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Lect# 7: Thermodynamics and Entropy

Lect# 7: Thermodynamics and Entropy. Reading: Zumdahl 10.2, 10.3 Outline: Isothermal processes (∆T = 0) Isothermal gas expansion and work(w) Reversible and irreversible processes. Isothermal Processes. Recall: Isothermal means D T = 0.

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Lect# 7: Thermodynamics and Entropy

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  1. Lect# 7: Thermodynamics and Entropy • Reading: Zumdahl 10.2, 10.3 • Outline: Isothermal processes (∆T = 0) Isothermal gas expansion and work(w) Reversible and irreversible processes

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

  3. Example: Isothermal Expansion Consider a mass (M) connected to a ideal gas confined by a piston. The piston is submerged in a constant T bath, so DT = 0.

  4. Initially, V = V1 P = P1 Pressure of gas is equal to that created by the pull of the hanging mass: P1 = force/area = M1g/A A = piston area g = gravitational acceleration (9.8 m/s2) Note: kg m-1 s-2 = 1 Pa

  5. One-Step Expansion: If we change the weight to M1/4, then the pressure becomes Pext = (M1/4)g/A = P1/4 The mass will be lifted until the internal pressure equals the external pressure, at which point Vfinal = 4V1 What work is done in this expansion? w = -PextDV = -P1/4 (4V1 - V1) = -3/4 P1V1

  6. What if we expand in two steps? 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

  7. 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 Note: wtotal,2 step > wtotal,1 step More work was done in the two-step expansion

  8. Graphically, we can envision this two-step process on a PV diagram: • Work is given by the area under the “PV” curve.

  9. Infinite Step Expansion Imagine that we perform a process in which we decrease the weight very slightly (∆M) between an infinite number of (reversible) expansions. Instead of determining the sum of work performed at each step to get wtotal, we can integrate:

  10. Graphically, see how much more work is done in the infinite-step expansion (red area) Reversible Two Step

  11. If we perform the integration from V1 to V2:

  12. Two Step Compression Now we will do the opposite, compress the fully expanded gas: Vinit = 4V1 Pinit = P1/4 Compress in two steps: first put on mass = M1/2, Then, in step two, replace mass M1/2 with a bigger mass M1

  13. 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)

  14. 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 Since isothermal (DT = 0) then, DE = 0 but, q = -w ≠ 0

  15. Let’s consider the four- step cycle illustrated: 1: Isothermal expansion 2: Const V cooling 3: Isothermal compression 4: Const V heating Entropy: A Thermodynamic Definition

  16. Step 1: Isothermal Expansion at T = Thigh from V1 to V2 DT = 0, so DE = 0 and q = -w Do this expansion reversibly, so that

  17. Step 2: Const V cooling to T = Tlow. DV = 0, therefore, w = 0 q2 = DE = nCvDT = nCv(Tlow-Thigh)

  18. Step 3: Isothermal compression at T = Tlow from V2 to V1. Since DT = 0, then DE = 0 and q = -w Do compression reversibly, then

  19. Step 4: Const-V heating to T = Thigh. DV = 0, so, w = 0, and q4 = DE = nCvDT = nCv(Thigh-Tlow) = -q2

  20. The thermodynamic definition of entropy (finally!)

  21. Calculating Entropy: summary DT = 0 DV = 0 DP = 0

  22. Calculating Entropy: simple example Example: What is DS for the heating of a mole of a monatomic gas @constant volume from 298 K to 350 K? 3/2R

  23. Connecting with Lecture 6 • From this lecture: • Exactly the same as derived in the previous lecture!

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