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 BT That means if there are no molecules in a sample of gas the number that will have energy E is n E no e E k BT This is is the Boltzmann distribution law There is a similar distribution for molecular speeds 3 2 m 2 mv 2 2k T B N v 4 N v e 2 k BT This is called the Maxwell Boltzmann speed distribution and looks like There are several ways to define a typical speed for a gas molecule One is the RMS speed which we found earlier to be vRMS 3k BT m 1 73 k BT m But we could also take the average of the MaxwellBoltzmann distribution v 8k BT 1 60 k BT m m or the most probable speed the peak of the MaxwellBoltzmann distribution vmp 2k BT m 1 41 k BT m Example Speed of the O2 molecules in the lecture room The room we re in now has a temperature of about 300K An O2 molecule contains 32 protons and neutrons so its mass is m 32mp 32 1 7 10 27 kg 5 4 10 26 kg That means that the various typical speeds are vRMS 3k BT 1 38 10 23 J K 300K m m 1 73 1 73 277 480 26 m s s 5 4 10 kg k BT m m v 1 60 1 60 277 440 m s s vmp k BT m m 1 41 1 41 277 391 m s 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 H2 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 2 v d 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 nV then there will be a collision when VnV 1 d 2 LnV 1 1 L d 2 nV L is the mean free path The number of collisions per second is v f d 2 vnV L Actually these expressions aren t quite right since all the molecules are moving The real answers are 1 L 2 d 2 nV f 2 d 2 vnV Example How long does it take an O2 molecule to cross the lecture room The size of an O2 molecule is about 10 10m Since the pressure in the room is about 1atm we have N P 1 013 105 Pa 25 3 nV 2 5 10 m V k BT 1 38 10 23 J K 300K So the mean free path is 1 1 L 2 2 10 2 d nV 2 10 m 2 5 1025 m 3 9 0 10 7 m The lecture room is about 30m across or 3 3 x 107 mean free paths But every time the O2 molecule collides it s direction changes in a random way That means on average it would take 3 3 x 107 2 1 1 x 1015 collisions to cross the room So the total distance the O2 molecule travels as it wanders from one side of the room to the other is D 1 1 1015 9 10 7 m 9 108 m Since the average velocity of a molecule is about 440m s it would take about 1 8 x 106s 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 Th energy Qh 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 Weng 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 Qc meaning this transfer takes place at low temperature The 1st Law of Thermodynamics tells us that Eint Q Weng 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 Qh and emits energy Qc So Q Qh Qc The 1st Law then says E int Qh Qc Weng 0 Schematically Th Weng Qh Qc Qh Engine Weng …