Thermodynamics General

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1 Thermodynamics General Lecture 5 Second Law and Entropy (Read pages 587-6; Physics for Scientists and Engineers (Third Edition) by Serway) Review: The first law of thermodynamics tells us the energy must be conserved. The second law helps establish which processes can and cannot occur in nature. I. Second Law of Thermodynamics A. The second law helps establish which processes can and cannot occur in nature and in which direction they should occur.. For example, the Second Law also tells us the maximum amount of heat that can be converted to useful work --- there is a theoretical limit to the amount of heat that can be converted to work.. For another more clear example, maybe, we know that heat flows from warm to cold, but never in erse. Why? The second law will help us explain why. 3. Additional examples, a. Exploding bomb cause a reaction that impossible for us to undo b. Salt dissolves into ions (NA+ and Cl-) and does not recombine without some external influence c. A dropped bowling ball does come back up into your hands without some external influence d. Pendulums slow owing to collisions with air molecules B. Reversible and irersible transformations. The examples above are examples of processes that occur naturally in one direction. None would occur in any other temporal order (at least probabilistically). These are irersible processes. All natural processes are irersible. All natural ones. a. The second law, by this statement, helps establish a direction of time. 3. Reversible Transformation a. A substance undergoes a ersible transformation if each state of the system is in equilibrium so that a ersal in the direction of an infinitesimal small change returns the substance back to its original state. C. Cyclic Process A sequence of changes in the state of substance in which its volume changes and it does external work with the substance returning to its original state.. The initial and final states must be identical.. If initial and final states are identical, internal energy is unchanged (du = 0) after the process. a. From the First Law of TD i. dq = dw ii. ii. q - w = du = 0 (q=w) The work done by a system is equal to the net heat absorbed by the substance.

2 D. Idealized Heat Engine. Q Q added heat > Engine > exhaust heat. The efficiency of the engine is Efficiency = (work done) / (heat added) = (Q - Q) / Q. If Q = 0 a Efficiency = b. no heat exhaust c. All added heat goes to useful work F. The Carnot Cycle T P B HEAT SOURCE C Note: T<T T A D Adiabat HEAT SINK. Start with T at A. Adiabatic compression to B. dq=0. du = -dw (work done on the system is manifested as an increase in u). a. du = cv(t-t) = -dw b. The work done on the system is cv (T-T) ---- u increases. Substance expands isothermally from B to C by absorbing heat Q. T=constant. du =0. dw pd nr * T dln nr * Tln c c c = = = = Q b b b 3. Further adiabatic expansion from C to D. dq = 0. a. du = -dw b. du = cv(t-t) = -dw c. The work done by system cv (T-T) --- u decreases 4. Substance compresses isothermally and returns back to original state (D->A). Heat exhausted is Q. du=0. dq = dw. dw pd nr * T dln nr * Tln a a a = = = = Q d d d

3 5. Cycle returns to initial state. Work done BY the system is Q-Q (work = net heat absorbed). Sum the work terms. a. wi + wii + wiii + wiv = Q-Q i. wi = -cv(t-t): adiabatic compression ii. wii = nr*t ln(c/b) = Q: isothermal expansion iii. wiii = cv(t-t): adiabatic expansion iv. wiv = nr*t ln(a/d) = Q = - nr*t ln(d/a) v. sum of w's = Q-Q NET WORK 6. Re-stating the Second Law of Thermodynamics Only by transferring heat from a warm body to a cold body, can heat be converted to work in a cyclic process. G. Entropy: What really is entropy?. Thermodynamics tells us about the macroscopic nature of T, u, s. But what about the microscopic nature. T is is the mean kinetic energy, u is the total molecular energy, but what is s???. What we know about entropy a. Entropy is maximized at equilibrium. b. Tds = dq for ersible processes and Tds > dq for irersible processes. c. An example. Consider a container of two different gases. Assume that the gases are isolated from each other initially by some partition. Equilibrium could be assumed to be a state when the gases are well mixed (each gas molecule is randomly distributed). This is the most likely configuration and also the state when entropy is maximized (a highly disordered state is more probable than the unmixed state). d. Another example might be with a container containing two gases; gas A with temperature T and the other B at T. Assume that T>T. Heat will flow from A to B via collision and conduction. It is much more probable (higher entropy) that heat will flow from A to B (flow from hot to cold) and that this will occur until U is spread equally through out the system so that thermal equilibrium is reached. Spreading energy equally between all molecules implies thermal equilibrium and maximum entropy. A system is most ordered in this case, from an energetics view when maximum entropy is reached. H. Entropy: More on entropy. Entropy is defined, mathematically, as ds = dq / T a. Q is the amount of heat that is added ersibly to a substance at a temperature T. b. Entropy is a function of State. i. Entropy on depends on the final and initial condition of a transformation.. The First Law of TD can be re-written in terms of Entropy ds = dq / T dq = T ds T ds = du + p dα 3

4 3. Integrating, the change in entropy from a transformation from state to state is dq s = s - s = T 4. Entropy in terms of potential temperature dq = cp dt - α dp dq / T = cp dt / T - α dp / T dq / T = cp dlnt - R dlnp; using α = RT / p θ = T (po/p) R/cp dlnθ = dlnt - (R/cp)dlnp cp dlnθ = cp dlnt - R dlnp dq / T = cp dq / θ ds = cp dlnθ Integrating, s = cp lnθ + constant a. Lines of constant entropy are also lines of constant potential temperature b. An isentropic transformation is a constant entropy (and constant potential temperature) transformation I. Generalized Second Law of TD. Real systems are typically irersible and spontaneous, and move in the direction of non-equilibrium to equilibrium.. Four postulates of the Second Law of TD a. There exists a function of state for a body called entropy b. Entropy may change by contact with a thermal reservoir or may undergo internal changes dse -> external entropy changes dsi -> internal entropy changes ds = dse + dsi c. dse = dq/t d. For ersible changes, dsi = 0. For irersible changes dsi > 0. i. dq is the heat added to a substance (per unit mass) in a ersible transformation. ii. dsi = 0 for ersible changes; dsi > 0 for irersible changes 4

5 J. Calculating entropy changes.. Entropy changes for ersible process dq s = s - s = T. Entropy changes for an irersible process s dqir / T 3. Since s depends only on the initial and final states for a ersible transformation, qir can be found from devising a series of ersible transformations that are equal to the irersible transformation a. Reversible adiabatic process dq = 0; ds = 0 b. Reversible phase change at constant p and T; With T = constant dq s = T T dq q = = T i. q is, for example, latent heat ii. With p = constant, q = h, so s = h / T = l m / T where l = latent heat (per unit mass) c. Reversible isothermal process (T=const); entropy is dq q s = s - s = T = T d. Reversible change of state of an ideal gas: Starting with First Law of TD: dq = du + dw = cvdt + pd = cvdt + nr * Tdln ds = dq = cv(t)dlnt + nr * dln T s = cv(t)dlnt + nr * dln s = cv(t)ln(t/t) + nr * ln(/) e. Irersible change of state of an Ideal gas. Change n moles of ideal gas from p, T, to p,, T: Consider ersible processes. i Expansion from at constant T. (e.g.; gas put into a frictionless piston-cylinder; place in constant temperature bath; slowly move piston out) ii Hold gas at, heat gas to temperature T until pressure of p. (Entropy is state function; s for irersible process is Σ s for the ersible process.) s = cv(t)dlnt + nr * ln(/) 5

6 f. Constant pressure heating (ersible process): dq = cpdt s = cp(t)dlnt i. if cp f(t); s = cp ln(t/t) g. Irersible phase change. We want entropy change for mol supercooled water at -0 C to mol of ice at -0 C at constant p i. Intermediate states consist of mixtures of water / ice at -0 C --- theses are not equilibrium states ii. Removal of small amount of heat will not result in freezing at -0 C to become supercooled water. h. isobaric heating dq s = = T water -----ir-----> ice (-0 ) (-0 ) 3 water > ice (0 C) (0 ) cp dlnt i. constant p phase change isobaric cooling s cp dlnt cp ln T/T = = ( ) i. Entropy change for free expansion of a gas i. A wrong conclusion be found if we start without thinking that dq s = T ii. First, T is constant, and q is constant, but is not, thus, dq s = = pd nr * T dln nr * ln(/) T = = 6

7 4. Entropy for Reversible and Irersible processes. a. suniv = ssystem + ssurround b. For a ersible process, heat flow must occur between the system and the surrounding with no finite temperature difference between the system and surrounding. dq -----> Tsys Tsurround Universe <----- dq dq dq dsuniv = dssys + dssurround = + Tsys Tsurround = 0 because Tsys = Tsurround i. suniv = 0 for ersible process ii. suniv > 0 for irersible process iii. suniv 0 in general iv. Entropy can't be destroyed. It can remain the same or increase. --- This is alternative way of stating the Second Law of Thermodynamics 5. Entropy and Equilibrium a. For an isolated - irersible process, s > 0. b. All real processes are irersible - entropy will be increasing i. Heat will flow from warm to cold bodies and s will increase. ii. Heat flow will continue until temperature of two bodies is equal -- entropy then will be a local maximum iii. Thermodynamic Equilibrium occurs when a systems entropy is a maximum c. For a ersible Carnot cycle, the net increase in entropy is i. s = Q/T - Q/T = 0 for Carnot cycle. ii. Q/T = the increase in entropy of working substance d. Q/T = the decrease in entropy of heat sink (cold reservoir) i. s = along adiabats ii. Q/T = Q/T for Carnot cycle iii. s = 0 for Carnot cycle. iv. No entropy change for a ersible process. 7

1. Second Law of Thermodynamics

1. Second Law of Thermodynamics 1. Second Law of hermodynamics he first law describes how the state of a system changes in response to work it performs and heat absorbed. he second law deals with direction of thermodynamic processes