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Understand pressure, specific heats, heat transfer, entropy, and thermodynamics laws. Explore mean free path, gas distribution, and degrees of freedom in ideal gases and solids. Learn about energy distribution and equipartition theorem.
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Lecture 26 Goals: • Chapter 18 • Understand the molecular basis for pressure and the ideal-gas law. • Predict the molar specific heats of gases and solids. • Understand how heat is transferred via molecular collisions and how thermally interacting systems reach equilibrium. • Obtain a qualitative understanding of entropy, the 2nd law of thermodynamics • Assignment • HW11, Due Tuesday, May 5th • For Tuesday, Read through all of Chapter 19
Macro-micro connectionMean Free Path If a “real” molecule has Ncollcollisions as it travels distance L, the average distance between collisions, which is called the mean free pathλ is The mean free path is independent of temperature The mean time between collisions is temperature dependent
And the mean free path is… • Some typical numbers
O2 at 25°C 1.4 1.2 1.0 O2 at 1000°C # Molecules 0.8 0.6 0.4 0 200 600 1000 1400 1800 Molecular Speed (m/s) Distribution of Molecular SpeedsA “Maxwell-Boltzmann” Distribution
Macro-micro connection • Assumptions for ideal gas: • # of molecules N is large • They obey Newton’s laws • Short-range interactions with elastic collisions • Elastic collisions with walls (an impulse…..pressure) • What we call temperature T is a direct measure of the average translational kinetic energy • What we call pressure p is a direct measure of the number density of molecules, and how fast they are moving (vrms)
(A) x1 (B) x1.4 (C) x2 Exercise • Consider a fixed volume of ideal gas. When N or T is doubled the pressure increases by a factor of 2. 1. If T is doubled, what happens to the rate at which a single molecule in the gas has a wall bounce? (A) x1.4 (B) x2 (C) x4 2. If N is doubled, what happens to the rate at which a single molecule in the gas has a wall bounce?
A macroscopic “example” of the equipartition theorem • Imagine a cylinder with a piston held in place by a spring. Inside the piston is an ideal gas a 0 K. • What is the pressure? What is the volume? • Let Uspring=0 (at equilibrium distance) • What will happen if I have thermal energy transfer? • The gas will expand (pV = nRT) • The gas will do work on the spring • Conservation of energy • Q = ½ k x2 + 3/2 n R T (spring & gas) • and Newton S Fpiston= 0 = pA – kx kx =pA • Q = ½ p (Ax) + 3/2 n RT • Q = ½ p V + 3/2 n RT (but pV = nRT) • Q = ½ nRT + 3/2 RT (25% of Q went to the spring) +Q ½ nRT per “degree of freedom”
Degrees of freedom or “modes” • Degrees of freedom or “modes of energy storage in the system” can be: Translational for a monoatomic gas (translation along x, y, z axes, energy stored is only kinetic) NO potential energy • Rotational for a diatomic gas (rotation about x, y, z axes, energy stored is only kinetic) • Vibrational for a diatomic gas (two atoms joined by a spring-like molecular bond vibrate back and forth, both potential and kinetic energy are stored in this vibration) • In a solid, each atom has microscopic translational kinetic energy and microscopic potential energy along all three axes.
Degrees of freedom or “modes” • A monoatomic gas only has 3 degrees of freedom (just K, kinetic) • A typical diatomic gas has 5 accessible degrees of freedom at room temperature, 3 translational (K) and 2 rotational (K) At high temperatures there are two more, vibrational with K and U • A monomolecular solid has 6 degrees of freedom 3 translational (K), 3 vibrational (U)
The Equipartition Theorem • The equipartition theorem tells us how collisions distribute the energy in the system. Energy is stored equally in each degree of freedom of the system. • The thermal energy of each degree of freedom is: Eth = ½ NkBT = ½ nRT • A monoatomic gas has 3 degrees of freedom • A diatomic gas has 5 degrees of freedom • A solid has 6 degrees of freedom • Molar specific heats can be predicted from the thermal energy, because
Exercise • A gas at temperature T is mixture of hydrogen and helium gas. Which atoms have more KE (on average)? (A) H (B) He (C) Both have same KE • How many degrees of freedom in a 1D simple harmonic oscillator? (A) 1 (B) 2 (C) 3 (D) 4 (E) Some other number
The need for something else: Entropy V1 You have an ideal gas in a box of volume V1. Suddenly you remove the partition and the gas now occupies a larger volume V2. • How much work was done by the system? (2) What is the final temperature (T2)? (3) Can the partition be reinstalled with all of the gas molecules back in V1 P P V2
ExercisesFree Expansion and Entropy V1 You have an ideal gas in a box of volume V1. Suddenly you remove the partition and the gas now occupies a larger volume V2. • How much work was done by the system? P P V2 (A) W > 0 (B) W =0 (C) W < 0
ExercisesFree Expansion and Entropy V1 You have an ideal gas in a box of volume V1. Suddenly you remove the partition and the gas now occupies a larger volume V2. (2) What is the final temperature (T2)? P P V2 (A) T2 > T1 (B) T2 = T1 (C) T2 < T1
Free Expansion and Entropy V1 You have an ideal gas in a box of volume V1. Suddenly you remove the partition and the gas now occupies a larger volume V2. (3) Can the partition be reinstalled with all of the gas molecules back in V1 (4) What is the minimum process necessary to put it back? P P V2
V1 P P V2 Free Expansion and Entropy You have an ideal gas in a box of volume V1. Suddenly you remove the partition and the gas now occupies a larger volume V2. (4) What is the minimum energy process necessary to put it back? Example processes: A. Adiabatic Compression followed by Thermal Energy Transfer B. Cooling to 0 K, Compression, Heating back to original T
ExercisesFree Expansion and the 2nd Law What is the minimum energy process necessary to put it back? Try: B. Cooling to 0 K, Compression, Heating back to original T Q1 = n CvDT out and put it where…??? Need to store it in a low T reservoir and 0 K doesn’t exist Need to extract it later…from where??? Key point: Where Q goes & where it comes from are important as well. V1 P P V2
Modeling entropy • I have a two boxes. One with fifty pennies. The other has none. I flip each penny and, if the coin toss yields heads it stays put. If the toss is “tails” the penny moves to the next box. • On average how many pennies will move to the empty box?
Modeling entropy • I have a two boxes, with 25 pennies in each. I flip each penny and, if the coin toss yields heads it stays put. If the toss is “tails” the penny moves to the next box. • On average how many pennies will move to the other box? • What are the chances that all of the pennies will wind up in one box?
2nd Law of Thermodynamics • Second law: “The entropy of an isolated system never decreases. It can only increase, or, in equilibrium, remain constant.” • The 2nd Law tells us how collisions move a system toward equilibrium. • Order turns into disorder and randomness. • With time thermal energy will always transfer from the hotter to the colder system, never from colder to hotter. • The laws of probability dictate that a system will evolve towards the most probable and most random macroscopic state Entropy measures the probability that a macroscopic state will occur or, equivalently, it measures the amount of disorder in a system Increasing Entropy
Entropy • Two identical boxes each contain 1,000,000 molecules. In box A, 750,000 molecules happen to be in the left half of the box while 250,000 are in the right half. In box B, 499,900 molecules happen to be in the left half of the box while 500,100 are in the right half. • At this instant of time: • The entropy of box A is larger than the entropy of box B. • The entropy of box A is equal to the entropy of box B. • The entropy of box A is smaller than the entropy of box B.
Entropy • Two identical boxes each contain 1,000,000 molecules. In box A, 750,000 molecules happen to be in the left half of the box while 250,000 are in the right half. In box B, 499,900 molecules happen to be in the left half of the box while 500,100 are in the right half. • At this instant of time: • The entropy of box A is larger than the entropy of box B. • The entropy of box A is equal to the entropy of box B. • The entropy of box A is smaller than the entropy of box B.
Reversible vs Irreversible • The following conditions should be met to make a process perfectly reversible: 1. Any mechanical interactions taking place in the process should be frictionless. 2. Any thermal interactions taking place in the process should occur across infinitesimal temperature or pressure gradients (i.e. the system should always be close to equilibrium.) • Based on the above answers, which of the following processes are not reversible? 1. Melting of ice in an insulated (adiabatic) ice-water mixture at 0°C. 2. Lowering a frictionless piston in a cylinder by placing a bag of sand on top of the piston. 3. Lifting the piston described in the previous statement by removing one grain of sand at a time. 4. Freezing water originally at 5°C.
Reversible vs Irreversible • The following conditions should be met to make a process perfectly reversible: 1. Any mechanical interactions taking place in the process should be frictionless. 2. Any thermal interactions taking place in the process should occur across infinitesimaltemperature or pressure gradients (i.e. the system should always be close to equilibrium.) • Based on the above answers, which of the following processes are not reversible? 1. Melting of ice in an insulated (adiabatic) ice-water mixture at 0°C. 2. Lowering a frictionless piston in a cylinder by placing a bag of sand on top of the piston. 3. Lifting the piston described in the previous statement by removing one grain of sand at a time. 4. Freezing water originally at 5°C.
Lecture 26 • Assignment rehash • HW11, Due Tuesday May 5th • For Tuesday, read through all of Chapter 19!