Classical Physics I PHY131 Lecture 36 Entropy and the Second Law of Thermodynamics Lecture 36 1
Recap: (Ir)reversible( Processes Reversible processes are processes that occur under quasi-equilibrium conditions: i.e. temperature and pressure differences are kept small, so that the path can be exactly reversed: We ve met such a situation already: a gas expanding from 1 L to 2 L, ending at the same temperature as it started at. The two extremes: 1. slow ISOTHERMAL expansion, ΔE int =0, moving a piston outwards while ADDING HEAT to the gas to counteract the cooling Work W = Q is Q 1 Q 2! done by the gas (piston moves out) This is a reversible process: extracting heat, while pressing the piston slowly in again will recover the original heat by doing the same work again 2. explosive expansion ( FREE EXPANSION ) into a vacuum space of equal volume (thereby doubling the original volume). NO HEAT is added, but still the temperature stays constant (as observed by experiment). In this case: W=0, Q=0, and ΔE int = 0. However, this process is IRREVERSIBLE, because to get back to the original state, we will have to do work as in 1, and thus take a DIFFERENT PATH back! the expanded gas will NEVER spontaneously regroup in one-half of the volume This can be traced back to the fact that the typical system contains an extremely large number of molecules (1 L at STP contains 2.7 10 22 ) Lecture 36 11
Randomness and Entropy S a typical system contains a very large number of molecules which make random motions (collisions with walls, each other). E.g. when a billiard ball is dropped onto the table, the kinetic energy of motion of the ball will be (partially) conveyed to the molecules in the table, which transmit this energy on by successive, and randomizing collisions: the originally organized macroscopic motion of the billiard ball is (partially) converted into additional random motion (internal energy) of molecules in the table Randomness is increased! the original energy is now in random motion energy and is (partially) unavailable for conversion back into useful energy E.g. in a small reversible and isothermal (const. T ΔE int =0) expansion dv of a gas, the randomness of the molecules increases, simply because their relative position uncertainty increases as dv/v: isothermal heat dq = pdv = nrtdv/v dq/t = nrdv/v we take ds dq/t as a measure of the increase in randomness! S is called the system s ENTROPY in a REVERSIBLE process: ΔS = S 2 S 1 = 1 2 dq/t in a REVERSIBLE ISOTHERMAL process ΔS = Q/T Entropy S is a STATE VARIABLE (as p, T, V, n, E int ) Lecture 36 12
Entropy Change Entropy change ds in an arbitrary reversible process for an ideal gas: dq = de + dw = nc dt + pdv int dq dt pdv dt dv ds = ncv + = ncv + nr T rev T T T V note, that the right-hand side can always be simply integrated (unlike dq!), because C V is (at worst) a function of T only. the entropy change ΔS = 1 2 ds is perfectly determined, and only depends on the endpoints (p 1,V 1,T 1 ), (p 2,V 2,T 2 ): i.e. S is a STATE VARIABLE! in any CYCLICAL reversible process, the ΔS of the system over a complete cycle is ZERO taking (for simplicity) C p = C V + R constant: 2 1 V T V Δ S = ds = nc + nr T V 2 2 V ln ln (expressed in, ) T1 V1 T2 T2p1 T 2 T2 p 2 = ncv ln + nr ln = ncv ln + nr ln ln T1 T1p2 T1 T1 p1 T p p V = nc nr T p = nc + nc V p 2 2 2 2 pln ln (in, ) V ln pln (in, ) T1 p1 p1 V1 Lecture 36 14
Entropy Examples 0.50 kg of 0 C ice heated to water of 100 C at 1 atm: ΔS f = ml f,water /T = ml water /273 = 612 J/K ΔS h = mc water dt/t = mc water ln(373/273) = 653 J/K a reversible isothermal process: (dq = pdv) ΔS = 1 2 pdv/t = 1 2 nrdv/v = nr ln(v 2 /V 1 ) = nr ln(p 1 /p 2 ) a reversible isobaric process: (dq = nc p dt) ΔS = nc p ln(t 2 /T 1 ) a reversible adiabatic process: (dq = 0) ΔS = 1 2 dq/t rev = 0 in the latter three processes the entropy change of the environment (typically) balances ΔS of the reversible process, so that the net entropy change of the universe is zero an IRREVERSIBLE adiabatic process (called free expansion ): we need a reversible path between 1 2 in order to calculate ΔS: a reversible isothermal process is the obvious choice (Note: Q is NOT a state variable!) ΔS = 1 2 nrdv/v = nr ln(v 2 /V 1 ), i.e. an increase in the entropy of the gas because no heat is exchanged with the environment (Q env = 0) the entropy of the environment is unchanged the net entropy of the universe increased in this irreversible process! Lecture 36 15
Molecular Interpretation of S Consider 4 indistinguishable molecules in a volume Label a each molecule by its position: U when it is in the Upper half of the volume, and D when it is in the Down half statistically, the molecules may be arranged in many permutations ( micro-states ) w = N!/(n U! n D!), each micro-state being equally likely: n U =4 n D =0 w 4,0 =1 UUUU n U =3 n D =1 w 3,1 =4 UUUD, UUDU, UDUU, DUUU n U =2 n D =2 w 2,2 =6 UUDD, UDUD, DUUD, UDDU, DUDU, DDUU peak n U =1 n D =3 w 1,3 =4 DDDU, DDUD, DUDD, UDDD n U =0 n D =4 w 0,4 =1 DDDD indeed, as we expect, already for a small set of only 4 molecules, the chance that all 4 are in the upper (lower) half is only (½) 4 = 1 / 16, whereas the chance that any 2 are up and 2 others are down is 6/16, i.e. almost half Lecture 36 16
Molecular Interpretation of S for larger number of molecules, the peak in the distribution around the middle rises quickly and sharpens up tremendously: the chance that a measurable difference exist between #molecules U and D becomes vanishingly small; using Stirling s approximation N! (N+½)ln(N+1) N : ln w n,n = ln[n!/(n U! n D!)] (N lnn N) (n U lnn U n U ) (n D lnn D n D ) = N lnn n U lnn U n D lnn D = N ln2 for n U = n D (ln w 50,50 = 66.8 to be exact w 50,50 =1.0 10 29 ) for 100 molecules, the chance that all molecules are in the upper half at any time is w 100,0 /w 50,50 1/10 29 =10 29 (1/2 100 10 30 ); even if the 100 molecules go at 1000 m/s between collisions and the average distance between collisions is 10 nm, the fastest travel time is 10 ps. The system moves to a new microstate at best 100x10 11 times per second, and would take us 10 29 /10 13 = 10 16 s = 3 10 8 yrs to have an even chance on all molecules to be Up. 3 10 8 yrs is about 2% of the current lifetime of the universe!! Lecture 36 17
Molecular Interpretation of S In thermodynamics the Entropy of a situation is defined as: S = k ln(w), w=#microstates giving the same situation =macrostate 4 molecules: S UUUU = k ln(1) = 0 J/K S 2U2D = k ln(6) = 8.28 10 23 J/K 100 molecules: w 50,50 = 1.0 10 29 S 50,50 = k ln(e 66.8 ) = 9.22 10 22 J/K w 40,60 = 0.14 10 29 i.e. 7x less probable w 30,70 = 0.0003 10 29 i.e. 3000x less probable 2x10 22 molecules: Prob 0,2x10 22= (1/2) 2x102 = (1/1024) 2x1021 (10 3 ) 2x1021 = 10 6x1021 S 10 22,10 22 k (2x10 22 ln2) 1 J/K even small fractional deviations from 10 22 will be EXTREMELY improbable a system will always go to the most probable macro-state, i.e. the macroscopic state which has the largest # of associated micro-states in free expansion from V 2V we derived from ds dq/t rev : ΔS = nrln(v 2 /V 1 ) = nrln2. Compare this to what we find here: (using N! NlnN N ) Δ S N! = ln ( wn 2, N 2 ) ln ( wn,0 ) = ln ln1 ln! 2 ln 2 = N ( N 2 )! 0 k 2! ( N ) N N N N R ln 2 ln = ln ln = ln2 Δ = ln2 = ln2 = nrln2 N N N A 2 2 2 N N N N 2 N N S Nk nn N A (or simply: ΔS = k ln(w tot /w 0,N ) = k ln(2 N ) = nrln2 ) Lecture 36 18
2 nd Law of Thermodynamics Entropy of a closed system can t decrease or: entropy in a closed system will stay either constant (for reversible processes) or increase (for irreversible processes)! Consequences: water at 0 C will never spontaneously freeze (a state of lower S); gas molecules will never get together in half the volume on their own a cyclic reversible process operating between two reservoirs, i.e. two isotherms at constant T Hot and T Cold, and two adiabats (CARNOT cycle): for the system: QH QC QC TC Δ S = + = 0 = T T Q T e.g. for an ENGINE (positive W Net ): H C H H WNet QH + QC QC TC ε = = = 1 + = 1 < 1 WNet < QH ; QC > 0 Q Q Q T H H H Carnot cycle is the most efficient possible! It exclusively uses reversible processes. Heat (Q H ) can never be converted completely into useful work W, and waste heat Q C is a necessary and unavoidable byproduct! REAL engines always do worse: friction, heat loss, H Lecture 36 19
Summary Thermodynamics expansion: L(T) = L 0 [1 +α(t T 0 )]; V(T) = V 0 [1 +β(t T 0 )] state variables: n, p, V, T, E int, S ideal gas: pv = nrt = nn A kt pv = nn A 2 / 3 K trans ; K trans = ½ mv 2 = ½ mv rms 2 1 st Law: Q = ΔE int + W = ΔE int + pdv heat: Q = mcδt or = ml f,v heat: Q = mc V ΔT or = mc p ΔT C V = 3 / 2 R (monatomic), 5 / 2 R (diatomic), or larger C p = C V + R internal energy: E int = nn A K tot = nn A 3 / 2 kt (for monatomic gas!) ΔE int = nc V ΔT Processes (reversible OR irreversible): isochor ΔV=0, isobar Δp=0, isotherm ΔT=0, adiabat Q=0 (pv γ =const.) efficiency: ε = W Net /Q H = 1+Q C /Q H ε Carnot = 1 T C /T H Entropy ds dq/t rev, ΔS 0 (closed system) Lecture 36 20