For solids and liquids, the volume change will be small and G = V P Such changes in free energy are very small. For gases, since volume change is large, G is large. 2 1 dg = 2 1 nrt/p dp G m o = nrt ln P 2 /P 1 This relation shows that G is (1) extensive and (2) a state function. G for a change 1 2 is the same whether the change of state is carried out reversibly or irreversibly. -Infinity P o G m (P) = G o m + RT ln P/P o
16 Gibb s Helmholtz equation G f values predict the feasibility of a reaction at 298 K. G values at any temperature can be calculated by Gibbs - Helmholtz equation. G = H T S ( G/ T) P = S ( G/ T) P = S G = H + T ( G/ T) P (1) G can be evaluated from emf measurement since G = nfe Where n = number of electrons evaluated, F = Faraday, E = potential of the cell. F= Coulombs/gm. equiv.
17 Divide eqn. 1 by T 2 G/T 2 + 1/T ( G/ T) P = H/T 2 Write 1/T 2 as / T (1/T) G [ / T (1/T)] P +1/T ( G/ T) P = H/T 2 {UdV + VdU = d(uv)} [ / T ( G/T)] P = H/T 2 Helmholtz equation: [ / T ( A/T)] P = U/T 2 ]
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