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Connecting Statistics and Entropy Thermal Expansion and Contraction

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Connecting Statistics and Entropy Thermal Expansion and Contraction Connecting statistics and entropy Thermal expansion and contraction Ron Reifenberger Birck Nanotechnology Center Purdue University April 3, 2013 S = kB ln (w) Lecture 11 1 Review: A reversible process does NOT produce entropy Processes that are idealized as reversible include: •Frictionless movement •Restrained compression or expansion •Energy transfer as heat due to infinitesimal temperature gradients •Electric current flowing through a zero resistance •Restrained chemical reaction •Mixing of two samples of the same substance in the same state Processes that are irreversible include: •Movement with friction •Unrestrained expansion •Energy transfer as heat due to large temperature gradients •Electric current flowing through a non zero resistance •Spontaneous chemical reaction •Mixing of matter of different composition or state 2 There is a strong connection between the 2nd Law of Thermodynamics, Probability and Statistics You can devise countless simple experiments to illustrate that “movement toward more disorder” is a law that nature follows. Highly ordered systems become improbable as the number of objects in a system increases. Essentially, every observed configuration of a large number of objects is highly disordered. This leads to the conclusion that nature has a strong tendency to move toward maximum multiplicity (and maximum entropy). 3 A simple example of states and multiplicities (pulling 4 marbles out of a hat) Large number of marbles Equal number of red & green Multiplicity or number of Macro microstates (w) States 1 4R , , , 4 1G,3R , , 2G,2R , , 6 3G,1R , , , 4 4G 14 16 The multiplicity of possible states when pulling N marbles from a jar Total Multiplicity w Microstates N=1 1 1 2 N=2 1 2 1 4 N=3 1 3 3 1 8 N=4 1 4 6 4 1 16 N=5 1 5 10 10 5 1 32 N=6 1 6 15 20 15 6 1 64 etc. a macrostate Multiplicity of the Central Configuration5 1 6 15 20 15 6 1 These numbers mean: 1 way to get (6R, 0G); 6 ways to get (5R, 1G); 15 ways to get (4R, 2G); 20 ways to get (3R,3G); 15 ways to get (2R, 4G); 6 ways to get (1R, 5G); 1 way to get (0R, 6G). In general: N ! N ! 6! 654321 w for example,15 w4,2 nn12, nn! ! 4!2!432121 nn12!! 12 wi 15 probability pi 0.234 6 wi 64 Which “state” has greatest multiplicity? Which “state” has highest entropy? Which “state” has lowest entropy? S = kB ln (w) e.g., the entropy of the (4R, 2G) state would be S=1.38x10-23 J/K x ln (15) = 3.7x10-23 J/K This simple thought experiment can be mapped into many other situations 7 Consider 10 Gas Atoms in a Container left half right half 23 7 10 6 5 4 8 9 order 1 probability increases decreases Probability that any one particular atom is in left side of box = ½ Probability that any two particular atoms are in left side of box = ½ x ½ - 5 Probability that any five particular atoms are in left side of box =( ½) = 1/32 - 10 Probability that ALL atoms are in left side of box = (½) = 1/1024 8 Working it out – Multiplicities or Probabilities? Sinit =kB ln (winit) Sfin =kB ln (wfin) Probinit =Pinit = winit / Wtot Probfin = Pfin = wfin / Wtot ΔS= Sfin – Sinit = kB ln (wfin)- kB ln (winit) = kB ln (Pfin) + kB ln (Wtot) – [kB ln (Pinit) + kB ln (Wtot)] = kB ln (Pfin) – kB ln (Pinit) 9 left half right half Initial: 23 7 10 6 5 4 8 1 9 left half right half Final: 236 9 5 4 7 1 10 8 ΔS= S10 left side – S5 left side = kB ln (1/1024) - kB ln (1/32) = kB [ -6.93 – (-3.46)] 10 = -3.46 kB Is this transition spontaneously possible for “N” gas atoms? N/2 N/2 Initial V1 N Final V2 1 VV 11 212 Calculate the probability that N molecules appear on one side The probability that a gas of N molecules spontaneously compress from original volume V1 to final volume V2 = ½V1 is 1 2 . N Pfin = (V2/V1) x (V2/V1) . x (V2/V1) N = (V2/V1) Pinit (for V2=V1)=1; ln (1)=0 So………………….ΔS= kB ln (Pfin) ln (Pfin) = N ln (V2/V1) = nNA ln (V2/V1); n=# moles so S = kBnNA ln (V2/V1) = nR ln (V2/V1) Reif, pg. 126 12 For instance, suppose we have three moles of an ideal gas. To spontaneously find all atoms in left half of a container, we have V2/V1 = ½ and 23 ln (Pfin) = nNA ln (½) = 3 x 6.02 x10 x (-.693) = -1.25 x 1024 then S = kB ln (Pfin) -17.28 J/K (same answer as in last lecture!) Since the entropy change is negative, the final entropy is LESS than initial entropy, so the process will not occur spontaneously because entropy must always increase. 13 Application to a Chemical Reaction CaCO3(s) + 2 HCl(aq) ⟶ CaCl2(aq) + CO2(g) + H2O (l) orange CaCl2 & red CaCO 3 purple CO & 2 & green HCl blue H2O BEFORE AFTER 3R 1G,2R , , 2G,1R , , 3 3G Multiplicity or number of 8 microstates (w) 3O orange CaCl2 & 1P,2O , , purple CO2 & 2P,1O , , blue H O 2 3P 8 3O 1P,1O,1B 1B,2O , , 2B,1O , , 3B 7 3 3P TOTAL Multiplicity or 1B,2P , , number of 2B,1P microstates , , (w)=24 3B 6 If Nature follows a path of increasing multiplicity, then this chemical reaction will occur spontaneously 16 Why Free Energy and Enthalpy? So far, we have discussed systems with known or zero energy exchange between boundaries – processes like gas expansions and the conversion of heat to work in heat engines. Under these circumstances, entropy is a good thermodynamic predictor of what processes are possible. For chemical processes inside test tubes placed in a laboratory heat bath, the work or heat flow is not a controlled quantity, but rather the temperature or pressure becomes the controlled quantity. This slight change in conditions requires the definition of new thermodynamic state variables like free energy (F=E-TS) and enthalpy (H=E+PV) rather than entropy. Systems held at constant temperature do not tend toward a state of maximum entropy. Instead, they tend toward states of minimum free energy. Enthalpy is useful to describe chemical reactions at constant pressure. Chemical reactions that occur in open beakers and test tubes often use Enthalpy as a state variable, since when P=constant, and Enthalpy is a measure of heat content. This is why chemists and biologists tend to focus on free energy and enthalpy rather than entropy. 17 The 2nd Law has both practical and philosophical implications Practical: Energy locked into non- renewable energy sources is constantly being used up. Philosophical: Nature has a built in hierarchy of more useful and less useful forms of energy. As useful energy (i.e. heat from high T reservoir) is converted into less useful forms of energy (i.e. heat from low T reservoir), we no longer can use the energy to do work. 18 The 2nd Law and Work Assuming you have an isolated system: If you take a system from a low entropy state to a higher entropy state, the system will produce a certain amount of work - Wout. If you then can somehow take the system from the higher entropy state back to the low entropy state, you will need to supply a certain amount of work – Win The 2nd Law guarantees that Win > Wout This ends our discussion of Entropy! 19 Thermal Expansion L+L L T T+T L = T L is a number – the coefficient of linear expansion 20 How much longer will a 1 cm rod of aluminum become if the o temperature increases by 1 C? For aluminum, 2.4 x 10-5 K-1 L = T L = 2.4 x 10-5 K-1 x 1 K x 0.01 m = 240 nm 21 Bimetallic Strip Unpinned end Pinned end T-T colder T 1 brass steel 2 T+T hotter 1 > 2 22 Thermal Expansion of a Plate with a Hole T Area of hole = Ao A 2T Ao T+T Expansion occurs in every dimension, and every dimension is increased in the same proportion 23 Volume expansion for isotropic solid L1 V = L1L2L3 L3 L2 dV dL dL dL = L L 3 + L L 2 + L L 1 dT 1 2 dT 1 3 dT 2 3 dT 1 dV 1 dL 1 dL 1 dL = 3 + 2 + 1 V dT L3 dT L2 dT L1 dT = 3 = β 24 Example V=VT o = 10 gal x (10 x 10-4/ oC) x (-25oC) T=0 C = - 0.25 gal G A S To o South T=25 C Park G A S 25 Water is anomalous! 26.
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