Homework Week The figure below depicts the isothermal compression of an ideal gas. isothermal du=0. δq rev = δw rev = P dv

Transcription

1 Statistical Molecular hermodynamics University of Minnesota Homework Week 6 1. he figure below depicts the isothermal compression of an ideal gas. Start from the First and Second Laws of thermodynamics and calculate the molar entropy change of the system, S sys, for a compression that reduces the volume by half, 1 2. a) 5.8 mol 1 K 1 b) 0.58 mol 1 K 1 c) 1.0 mol 1 K 1 d) 5.8 mol 1 K 1 e) 0.0 mol 1 K 1 From the first law, du δq rev + δw rev And from the second law, isothermal du0 δq rev δw rev P d ds δq rev S 2 δq 2 rev P d From the ideal gas law, P nr S 2 nr d nr ln 2 ) We can now substitute, 2 S nr ln 2 ) 1 nr ln 2 2

2 o get the molar change, divide both sides by n, S 1 R ln 2 And now we can plug in the numbers. S mol 1 K 1) 0.693) mol 1 K 1 2. he figure below depicts the isothermal compression of an ideal gas. If the compression is done along a reversible path, calculate the change in the entropy of the surroundings, S surr, for a compression that reduces the volume by half, 1 2. a) 5.8 mol 1 K 1 b) 0.58 mol 1 K 1 c) 1.0 mol 1 K 1 d) 5.8 mol 1 K 1 e) 0.0 mol 1 K 1 We know from the previous problem that we can compute the entropy change of the system as follows: From the first law, du δq rev + δw rev And from the second law, From the ideal gas law, We can now substitute, ds δq rev P nr isothermal du0 q rev δw rev P d S S S nr ln δq rev δw rev P d δq rev 2 nr d nr ln 2 P d 2 ) ) 1 nr ln 2 2

3 o get the molar change, divide both sides by n, S 1 R ln 2 And now we can plug in the numbers. S mol 1 K 1) 0.693) mol 1 K 1 he process is reversible, so S total 0 S system + S surroundings S surroundings S system mol 1 K 1 3. You have one mole of distinguishable molecules, and at 0 K each molecule has 3 equally probable states each molecule has a degeneracy of 3 in its lowest energy level). What is the entropy of this system at 0 K? a) S 8.2 mol 1 K 1 b) S 3.5 mol 1 K 1 c) S 0.0 mol 1 K 1 d) S 3.1 mol 1 K 1 e) S 6.2 mol 1 K 1 f) S 9.1 mol 1 K 1 g) S 12 mol 1 K 1 h) S 16 mol 1 K 1 i) S 24 mol 1 K 1 S k B ln Ω 1) k B ln 3 N A 2) N A k B ln 3 3) R ln 3 4) mol 1 K 1) 1.099) 5) mol 1 K 1 6)

4 4. Given an ideal gas, determine q rev and S for path I in the, P plane illustrated in the figure below. a) q rev 4 1 C d and S 3 2 C d b) q rev 4 1 C d and S 4 C d 1 c) q rev 4 1 C d and S 4 C d 1 d) q rev 4 1 C d and S 4 C d 1 + nr Figure 1: For path I, P 1,, 1 P 2,, 4. As this is a constant volume process, and hence no work is done, we will start from the constant volume heat capacity, δu δq C d d From that we can write, δq C d. 7) Since we are dealing with a reversible process, we can write, δq rev C d. So, for the transformation we re considering, q rev is given by the path integral, q rev C d ) 8) Path I

5 in which everything in the integral depends only on, and therefore q rev 4 o calculate S, recall that by definition, 1 C d 9) From our previous considerations, S f. state i. state δq rev. 10) S 4 1 C d 11)

6 5. Given an ideal gas, determine q rev and S for path II in the, P plane illustrated in the figure below. Figure 2: a) q rev 1 4 C d + nr 1 4 ) and S 1 C d 4 b) q rev 1 4 C d + nr 1 4 ) and S 4 c) q rev 1 4 d) q rev 1 4 e) q rev nr ln C d C d ) and S C d and S 4 C d 1 and S 4 C d 1 + nr C d + nr ) Let s start by applying the First Law to our process, δq rev du δw rev 12) du P d ) 13) du + P d 14) C d + P d 15)

7 And integrating over the path: q rev C d + P d ) 16) Path II C d + P 2 1 d 17) 4 C d + P 2 ) 18) 1 4 C d + nr 1 4 ) 19) For the entropy change, S f. state i. state Path II δq rev C d C d + C d 20) + P d ) 21) 2 P d 2 ) 22) 23)

8 6. Given an ideal gas, determine q rev and S for proceeding first along path I and then along path II in the, P plane illustrated in the figures for Problems 4 and 5, resulting in a net change from P 1,, 1 P 2,, 1. a) q rev nr 1 4 ) and S 4 b) q rev nr 1 ln c) q rev 1 4 d) C d and S nr ln 1 C d q rev nr 1 4 ) and S nr ln e) q rev nr ln and S 4 C d 1 + nr 1 and S 4 C d 1 + nr 1 Since S is a state function, we can calculate the total entropy change from P 1,, 1 ) to P 2,, 1 ) independent of path. As the two endpoints are the same temperature, let us follow a reversible, isothermal path, and then apply the second law, S q, to determine the entropy change. o do this, we need to know q for this process, of course. Recall that the internal energy of an ideal gas is dependent only upon its temperature, and we are considering an isothermal process, so, from the first law, U q + w 0 q + w q w We can calculate the work done from w P d, which for an ideal gas is, 2 w 1 P d nr 1 d nr 1 ln, and since q w, q nr 1 ln. Substituting into S q, S q S nr 1ln 1 S nrln

9 Of course, we could determine the changes in heat and entropy more quickly simply by adding the corresponding quantities from the previous two problems for the changes over paths I and II, viz., q rev,total 4 1 C d C d + nr 1 4 ) 24) nr 1 4 ) 25) and 4 1 C d S total + 2 nr ln 1 4 C d 2 ) 26) 27) It s instructive to have done the calculation both ways, however, notice that q rev,total is not the same for the sum of paths 1 and 2 as it was for the isothermal path considered above. hat should come as no surprise as we know that q is a path function, not a state function. However, S is a state function, and, as it must, it has the same value irrespective of the path taken between the endpoints. Note that the difference in q will be balanced by a difference in w, were we to compare that along the different paths, since U is also a state function and one that has a net change of zero given the same temperature 1 at both endpoints.)

10 7. You are given two different gases that are at the same conditions and you mix them in equal volumes. Determine the molar entropy of mixing, S mix. a) ln 2 b) 2R c) 0.25 d) R ln 2 e) 2.73 f) R 2 g) R ln 4 h) none of the above We are mixing equal numbers of moles of the two gases and so each has a mole fraction of 0.5. We can use the formula for S mix discussed in lecture video 6.5, S mix R i y i lny i with y 1 1/2 and y 2 1 y 1 1/2. S mix R [y 1 ln y y 1 ) ln1 y 1 )] 28) [ 1 R 2 ln ) ln1 1 ] 2 ) 29) R ln ) R ln 2 31) 8. Consider the reversible and isothermal expansion of an ideal gas from a pressure of 10 bar to a pressure of 2 bar. Defining the system as the gas and the surroundings as everything else, S total S system + S surroundings the change in the entropy of the system at 300 K is S system 13.4 mol 1 K 1. Which of the following statements is RUE if you carry out the same isothermal expansion, but you carry it out irreversibly against an external pressure of 0 bar? a) he entropy change for the system, S system, will be larger than in the reversible case. b) he entropy change for the surroundings, S surroundings, will be 13.4 mol 1 K 1. c) he total entropy change will be larger than in the reversible case. d) all of the above

11 he gas is expanded isothermally in both cases, so we know U is equal to zero, and S system nrln for either expansion because S is a state function). In the case of the reversible expansion, the OAL entropy change is equal to zero, i.e., the entropy change of the system is exactly equal to that of the surroundings since heat flows reversibly from the surroundings to the system to maintain the gas temperature). S total S system + S surroundings nrln nrln 0 In the case of the irreversible expansion, P ext 0, and therefore w irreversible 0. But we know that U 0 for an ideal gas for an isothermal change. his means that q irreversible 0. here is no heat transferred from the surroundings to the system, so S surroundings 0. Now we can determine the OAL entropy change, S total S system + S surroundings nrln + 0 nrln he correct answer is therefore that the total entropy change for the irreversible process is larger than that for the reversible process. he entropy change of the surroundings is zero; it is not negative as in the irreversible case. he entropy change for the system is the same in both the irreversible and reversible case because S is a state function).

12 9. In the lecture videos associated with week 4, we saw that the macroscopic partition function Q for an ideal gas could be written as Q qn N!. 32) where q is the molecular partition function and N is the number of molecules in the macroscopic sample. In lecture video 6.6, the entropy was shown to be computed from the partition function, ln Q S k B + k B ln Q 33) k B k B N, ln qn N! ) N, [ N ln q ln N!) N, + k B ln qn N! ] 34) + k B N ln q ln N! 35) Nk B ln q + k BN ln q N ln N + N) 36) Where we have used Stirling s formula for the final term; thus, 37) S Nk B ln q + Nk B ln q N + Nk B 38) Determine the molar entropy of oxygen gas at K and 1 bar given a molecular mass of kg, Θ vib 2256 K, Θ rot 2.07 K, σ 2, and g e1 3 oxygen has an interesting triplet ground electronic state). a) b) 7R 2 c) K mol K mol K mol d) K mol e) K mol We need to find the expression for the molecular partition function for O 2 and do some algebra. Recall from lecture video 4.4 that q q trans q rot q vib q elec, 39) which for a homonuclear diatomic molecular gas is see lecture video 4.6), q 2πMkB h 2 ) 3 2 e Θ vibr/2 σθ rot 1 e g e1e De/kB. 40) Θ vibr/

13 We can first compute ln q N : ln q N ln q trans N + ln q rot + ln q vib + ln q elec [ 2πMkB ) ] 3 2 ln + ln Θ vib h 2 N σθ rot 2 ln1 e Θ vib/ ) + ln g e1 + D e k B and then ln q : ln q Θ Θvib vib 2 + e Θ vib/ e D e 41) Θ vib/ k B 2 Plugging these expressions into the final equation given in the problem and doing some algebra using N N A, one has, [ S R 7 2πMkB ) ] ln 2 + ln ln1 e Θvib/ ) + Θ vib/ h 2 N σθ rot e Θ vib/ 1 + ln g e1. For O 2 we have the values: M kg, Θ vib 2256 K, Θ rot 2.07 K, σ 2, and g e1 3. So each term becomes: 2πMkB h 2 ) 3 2 2π kg /K K m s) 2 ) 3 2 R N P R N A N A P L bar K mol K mol 1 1 bar L m 3 ln ln K σθ rot K ln1 e Θ vib/ ) Θ vib / 2256 K/ K e Θ vib/ 1 e 2256 K/ K 1 ln g e1 ln3) Using these many values, we get: S K mol [7 2 + ln m m 3 ) ] K mol

14 Note that most of the entropy is due to terms related to the translational, and to a lesser degree, rotational and electronic degrees of freedom. Experiment measures an entropy of 205.2, which certainly must be regarded as stunningly good agreement. K mol 10. What is the maximum efficiency of a steam engine operating between 20 C and the boiling point of water, 100 C, at one atmosphere? a) 1 b) 21 % c) 0.21 % d) 42 % e) 0.80 % f) 80 % g) 78 % From lecture video 6.8 we know that the maximum efficiency of a heat engine is equal to 1 cold / hot ; therefore, Max Efficiency K 373 K %