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Lecture 7: Thermo and Entropy Reading: Zumdahl 10.2, 10.3 Outline –Isothermal processes –Isothermal gas expansion and work –Reversible Processes

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

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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 T = 0.

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

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Isothermal Expansion (cont.) One-Step Expansion. We change the weight to M 1 /4, then P ext = (M 1 /4)g/A = P 1 /4 The mass will be lifted until the internal pressure equals the external pressure. In this case V final = 4V 1 w = -P ext V = -P 1 /4 (4V 1 - V 1 ) = -3/4 P 1 V 1

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Two Step Expansion 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

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Two Step Expansion (cont.) 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 w total,2 step > w total,1 step

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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.

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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 w total, we integrate:

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Infinite Step Expansion (cont.) Graphically: Two StepReversible

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Infinite Step Expansion (cont.) If we perform the integration from V 1 to V 2:

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Two Step Compression Now we will do the opposite….take the gas and compress: V init = 4V 1 P init = P 1 /4 Compression in two steps: first place on mass = M 1 /2 second, replace mass with one = M 1

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Two Step Compression (cont.) 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

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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. Now, E = 0 (state fxn) But, q = -w ≠ 0

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

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Defining Entropy (cont) Step 1: Isothermal Expansion at T = T high from V 1 to V 2 Now T = 0; therefore, E = 0 and q = -w Do expansion reversibly. Then:

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Defining Entropy (cont) Step 2: Isochoric Cooling to T = T low. Now V = 0; therefore, w = 0 q 2 = E = nC v T = nC v (T low -T high )

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Defining Entropy (cont) Step 3: Isothermal Compression at T = T low from V 2 to V 1. Now T = 0; therefore, E = 0 and q = -w Do compression reversibly, then

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Defining Entropy (cont) Step 4: Isochoric Heating to T = T high. Now V = 0; therefore, w = 0 q 4 = E = nC v T = nC v (T high -T low ) = -q 2

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Defining Entropy (cont)

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

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Calculating Entropy T = 0 V = 0 P = 0

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

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

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