Thermodynamics Boltzmann (Gibbs) Distribution Maxwell-Boltzmann Distribution Second Law Entropy

Transcription

1 Thermodynamics Boltzmann (Gibbs) Distribution Maxwell-Boltzmann Distribution Second Law Entropy Lana Sheridan De Anza College May 8, 2017

2 Last time modeling an ideal gas at the microscopic level pressure, temperature, and internal energy from microscopic model equipartition of energy rms speed of molecules heat capacities for ideal gases

3 Overview the Boltzmann distribution (distribution of energies) the Maxwell-Boltzmann distribution (distribution of speeds) the Second Law of thermodynamics entropy

4 Adiabatic Process in Ideal Gases For an adiabatic process (Q = 0): PV γ = const. and: TV γ 1 = const.

5 Adiabatic Process in Ideal Gases For an adiabatic process (Q = 0): PV γ = const. and: TV γ 1 = const. (Given the first one is true, the second follows immediately from the ideal gas equation, P = nrt V.)

6 Example Based on problem 28, Chapter 21. How much work is required to compress 5.00 mol of air at 20.0 C and 1.00 atm to one-tenth of the original volume in an adiabatic process? Assume air behaves as an ideal diatomic-type gas. 1 Serway & Jewett, page 647.

7 Example Based on problem 28, Chapter 21. How much work is required to compress 5.00 mol of air at 20.0 C and 1.00 atm to one-tenth of the original volume in an adiabatic process? Assume air behaves as an ideal diatomic-type gas. One way: and PV γ = P i V γ i W = P dv Another way: and T i V γ 1 i = T f V γ 1 f 0 W = E int Q = nc V T 1 Serway & Jewett, page 647.

8 Example Based on problem 28, Chapter 21. How much work is required to compress 5.00 mol of air at 20.0 C and 1.00 atm to one-tenth of the original volume in an adiabatic process? Assume air behaves as an ideal diatomic-type gas. One way: and PV γ = P i V γ i W = P dv Another way: and W = 46.0 kj T i V γ 1 i = T f V γ 1 f 0 W = E int Q = nc V T 1 Serway & Jewett, page 647.

9 Weather and Adiabatic Process in a Gas On the eastern side of the Rocky Mountains there is a phenomenon called chinooks. These eastward moving wind patterns cause distinctive cloud patterns (chinook arches) and sudden increases in temperature.

10 Weather and Adiabatic Process in a Gas As the air rises from the ocean it expands in the lower pressure at altitude and cools. The water vapor condenses out of the air and falls as precipitation. As the air passes over the mountain it absorbs the latent heat from the water condensation, then it stops cooling. As it descends, it is compressed (nearly) adiabatically as the ambient pressure increases. The air temperature rises!

11 Temperature and the Distribution of Particles Energies In a gas at temperature T, we know the average translational KE of the molecules. However, not all of the molecules have the same energy, that s just the average. How is the total energy of the gas distributed amongst the molecules?

12 Temperature and the Distribution of Particles Energies Ludwig Boltzmann first found the distribution of the number of particles at a given energy given a thermodynamic system at a fixed temperature. Assuming that energy takes continuous values we can say that the number of molecules per unit volume with energies in the range E to E + de is: Where N [E,E+dE] = E+dE E n V (E) de n V (E) = n 0 e E/k BT and n 0 is a constant setting the scale: when E = 0, n V (E) = n 0.

13 The Boltzmann Distribution This particular frequency distribution: n V (E) e E/k BT is called the Boltzmann distribution or sometimes the Gibbs distribution (after Josiah Willard Gibbs, who studied the behavior of this distribution in-depth). This distribution is even easier to understand for discrete energy levels.

14 The Boltzmann Distribution This particular frequency distribution: n V (E) e E/k BT is called the Boltzmann distribution or sometimes the Gibbs distribution (after Josiah Willard Gibbs, who studied the behavior of this distribution in-depth). This distribution is even easier to understand for discrete energy levels. The probability for a given particle to be found in a state with energy E i drawn from a sample at temperature T : p(e i ) = 1 Z e E i /k B T where Z is simply a normalization constant to allow the total probability to be 1. (The partition function.)

15 The Boltzmann Distribution p(e i ) = 1 Z e E i /k B T If we know the energies of two states E 1 and E 2, E 2 > E 1, we can find the ratio of the number of particles in each: n V (E 2 ) n V (E 1 ) = e (E 2 E 1 )/k B T States with lower energies have more particles occupying them.

16 The Boltzmann Distribution Lower temperature Higher temperature 1 Figure from the website of Dr. Joseph N. Grima, University of Malta.

17 Aside: Lasers Lasers emit coherent light. One photon interacts with an atom and causes another to be emitted in the same state. This starts a cascade. Inside a laser cavity there are atoms that are in a very strange state: a higher energy level is more populated than a lower one. This is called a population inversion.

18 Aside: Lasers This is necessary for the photon cascade. Since: n V (E 2 ) n V (E 1 ) = e (E 2 E 1 )/k B T, E 2 > E 1 we can associate a negative temperature, T, to these two energy states in the atoms.

19 Maxwell-Boltzmann speed distribution The Boltzmann distribution for energy can be leveraged to find a distribution of the speeds of the molecules. This is the Maxwell-Boltzmann speed distribution. The number of molecules with speeds between v and v + dv is v+dv v N v dv = v+dv v 4πN ( ) 3/2 m0 v 2 e m 0v 2 /2k B T dv 2πk B T

20 Maxwell-Boltzmann speed distribution The number of molecules with speeds between v and v + dv is v+dv v+dv ( ) 3/ Distribution of Molecular Speeds m0 641 N v dv = 4πN v 2 e m 0v 2 /2k B T dv 2πk B T of speeds of N gas v (21.41) N v v The number of molecules having speeds ranging from v to v dv equals the area of the tan rectangle, N v dv. tant, and T is the ann factor e 2E/k BT v mp v avg v rms at lower than the distribution curve N v (21.42) dv v

21 Maxwell-Boltzmann speed distribution The energy of a molecule can be written: where E = K trans + ɛ + U translational kinetic energy, K trans = p2 2m 0 ɛ includes any rotational or oscillational energy U is potential energy (if relevant) that depends on the location of the molecule Since we only want to know about the distribution of speeds, we will need to get rid of any dependence on ɛ and U.

22 Aside: Reminder about probability distributions Suppose I have a probability distribution over two variables, x and y: p(x, y) If the two variables are independently distributed then: p(x, y) = p(x)p(y)

23 Aside: Reminder about probability distributions Suppose I have a probability distribution over two variables, x and y: p(x, y) If the two variables are independently distributed then: p(x, y) = p(x)p(y) We can eliminate the dependence on x by just summing over x: p(x, y) = x x 1 p(x)p(y) = p(y) p(x) = p(y) x This is how we eliminate rotational and vibrational motion.

24 Maxwell-Boltzmann speed distribution Put this expression in the Boltzmann distribution: p(r, p, ɛ) d 3 r d 3 p dɛ = Ae E/k BT d 3 r d 3 p dɛ Eliminate dependence on position, rotation, and oscillation: p(r, p, ɛ) d 3 r d 3 p dɛ ɛ p ) ) =Ce p2 /2m 0 k B T dp (C e ɛ/kbt dɛ (C e U/kBT d 3 r ɛ r

25 Maxwell-Boltzmann speed distribution Put this expression in the Boltzmann distribution: p(r, p, ɛ) d 3 r d 3 p dɛ = Ae E/k BT d 3 r d 3 p dɛ Eliminate dependence on position, rotation, and oscillation: p(r, p, ɛ) d 3 r d 3 p dɛ ɛ p =Ce p2 /2m 0 k B T dp 1 (C ) e ɛ/kbt dɛ 1 ) (C e U/kBT d 3 r ɛ r

26 Maxwell-Boltzmann speed distribution Put this expression in the Boltzmann distribution: p(r, p, ɛ) d 3 r d 3 p dɛ = Ae E/k BT d 3 r d 3 p dɛ Eliminate dependence on position, rotation, and oscillation: p(r, p, ɛ) d 3 r d 3 p dɛ ɛ p =Ce p2 /2m 0 k B T dp 1 (C e ɛ/kbt dɛ ) ) 1 (C e U/kBT d 3 r ɛ r p(p) d 3 p = Ce p2 /2m 0 k B T d 3 p

27 Maxwell-Boltzmann speed distribution Put this expression in the Boltzmann distribution: p(r, p, ɛ) d 3 r d 3 p dɛ = Ae E/k BT d 3 r d 3 p dɛ Eliminate dependence on position, rotation, and oscillation: p(r, p, ɛ) d 3 r d 3 p dɛ ɛ p =Ce p2 /2m 0 k B T dp 1 (C e ɛ/kbt dɛ ) ) 1 (C e U/kBT d 3 r ɛ r p(p) d 3 p = Ce p2 /2m 0 k B T d 3 p replace momentum with velocity components: p(v) d 3 v = Ce m 0(v 2 x +v 2 y +v 2 z )/2k B T dv x dv y dv z

28 Maxwell-Boltzmann speed distribution We can find C. The total probability must equal one. Ce m 0(v 2 x +v 2 y +v 2 z )/2k B T dv x dv y dv z = 1 Using the identity: e x2 dx = π the three integrals can be evaluated separately: e m 0vx 2 /2k B T 2πkB T dv x = There are three integrals, so C = ( m0 ) 3/2 2πk B T m 0

29 Maxwell-Boltzmann speed distribution Now our distribution is: p(v) d 3 v = ( ) 3/2 m0 e m 0(vx 2 +vy 2 +vz 2 )/2k B T dv x dv y dv z 2πk B T Lastly, we want an expression for how many molecules have speeds between v and v + dv. This means we need to get rid of the direction dependence transform to spherical coordinates. dv x dv y dv z = v 2 sin θ dv dθ dφ

30 Maxwell-Boltzmann speed distribution φ θ = p(v) d 3 v ( ) 3/2 2π m0 v 2 e m 0v 2 /2k B T dv dφ 2πk B T 0 ( ) 3/2 m0 = 4π v 2 e m 0v 2 /2k B T dv 2πk B T π 0 sin θdθ This is the probability density for 1 molecule. For N molecules: N v dv = 4πN ( ) 3/2 m0 v 2 e m 0v 2 /2k B T dv 2πk B T

31 both on mass and on temperature. At a given temperature, the fraction of molecules with speeds exceeding a fixed value increases as the mass decreases. Hence, The Shape of the Maxwell-Boltzmann distribution The total area under either curve is equal to N, the total number of molecules. In this case, N Note that v rms v avg v mp. 200 T 300 K N v [molecules/(m/s)] v mp v avg v rms T 900 K 0 0 From previous lecture: v (m/s) 3 For the derivation of this expression, see an advanced textbook on thermodynamics. v rms = 3kB T kb T = 1.73 m 0 m 0 Fig but mo

32 The Shape of the Maxwell-Boltzmann distribution Average speed, can find by integrating over all speeds, then dividing by the number of particles. v avg = 1 v N v dv N 0 ( ) 3/2 m0 = 4π v 3 e m 0v 2 /2k B T dv 0 2πk B T ( ) 3/2 m0 = 4π v 3 e m 0v 2 /2k B T dv 2πk B T 0 ( ) 3/2 ( ) m0 1 2 m0 = 4π 2πk B T 2 2k B T v avg = 8 π kb T m 0 = 1.60 kb T m 0

33 The Shape of the Maxwell-Boltzmann distribution To find the most probable speed (peak of the distribution), can find the value of v for which the derivative of the particle number distribution is zero.

34 The Shape of the Maxwell-Boltzmann distribution To find the most probable speed (peak of the distribution), can find the value of v for which the derivative of the particle number distribution is zero. Set dn v dv = 0 Then we find: v mp = 2kB T m 0 = 1.41 kb T m 0

35 The Shape of the Maxwell-Boltzmann distribution To find the most probable speed (peak of the distribution), can find the value of v for which the derivative of the particle number distribution is zero. Set dn v dv = 0 Then we find: v mp = 2kB T kb T = 1.41 m 0 m 0 So, v rms > v avg > v mp

36 21.5 Distribution of Molecular Speeds 641 The Shape of the Maxwell-Boltzmann distribution of speeds of N gas (21.41) N v The number of molecules having speeds ranging from v to v dv equals the area of the tan rectangle, N v dv. tant, and T is the ann factor e 2E/k BT v mp v avg v rms at lower than the distribution curve N v (21.42) dv v (21.43) Figure v rms > v avg The > speed v mp distribution of gas molecules at some 0 Graph from Serway temperature. & Jewett, page The 641. function N v

37 Speed Distribution and Evaporation We can understand evaporation as a change of some of our system from the liquid to the gaseous state at the surface of the liquid. Even well below the boiling point there are some molecules with very high translational KE. These molecules move fast enough to overcome the strength of the liquid bonds. Slower moving molecules are left behind, so the remaining liquid is cooler.

38 The Second Law of Thermodynamics We will state this law in several different ways. First an intuitive statement: 2nd Law Unless work is done on a system, heat in the system will flow from a hotter body in the system to a cooler one. This is obvious from experience, but it s not obvious why this should happen. It also indicates there is are processes in the physical world that seem not to happen in the same way if time is reversed.

39 The Second Law of Thermodynamics and Reversibility Scientists and engineers studying and designing steam engines wanted to make them as efficient as possible. They noticed there were always losses. There seemed to be more to it. Energy seems to always spread out. Heat goes from hotter to colder objects. Energy is lost as heating in friction. These things do not happen in reverse.

40 Isolated, Closed, and Open Systems Isolated system does not exchange energy (work, heat, or radiation) or matter with its environment. Closed system does not exchange matter with its environment, but may exchange energy. Open system can exchange energy and matter with its environment.

41 Reversible and Irreversible Processes Reversible process a process that takes a system from an initial state i to a final state f through a series of equilibrium states, such that we can take the same system back again from f to i along the same path in a PV diagram. Irreversible process any process that is not reversible. In real life, all processes are irreversible, but some are close to being reversible. We use reversible processes as an idealization.

42 20-3 CHANGE IN ENTROPY Irreversible Process Example not obey a conservation law. lways remains constant. For always increases. Because of alled the arrow of time. For rn kernel with the forward e backward direction of time o the exploded popcorn red process would result in an System Stopcock closed Vacuum 537 ange in entropy of a system: ergy the system gains or loses toms or molecules that make roach in the next section and Insulation (a) Initial state i Irreversible process Stopcock open by looking again at a process ree expansion of an ideal gas. state i, confined by a closed d container. If we open the, eventually reaching the final irreversible process;all the of the container. ws the pressure and volume (b) Final state f

43 is nd Irreversible Process Example of re, e a the old a py ay the cts on This process has well-defined initial and final equilibrium states, but during the expansion of the gas is not in equilibrium. Pressure i Volume Fig A p-v diagram showing the It cannot be plotted on a PV diagram. Also, no work is done on the gas initial thisstate process. i and the final state f of the free f

44 Figure 20.7 Gas in a cylinder. (a) The gas is in contact with an energy reservoir. The walls of the base in contact with the reservoir is conducting. (b) The gas expands slowly to a larger volume. (c) T Reversible Counterpart Allow gas to expand very slowly through equilibrium states at 20.5 The First Law constant temperature. The gas is initially at temperature T i. The hand reduces its downward force, allowing the piston to move up slowly. The energy reservoir keeps the gas at temperature T i. Th in te T i co by m wi ab Energy reservoir at T i a Energy reservoir at T i b c

45 Reversible Counterpart 606 Chapter 20 The First Law of Thermo We can plot this isothermal expansion: P i P f P i Isotherm PV = constant The curve is a hyperbola. f Isothermal Ex Suppose an ideal This process is d hyperbola (see A stant indicates th Let s calculate The work done on process is quasi-st V i V f V Figure 20.9 The PV diagram Negative work is done Because T is con for an on isothermal gas, expansion heat is transferred of in, and the internal energy and ideal thegas temperature from an initial remain state constant. to a n and R: final state.

46 Comparing the Processes In both of these processes the gas expands into a region it was not in previously. The energy of the system spreads out. This corresponds to a change of state, but it is not captured by the internal energy of the gas system, which does not change in either process.

47 Comparing the Processes In both of these processes the gas expands into a region it was not in previously. The energy of the system spreads out. This corresponds to a change of state, but it is not captured by the internal energy of the gas system, which does not change in either process. Something does change in these processes and we call it entropy.

48 State Variables State variables of a thermodynamics system are variables that are determined if the system is in thermodynamic equilibrium and you know the system s state. Examples: pressure, volume, internal energy, temperature. Also, entropy.

49 State Variables State variables of a thermodynamics system are variables that are determined if the system is in thermodynamic equilibrium and you know the system s state. Examples: pressure, volume, internal energy, temperature. Also, entropy. Each variable on it s own is not enough to determine the state of the system. (Many systems in different states might have the same volume.) Once the current state is known, we do know all of these variables values. How the system arrived at its current state, does not affect these values. Heat and work are not state variables.

50 Entropy Sadi Carnot discovered that the most efficient possible engine must be reversible (more on this to come). Rudolph Clausius interpreted this as being due to the behavior of a new quantity (entropy). The change in entropy moving between two states i and f is: S = f i dq r T where dq r is an infinitesimal heat transfer when the system follows a reversible path. (T can be a function of Q!)

51 Entropy When a reversible path is followed: S r = f i dq T Can we find the entropy change for an irreversible process?

52 Entropy When a reversible path is followed: S r = f i dq T Can we find the entropy change for an irreversible process? Yes! Since entropy is a state variable, we can consider the entropy change in any reversible process with the same initial and final states. Then: S irr = f i dq r T The entropy change in that process will give us the entropy difference between those two states, regardless of the process.

53 Entropy Change Consider an initial state i and final state f. We can find the entropy change moving between those two states (for any process, reversible or irreversible), by finding the entropy change along an arbitrary reversible path. First law, de int = dw + dq r Rearranging, we can find an expression for dq r, using E int = nc V T : dq r = nc V dt +P dv To find entropy, we multiply by 1/T and integrate: S = f i dq f r T = i nc f V T dt + i P T dv

54 Entropy Change S = f i Finally, replace P T = nr V dq f r T = i (ideal gas): nc f V T dt + i P T dv S = f The entropy difference is: i dq f r T = i nc V T f dt + nr i ( ) ( ) Tf Vf S = nc v ln + nr ln T i V i 1 V dv

55 Question Quick Quiz True or False: The entropy change in any adiabatic process must be zero because Q = 0. (A) True (B) False 1 Serway & Jewett, page 673.

56 Example What is the entropy change during an adiabatic free expansion of an isolated gas of n moles going from volume V i to volume V f? (Note: such a process is not reversible.)

57 Example Exercise for you: What is the entropy change the same n moles of gas (a diatomic gas around room temperatures) in an constant volume process, with temperature going T i to T f? What is the entropy change when the pressure is constant and the volume goes V i to V f?

58 Question Quick Quiz An ideal gas is taken from an initial temperature T i to a higher final temperature T f along two different reversible paths. Path A is at constant pressure, and path B is at constant volume. What is the relation between the entropy changes of the gas for these paths? (A) S A > S B (B) S A = S B (C) S A < S B 1 Serway & Jewett, page 673.

59 Clausius Equality Clausius found that the entropy change around any reversible cycle (closed path) is zero. This is called the Clausius Equality: S = dqr T = 0 And moreover that since real engines can never quite conform to this ideal, for all engines the Clausius Inequality holds: dqr S = T 0

60 Entropy in an isolated system This gives us another way to state the second law: 2nd Law In an isolated system, entropy does not decrease. In a non-isolated system (either closed or open) entropy can decrease, but only by increasing the entropy of the environment at least as much.

61 Summary Boltzmann distribution (energies) Maxwell-Boltzmann distribution (speeds) the second law entropy Test Monday, May 15. Homework Serway & Jewett: new: Ch 21, onward from page 644. Probs: 33, 37, 41, 42, 43, 52, 57, 73 new: Ch 22, onward from page 679. OQs: 9, 11; CQs: 9; Probs: 39, 43, 45, 47, 49, 53