du = δq + δw = δq rev + δw rev = δq rev + 0

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1 Chem 4501 Introduction to hermodynamics, 3 Credits Kinetics, and Statistical Mechanics Module Number 6 Active Learning Answers and Optional Problems/Solutions 1. McQuarrie and Simon, 6-6. Paraphrase: Compute q rev and ΔS for a reversible cooling of one mole of an ideal gas at a constant volume V 1 from P 1, V 1, 1 to P 2, V 1, 4 followed by a reversible expansion at constant pressure P 2 from P 2, V 1, 4 to P 2,, 1. In the first step, there is no volume change, so there is no work. From the First Law we thus have du = δq + δw = δq rev + δw rev = δq rev + 0 = C V ( )d he change from general heat and work to reversible heat and work follows from U being a state function. We were told that this process is reversible, so we could have just started with a subscript reversible on everything, but it s useful to emphasize the relationship between the two possibilities when a state function like U is involved. he final equality follows because we have an ideal gas, whose energy depends only on and does so in the matter shown. o determine qrev we simply need to integrate the differential q (which is exact in this case, since the differential work is zero) and to determine the entropy change, we integrate q/. hus

2 q rev = C V ( )d =? ΔS = 1 4 ( ) C V 4 d =? 1 he question marks are there because we have not been told that CV() is independent of, so we cannot remove it from the integral. But, let s go on to the second step, the reversible expansion. In this case, there is a volume change, and so work, and the first law gives us du = δq +δw = δq rev +δw rev = δq rev PdV Rearranging to solve for differential q we have δq rev = du + PdV = C V ( )d + PdV he differential in this case is exact (no mixing of V dependence in the first term or dependence in the second term), so we can solve for the heat transfer first as 1 q rev = C V ( )d + P 2 dv 4 4 = C V ( )d + P 2 dv 1 4 = C V ( )d + P 2 V 1 1 V 1 V 1 ( ) If we consider the two steps together, we see that the unknown integral including the heat capacity cancels out in the two steps, so that the change in heat is P2(V2 V1). As V2 is larger than V1, the heat is positive as it should be given that the system did negative work (expansion) but the change in U was zero

3 (isothermal ideal gas process). Don t be confused if that seems a paradox (how can you gain heat and stay at the same temperature?) he heat that keeps the ideal gas at the same temperature must be provided from outside (a bath). As for the entropy change, while P2 is a constant over the expansion, P2/ is not, and we need the ideal gas equation of state (for one mole of gas) to compute ΔS = 1 C V ( ) d C V ( ) V 1 P 2 dv R V dv = d + 1 V 1 4 C = V ( ) d + Rln 1 V 1 Again, the unknown integral involving CV cancels when summed with the entropy change from the first step and the net entropy change is Rln(V2/V1). Comparing to the examples of course video 6.2, we see that this new two-step process, from the same start point to the same end point as others, gives the same entropy change (as it must: entropy is a state function), but a different heat change, as the heat change depends on path. 2. McQuarrie and Simon, 6-9. Paraphrase: What is the entropy change for one mole of an ideal gas expanded reversibly and isothermally from 1 bar to 0.1 bar? Since the process is isothermal, the internal energy does not change (ideal gas) and the change in heat is the opposite of the change in work, or δqrev = δwrev = ( PdV) = PdV. But, we

4 aren t told the change in volume, we re told the change in pressure. So, we need to use the ideal gas equation of state to express dv as dp. We proceed V = R P so dv = R P 2 dp (we were told the process was isothermal, so we don t need to worry about d). Now, to compute entropy, we proceed ΔS = δq rev PdV = P& = ( R ' P 2 dp ) + * 0.1dP = R = R ln P ( ) =19.1 J K 1 he sign is positive because our gas is expanding (more disorder for a gas in more volume compared to less volume). 3. McQuarrie and Simon, Paraphrase: Derive the entropy change for a constant pressure process if C P is independent of and apply it to 2 moles of liquid water heated from 10 to 90 C at constant pressure with molar C P = 75.2 J K 1 mol 1. For a constant pressure process, the change in heat is equal to the change in enthalpy (see M&S eq. 5.37). And the constant pressure heat capacity is defined from dh = CPd, so we have ΔS = δq rev = C P d % = C P ln ( f ' & * i )

5 where i and f are the initial and final temperatures, respectively. If we plug in the relevant values given for water (noting that we have 2 moles of water) we determine an entropy change of 37.4 J K 1. Positive, as expected (more entropy at higher temperature as more states can be accessed). 4. McQuarrie and Simon, Paraphrase: If two systems are in contact with one another at the same temperature, but different pressures, and the wall between them can move, what is the entropy change for some move? his is kind of a weird problem, since the way it is worded, the pressure in each system does not change, even though the volume is changing (so, given enough time, we could evidently completely collapse the system at lower pressure in favor of the one at higher pressure, which would not be possible for a system composed of particles having non-zero volume). But, to solve the problem we note that for each case ds i = δq i = du i δw i = du i + P idv i Given the pair of systems, i = 1, 2, we have ds = ds 1 + ds 2 = du 1 + P 1 dv 1 + du 2 + P 2 d = du 1 + P 1 dv 1 = ( P 1 P 2 ) dv 1 du 1 P 2 dv 1 where conservation of energy ensures that du2 = du1 and the nature of the flexible wall ensures that dv2 = dv1. Note that if P1 > P2, then the change in V1 will be positive (expansion of the

6 system under greater pressure) and the entropy change is positive. If P1 < P2, the change in V1 will be negative, and again, entropy change will be positive. 5. McQuarrie and Simon, 6-25 and Paraphrase: What is the entropy change in the system and in the surroundings if one mole of an ideal gas is expanded isothermally and reversibly from a pressure of 10 bar to 2 bar at 300 K. And, what if the expansion is into a vacuum and not reversible? his is an ideal gas process that is isothermal, so we know that du = 0 (U for ideal gas depends only on ). hus, we have that δqrev,sys = δwrev,sys = ( PdV) = PdV. In the solution to M&S 6-6 (problem #1 above), we already determined that the entropy change for this process is Rln(V2/V1). Since the process is isothermal, we can use the ideal gas equation of state to write the entropy change as Rln(P1/P2), or Rln5, which is 13.4 J K 1. As expected, the entropy change for the gas is positive (it expanded). Since the process is reversible, the entropy change of the universe is zero (the universe is the ultimate isolated system and ds = 0 for a reversible process in an isolated system), so the entropy change of the surroundings must then be 13.4 J K 1. (Another way of looking at it is that the reversible change in heat of the system is equal and opposite to the reversible change in heat of the surroundings, that is, δqrev,sys = δqrev,surr, and thus, given a constant, the two entropies derived from integrating over the reversible path are equal and opposite). If the expansion is against a vacuum, no work is done, but, since entropy is a state function, the change in entropy of the system is the same, i.e., 13.4 J K 1. Since there was no work done, and the process was isothermal, both dusys and δwsys are zero, and thus so is δqsys and, by extension, δqsurr. he entropy of the surroundings has not changed, so the net entropy change of the universe is 13.4 J K 1. Note that it should not seem a paradox

7 that the entropy change is positive even though all heat changes were zero. he Second Law says ds δq where the equality holds true only for reversible processes. In all other processes, we don t necessarily have a way to compute the entropy change (unless we know the entropy of the initial and final states from some other measurement or calculation entropy is a state function), but we will be able to set a lower limit from knowledge of the reversible process. 6. McQuarrie and Simon, Paraphrase: What is the change in entropy for mixing equal volumes of two ideal gases under identical conditions? Using the final result of Example 6-3, we have that the entropy of mixing of two ideal gases is ΔS = R( x 1 ln x 1 + x 2 ln x 2 ) in the case of equal volumes of ideal gases under the same conditions, the number of particles must be equal, so the mole fraction of each gas must be 0.5. In that case, we have Q.E.D. ΔS = R( 0.5ln ln 0.5) = R ln0.5 = R ln 2 7. McQuarrie and Simon, Paraphrase: Determine the Maxwell relation for enthalpy of an ideal gas and use it to compute the molar entropy

8 change on going from one,p state point to another if the molar constantpressure heat capacity is temperature independent. We begin with the derivation of dh as an exact differential. dh = d( U + PV ) = du + PdV +VdP = δq rev + δw rev + PdV +VdP = δq rev PdV + PdV +VdP = ds +VdP where in the last line we ve used the relationship ds = qrev/. Now solving for a change in entropy we rearrange to ds = 1 dh V dp For an ideal gas, H depends only on according to dh = CPd (we ve been told CP doesn t depend on ) and we can also use the ideal gas equation of state to relate V/ to P and arrive at ds = C P d nr dp P Upon integrating from an initial (i) to a final (f) state point, we have d ΔS = C f dp P nr i P f P i P % = C P ln ( % f ' & * i ) nr ln P ( f ' & P * i )

9 In the case of molar quantities, n = 1 giving " ΔS = C P ln f $ # i % " ' Rln P f $ & # P i % ' & 8. When is S = k B ln Ω, where Ω is the degeneracy of a level having energy E? (a) After an isothermal expansion (d) When H > U (b) For an isolated system at (e) (b) and (c) equilibrium (c) When every state of the level is equally populated (f) Never

Chapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc.

Chapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc. Chapter 3 Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc. Concepts Energy functions F and G Chemical potential, µ Partial Molar properties