2. 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

3. Isothermal Processes Recall: Isothermal means  T = 0. Since  E = nC v  T, then  E = 0 for an isothermal process. Since  E = q + w, then q = -w (isothermal process)

4. 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  T = 0.

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

6. One-Step Expansion: If we change the weight to M 1 /4, then the pressure becomes P ext = (M 1 /4)g/A = P 1 /4 The mass will be lifted until the internal pressure equals the external pressure, at which point V final = 4V 1 What work is done in this expansion? w = -P ext  V = -P 1 /4 (4V 1 - V 1 ) = -3/4 P 1 V 1

7. What if we expand in two steps? In this expansion we go in two steps: Step 1: M 1 to M 1 /2 Step 2: M 1 /2 to M 1 /4 In first step: P ext = P 1 /2, V final = 2V 1 w 1 = -P ext  V = -P 1 /2 (2V 1 - V 1 ) = -1/2 P 1 V 1

8. In Step 2 (M 1 /2 to M 1 /4 ): P ext = P 1 /4, V final = 4V 1 w 2 = -P ext  V =- P 1 /4 (4V 1 - 2V 1 ) = -1/2 P 1 V 1 w total = w 1 + w 2 = -P 1 V 1 /2 - P 1 V 1 /2 = -P 1 V 1 Note: w total,2 step > w total,1 step More work was done in the two-step expansion

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

10. 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 w total, we can integrate:

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

12. If we perform the integration from V 1 to V 2:

13. Two Step Compression Now we will do the opposite, compress the fully expanded gas: V init = 4V 1 P init = P 1 /4 Compress in two steps: first put on mass = M 1 /2, Then, in step two, replace mass M 1 /2 with a bigger mass M 1

14. In first step: w 1 = -P ext  V = -P 1 /2 (2V 1 - 4V 1 ) = P 1 V 1 w total = w 1 + w 2 = 2P 1 V 1 (see table 10.3) In second step: w 2 = -P ext  V = -P 1 (V 1 - 2V 1 ) = P 1 V 1

15. Compression/Expansion In two-step example: w expan. = -P 1 V 1 w comp. = 2P 1 V 1 w total = P 1 V 1 q total = -P 1 V 1 We have undergone a “cycle” where the system returns to the starting state. Since isothermal (  T = 0) then,  E = 0 but, q = -w ≠ 0

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

17. Step 1: Isothermal Expansion at T = T high from V 1 to V 2  T = 0, so  E = 0 and q = -w Do this expansion reversibly, so that

18. Step 2: Const V cooling to T = T low.  V = 0, therefore, w = 0 q 2 =  E = nC v  T = nC v (T low -T high )

19. Step 3: Isothermal compression at T = T low from V 2 to V 1. Since  T = 0, then  E = 0 and q = -w Do compression reversibly, then

20. Step 4: Const-V heating to T = T high.  V = 0, so, w = 0, and q 4 =  E = nC v  T = nC v (T high -T low ) = -q 2

22. The thermodynamic definition of entropy(finally!)

23. Calculating Entropy: summary  T = 0  V = 0  P = 0

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

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