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 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

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

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.

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) = Nk ln(Wfinal/Winitial) for N molec. = Nkln(Vfinal/Vinitial)

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)

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.

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)

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

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.

Two Step Expansion • w1 = -PextDV = -P1/2 (2V1 - V1) = -1/2 P1V1 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

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

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

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:

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

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

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)

Compression/Expansion 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. Now, DE = 0 (state fxn) But, q = -w ≠ 0

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

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

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

Defining Entropy (cont)

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

Calculating Entropy DT = 0 DV = 0 DP = 0

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

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