Lecture 9: Kinetic Theory of Gases, Part 4, and Heat Engines We now know that the temperature of a gas is proportional to the average energy of each molecule But we also know that all the molecules don t have the same energy some are moving faster and some slower at any moment For any given molecule, we can find the probability for that molecule to have a certain energy E: p( E) = e! E/k B T That means if there are n o molecules in a sample of gas, the number that will have energy E is: n E ( ) = n o e! E/k B T This is is the Boltzmann distribution law
There is a similar distribution for molecular speeds: N ( v) = 4! N " $ # This is called the Maxwell-Boltzmann speed distribution, and looks like: m 2!k B T % ' & 3 2 v 2 e ( mv2 / 2k B T
There are several ways to define a typical speed for a gas molecule One is the RMS speed, which we found earlier to be: v RMS = 3k B T m = 1.73 k B T m But we could also take the average of the Maxwell- Boltzmann distribution: v = 8k T B!m = 1.60 k T B m or the most probable speed (the peak of the Maxwell- Boltzmann distribution): v mp = 2k B T m = 1.41 k B T m
Example: Speed of the O 2 molecules in the lecture room The room we re in now has a temperature of about 300K An O 2 molecule contains 32 protons and neutrons, so its mass is: m = 32m p = 32!1.7 "10 #27 kg = 5.4 "10 #26 kg That means that the various typical speeds are: v RMS = 3k B T m = 1.73 1.38!10"23 J/K #300K 5.4!10 "26 kg = 1.73#277 m s = 480 m s v = 1.60 k B T m = 1.60! 277 m s = 440 m s v mp = 1.41 k B T m = 1.41!277 m s = 391m s
What Does This Tell Us About the Atmosphere? Escape velocity for any object to leave the Earth is ~11000m/s Let s compare this velocity to the Maxwell-Boltzmann Distribution for various gases at 273K: So some H 2 molecules will escape that means eventually, they all will!
Mean Free Path You know from experience that gas molecules don t move across a room nearly as quickly at their average speed would suggest when someone opens a bottle of perfume on one side of a room, you don t instantly smell it on the other side That s because the molecules are constantly bouncing in to one another So, how far does a molecule typically get before hitting another one, anyway? this is called the mean free path for a molecule To estimate, we pretend that a gas molecule is a hard sphere with diameter d
As the molecule moves, it passes through a cylinder: d d/2 v If there s another molecule with its center in the bigger cylinder, there will be a collision L If the number of molecules per unit volume is n V, then there will be a collision when: Vn V = 1!d 2 Ln V = 1 L = 1!d 2 n V L is the mean free path
The number of collisions per second is: Actually, these expressions aren t quite right since all the molecules are moving The real answers are: f = v L =!d2 vn V L = 1 2!d 2 n V f = 2!d 2 vn V
Example: How long does it take an O 2 molecule to cross the lecture room? The size of an O 2 molecule is about 10-10 m Since the pressure in the room is about 1atm, we have: N V = n = P V k B T = 1.013!10 5 Pa 1.38!10 "23 J/K #300K = 2.5!1025 m -3 So the mean free path is: L = 1 = 1 2!d 2 n V = 9.0 #10 "7 m 2! ( 10 "10 m) 2 2.5#10 25 m -3 The lecture room is about 30m across, or 3.3 x 10 7 mean free paths But every time the O 2 molecule collides, it s direction changes in a random way
That means, on average, it would take (3.3 x 10 7 ) 2 = 1.1 x 10 15 collisions to cross the room So the total distance the O 2 molecule travels as it wanders from one side of the room to the other is: D = 1.1!10 15 "9!10 #7 m = 9!10 8 m Since the average velocity of a molecule is about 440m/s, it would take about 1.8 x 10 6 s for the molecule to cross the room! that s over 20 days! Of course, we know that gas molecules really move around much more quickly we don t have to wait a month to smell perfume But that s due to convection -- motion of a mass of air due to little wind currents in the room
Heat Engines and the Second Law of Thermodynamics So far in both 141 and 142 you ve learned about the laws of physics, which tell us what kind of processes can happen But there are many processes that obey Newton s Laws and the first law of thermodynamics that actually don t happen for example, a car that plows into a wall ends up resting against the wall with a crumpled front end but that car never spontaneously uncrumples and leaves the wall in reverse at the same speed that it came in This is obviously true, and yet neither process would violate any of the laws of physics we ve learned so far! We need another law to account for these unobserved processes
Heat Engine To formulate this new law of physics, we start by considering heat engines any device that converts heat energy to mechanical work in a cyclic process is a heat engine there are some familiar examples: the gasoline engine in a car steam engine in a railroad locomotive or ship Let s look at the steam locomotive: first, a fuel (wood, coal, or oil) is burned, forming a hightemperature area (T h ) energy Q h is then transferred by heat to water in the boiler water is the working substance for this engine the water boils, producing steam, which then does mechanical work W eng by pushing against a piston
as the steam expands, it cools and condenses to liquid, which is then fed back into the boiler this completes the cycle for the working substance note that the water leaving the piston still has some internal energy (its temperature is not absolute zero!), which is carries away when it returns to the boiler This is an example of energy transfer by heat (convection), and the amount of energy is Q c (meaning this transfer takes place at low temperature) The 1st Law of Thermodynamics tells us that:!e int = Q "W eng = 0 Minus sign because the water does work on the engine Zero because the process is cyclic
What about Q? the engine takes in energy Q h, and emits energy Q c So, Q = Q h! Q c The 1st Law then says: Schematically: = Q!E int h " Q c "W eng = 0 W eng = Q h " Q c T h Q h Engine W eng Q c T c
Efficiency of the Heat Engine We define efficiency as the ratio of the energy supplied to the engine to the work done by it: e = W eng Q h = Q h! Q c Q h = 1! Q c Q h For the steam locomotive, the efficiency is not 100% because Q c is not zero It turns out that that s true for any type of heat engine a heat engine that gets 50% efficiency is really good! This experimental fact forms the 2nd Law of Thermodynamics: No cyclic heat engine can convert heat to work at 100% efficiency
Heat Pumps and Refrigerators Doing mechanical work is not the only way we d like to manipulate heat imagine it s 95 o F outside, and we have a can of soda that s 85 o F we d prefer the soda to be even colder before we drink it in other words, we d like to transfer energy by heat from the soda to the warmer atmosphere But that s an example of a process that never happens! if we wait, we d observe energy transfer in the other direction, and the soda would warm up to 95 o F This is the 2nd Law in action again. In fact, we can write the 2nd Law as: Heat will never spontaneously flow from a cold object to a warmer one
Even though the two ways of writing the 2nd Law sound very different, they are in fact equivalent as long as one is true, the other must be as well For example, imagine we had an imperfect heat engine but a perfect heat pump: Realistic heat engine: Perfect heat pump: T h T h Q h Q h Engine W eng Pump Q c Q c T c T c
We could then combine the two devices to get: T h T h Q h Q h Q h = W eng W eng Q c Q c T c T c A perfect heat engine! So if we were allowed to break the 2nd form of the 2nd Law, the first form would be broken as well
Refrigerators Of course, it is possible to cool down our can of soda by placing it in a refrigerator Due to the 2nd Law of thermodynamics, though, we must supply energy (in the form of work) to make the heat flow in the desired direction: T h Q h Real refrigerator: W Q c T c
A good refrigerator will remove a lot of heat for a little amount of external work Q The coefficient of performance is defined as h a higher number means a better refrigerator W typically can achieve values of 5-6