Entropy A measure of molecular disorder

Transcription

1 Entropy A measure of molecular disorder Second Law uses Entropy, S, to identify spontaneous change. Restatement of Second Law: The entropy of the universe tends always towards a maximum (S universe > 0 for a spontaneous process)

2 Second Law of Thermodynamics Thermodynamic Definition of Entropy No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work During a spontaneous change in an isolated system, the total energy of the system is conserved, but the total energy is rearranged is such a way as to increase the random thermal motion of the particles that make up the isolated system. Second Law: The entropy of an isolated system increases in a spontaneous change. dqrev ds T Not useful! S f i dq T rev

3 Heat is Never Completely Converted into Work Some Heat Dissipated to Surroundings Not 100% Conversion

4 Thermodynamic Definition of Entropy To make progress with the thermodynamic definition of S, we need to break the isolated system (Universe) into two components: system and surroundings. Heat can be transferred between the two components. surroundings system ds surr, rev = dq surr, rev /T ds sys = dq sys, rev /T ds total = ds surr +ds sys dq f surr, rev S T i f i dq sys, rev T

5 Sample Problem Calculate the entropy change in the surroundings when 1.0 mole of H 2 O(l) is formed from its elements under standard conditions at 25 C ( K). H 2 (g) + 1/2 O 2 (g) H 2 O (l) f H = -286 kj/mol surr H 2 + O 2 sys Solution: all processes are at one temperature, and so (if reversible) 286 kj/mol is transferred to the surroundings in this process. Thus ΔS surroundings = ΔS surr =dq surr /T= (+286 kj/mol)/298 K = kj/mol K = 960 J/K No shocking new insights here!

6 Sample Problem 2 Calculate the entropy change when a sample of an ideal gas at temperature T expands isothermally from V i to V f. The gas is the system. Vf dq S sys V i Vf sys, rev 1 qrev dqsys, rev T T T V i = since constant T So we must now choose a reversible path, and evaluate dq rev along that path. We carry out a reversible isothermal expansion, and we obtain U 0 q w rev rev V f q w nrt ln rev rev V i V f S nrln sys V i Expansion, so S > 0. Spontaneous?? We have looked at S sys and S surr, but not S total. Issue?

7 Entropy The 2 nd Law involves something called entropy. Entropy, S, is a STATE Function, like U and H. Its meaning, however, is really difficult to really understand. The text follows the traditional (historical) route in describing entropy in terms of macroscopic thermodynamics and then defined entropy microscopically. We will take the opposite approach.

8 Statistical Definition of Entropy S = k B ln Ω Ω is the number of ways to assign a POSITION and a VELOCITY to EACH individual particle in the system (i.e., a state to each individual particle), and still have the energy of all of the particles add up to the total energy of the system.

9 Example To simplify, suppose the state of a particle is given only by its 2D position in a rectangle, rather than both its 3D position and velocity. Further, assume there are 3 particles and 63 positions. Each box below represents a position and a different individual microstate. The red line is a partition restricting the particles to the left side. With the partition in place, the number of available microstates is given by 21! 21 3!

10 Example (2) 21! ! Now remove the partition: Spontaneous expansion and S increases. 63! 63 3! microstates, most looking a lot like the one above, with the particles spread all over the box.

11 ConcepTest 1 21! 7980 compressed 21 3! 63! 63 3! ful lbox Consider a compressed condition, where all three particles are in the left three columns? We allow this system to evolve for an extended period of time, and take 1000 snapshots of the gas. In approximately how many of the snapshots will all three particles be found in the left three columns? a. 3 d. 500 b. 17 e. Not enough information to obtain answer c. 35

12 ConcepTest 1 21! 7980 compressed 21 3! 63! 63 3! ful lbox Consider a compressed condition, where all three particles are in the left three columns? We allow this system to evolve for an extended period of time, and take 1000 snapshots of the gas. In approximately how many of the snapshots will all three particles be found in the left three columns? a. 3 d. 500 b. 17 e. Not enough information to obtain answer c. 35 Key assumption: ergodic hypothesis!

13 Example (2) Even more unlikely: all three in the top left 2 x 2 rectangle. 4! microstates 4 3! Compare with 63! 63 3! microstates in the full box. Now if we take snapshots, we expect to see 1 1 snapshots where all three particles are in the upper corner. Where did the 1 come from? Poisson statistics (P Chem Lab)

14 Thermodynamic Entropy Definition: The thermodynamic definition of entropy is dqrev ds T dq change in heat of the corresponding reversible process rev The processes that we are interested in may be irreversible, but we still calculate S for the corresponding reversible process.* *Any reversible process that starts at the same initial state and ends at the same final state will do, as we will see that S is a state function.

15 Another Example Calculate the entropy change in an isothermal compression of an ideal gas at temp T from V to 0.5V. Reversible: 2P du = C P dt = 0 for an isothermal process 0 = đq + đw P đq = -đw = PdV Rev dq P nrt nr ds dv dv dv T T VT V V nr 0.5V 1 2 S dv nrln nrln nrln 2 V V V 2 S nrln 2 Irreversible: Instantly jump P to 2P but keep T constant as V goes to 0.5 V du = 0, so q = -w = 2PΔV = PV Still, however, ΔS = -nrln(2), as S is a STATE FUNCTION Are these processes spontaneous or not? P Irrev 0.5V V V

16 Calculate the entropy change in an isothermal compression of an ideal gas at temp T from V to 0.5V. S nrln 2 for either path Are these processes spontaneous or not? S S S univ sys surr Another Example (2) We know ΔS sys for both paths. We need to determine ΔS surr. P 2P P Rev Irrev 0.5V V V It is very tempting to say that we know that q sys = -q surr and thus for either path that it should be that ΔS sys = -ΔS surr But we would be wrong!!!