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Lecture 6: Thermo and Entropy

Lecture 6: Thermo and Entropy . Reading: Zumdahl 10.2, 10.3 Outline Isothermal processes Isothermal gas expansion and work Reversible Processes. Weight and Entropy. The connection between weight ( W ) and entropy (S) is given by Boltzmann’s Formula: S = k ln W.

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Lecture 6: Thermo and Entropy

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  1. Lecture 6: Thermo and Entropy • Reading: Zumdahl 10.2, 10.3 • Outline • Isothermal processes • Isothermal gas expansion and work • Reversible Processes

  2. Weight and Entropy • The connection between weight (W) and entropy (S) is given by Boltzmann’s Formula: S = k lnW k = Boltzmann’s constant = R/Na = 1.38 x 10-23 J/K • The dominant configuration will have the largest W; therefore, S is greatest for this configuration

  3. Example: Crystal of CO • For a mole of CO: W = Na!/(Na/2!)2 = 2Na • Then, S = k ln(W) = k ln (2Na) = Nak ln(2) = R ln(2) = 5.64 J/mol.K

  4. Another Example: Expansion • What is DS for the expansion of an ideal gas from V1 to 2V1? • Focus on an individual particle. After expansion, each particle will have twice the number of positions available.

  5. Expansion (cont.) • Note in the previous example that weight was directly proportional to volume. • Generalizing: DS = k ln (Wfinal) - kln(Winitial) = k ln(Wfinal/Winitial) = k ln(Wfinal/Winitial) = Nk ln(Wfinal/Winitial) for N molec. = Nkln(Vfinal/Vinitial)

  6. Isothermal Processes • Recall: Isothermal means DT = 0. • Since DE = nCvDT, then DE = 0 for an isothermal process. • Since DE = q + w: q = -w (isothermal process)

  7. Example: Isothermal Expansion • Consider a mass connected to a ideal gas contained in a “piston”. Piston is submerged in a constant T bath such that DT = 0.

  8. Isothermal Expansion (cont.) • Initially, V = V1 P = P1 • Pressure of gas is equal to that created by mass: P1 = force/area = M1g/A where A = piston area g = gravitational acceleration (9.8 m/s2)

  9. Isothermal Expansion (cont.) • One-Step Expansion. We change the weight to M1/4, then Pext = (M1/4)g/A = P1/4 • The mass will be lifted until the internal pressure equals the external pressure. In this case Vfinal = 4V1 • w = -PextDV = -P1/4 (4V1 - V1) = -3/4 P1V1

  10. Two Step Expansion (cont.) • Graphically, we can envision this two-step process on a PV diagram: • Work is given by the area under the “PV” curve.

  11. Two Step Expansion • In this expansion we go in two steps: Step 1: M1 to M1/2 Step 2: M1/2 to M1/4 • In first step: Pext = P1/2, Vfinal = 2V1 • w1 = -PextDV = -P1/2 (2V1 - V1) = -1/2 P1V1

  12. Two Step Expansion (cont.) • In Step 2 (M1/2 to M1/4 ): Pext = P1/4, Vfinal = 4V1 • w2 = -PextDV =- P1/4 (4V1 - 2V1) = -1/2 P1V1 • wtotal = w1 + w2 = -P1V1/2 - P1V1/2 = -P1V1 • wtotal,2 step > wtotal,1 step

  13. Infinite Step Expansion (cont.) • Graphically: Two Step Reversible

  14. Infinite Step Expansion • Imagine that we perform a process in which we change the weight “infinitesimally” between expansions. • Instead of determining the sum of work performed at each step to get wtotal, we integrate:

  15. Infinite Step Expansion (cont.) • If we perform the integration from V1 to V2:

  16. Two Step Compression • Now we will do the opposite….take the gas and compress: Vinit = 4V1 Pinit = P1/4 • Compression in two steps: first place on mass = M1/2 second, replace mass with one = M1

  17. Two Step Compression (cont.) • In first step: w1 = -PextDV = -P1/2 (2V1 - 4V1) = P1V1 • In second step: w2 = -PextDV = -P1 (V1 - 2V1) = P1V1 • wtotal = w1 + w2 = 2P1V1 (see table 10.3)

  18. In two step example: wexpan. = -P1V1 wcomp. = 2P1V1 wtotal = P1V1 qtotal = -P1V1 We have undergone a “cycle” where the system returns to the starting state. Compression/Expansion • Now, DE = 0 (state fxn) • But, q = -w ≠ 0

  19. Let’s consider the four-step cycle illustrated: 1: Isothermal expansion 2: Isochoric cooling 3: Isothermal compression 4: Isochoric heating Defining Entropy

  20. Step 1: Isothermal Expansion at T = Thigh from V1 to V2 Now DT = 0; therefore, DE = 0 and q = -w Do expansion reversibly. Then: Defining Entropy (cont)

  21. Step 2: Isochoric Cooling to T = Tlow. Now DV = 0; therefore, w = 0 q2 = DE = nCvDT = nCv(Tlow-Thigh) Defining Entropy (cont)

  22. Step 3: Isothermal Compression at T = Tlow from V2 to V1. Now DT = 0; therefore, DE = 0 and q = -w Do compression reversibly, then Defining Entropy (cont)

  23. Step 4: Isochoric Heating to T = Thigh. Now DV = 0; therefore, w = 0 q4 = DE = nCvDT = nCv(Thigh-Tlow) = -q2 Defining Entropy (cont)

  24. Defining Entropy (cont)

  25. Defining Entropy (end) The thermodynamic definition of entropy (finally!)

  26. Calculating Entropy DT = 0 DV = 0 DP = 0

  27. Calculating Entropy • Example: What is DS for the heating of a mole of a monatomic gas isochorically from 298 K to 350 K? 3/2R

  28. Connecting with Dr Boltzmann • From this lecture: • Exactly the same as derived in the previous lecture!

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