Speed Distribution at CONSTANT Temperature is given by the Maxwell Boltzmann Speed Distribution

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1 Temperature ~ Average KE of each particle Particles have different speeds Gas Particles are in constant RANDOM motion Average KE of each particle is: 3/2 kt Pressure is due to momentum transfer Speed Distribution at CONSTANT Temperature is given by the Maxwell Boltzmann Speed Distribution

volume (N/V) and to the average translational kinetic energy of the molecules.

2 Pressure and Kinetic Energy Assume a container is a cube with edges d. Look at the motion of the molecule in terms of its velocity components and momentum and the average force Pressure is proportional to the number of molecules per unit volume (N/V) and to the average translational kinetic energy of the molecules. This equation also relates the macroscopic quantity of pressure with a microscopic quantity of the average value of the square of the molecular speed One way to increase the pressure is to increase the number of molecules per unit volume The pressure can also be increased by increasing the speed (kinetic energy) of the molecules P 2 N 1 2 m v 3 V 2 o

3 Molecular Interpretation of Temperature We can take the pressure as it relates to the kinetic energy and compare it to the pressure from the equation of state for an ideal gas 2 N 1 2 P mv nrt NkBT 3 V 2 Temperature is a direct measure of the average molecular kinetic energy

4 Molecular Interpretation of Temperature Simplifying the equation relating temperature and kinetic energy gives mo v kbt 2 2 This can be applied to each direction, mv x kbt 2 2 with similar expressions for v y and v z

5 Total Kinetic Energy The total kinetic energy is just N times the kinetic energy of each molecule Ktot trans N mv NkBT nrt If we have a gas with only translational energy, this is the internal energy of the gas This tells us that the internal energy of an ideal gas depends only on the temperature

6 Kinetic Theory Problem A 5.00-L vessel contains nitrogen gas at 27.0 C and 3.00 atm. Find (a) the total translational kinetic energy of the gas molecules and (b) the average kinetic energy per molecule.

If you apply the same flame for the same time to 2 liters of water, by

7 Hot Question Suppose you apply a flame to 1 liter of water for a certain time and its temperature rises by 10 degrees C. If you apply the same flame for the same time to 2 liters of water, by how much will its temperature rise? a) 1 degree b) 5 degrees c) 10 degrees d) zero degrees

8 Ludwig Boltzmann or Dean Gooch? Austrian physicist Contributed to Kinetic Theory of Gases Electromagnetism Thermodynamics Pioneer in statistical mechanics

9 Distribution of Molecular Speeds The observed speed distribution of gas molecules in thermal equilibrium is shown at right N V is called the Maxwell-Boltzmann speed distribution function m o is the mass of a gas molecule, k B is Boltzmann s constant and T is the absolute temperature m 3 / 2 o 2 mv NV 4 N v e 2 kt B 2 /2k T B

10 Molecular Speeds and Collisions

11 Speed Summary Root mean square speed v rms 3kBT m o 1.73 kbt m o The average speed is somewhat lower than the rms speed v avg 8kBT m The most probable speed, v mp is the speed at which the distribution curve reaches a peak o 1.60 kbt m o v rms > v avg > v mp v mp 2kBT m 1.41 kbt m

12 Some Example v rms Values At a given temperature, lighter molecules move faster, on the average, than heavier molecules

13 The peak shifts to the right as T increases This shows that the average speed increases with increasing temperature The asymmetric shape occurs because the lowest possible speed is 0 and the highest is infinity Speed Distribution

14 The Kelvin Temperature of an ideal gas is a measure of the average translational kinetic energy per particle: 3/ 2kT KE 1/ 2mv 2 rms k =1.38 x J/K Boltzmann s Constant Root-mean-square speed: v rms v 2 3 kt m

15 Kinetic Theory Problem Calculate the RMS speed of an oxygen molecule in the air if the temperature is 5.00 C. The mass of an oxygen molecule is u (k = 1.3 8x J/K, u = 1.66 x kg) v rms 3kT m What is m? m is the mass of one oxygen molecule in kg. What is u? How do we get the mass in kg?

16 Kinetic Theory Problem Calculate the RMS speed of an oxygen molecule in the air if the temperature is 5.00 C. The mass of an oxygen molecule is u (k = 1.3 8x J/K, u = 1.66 x kg) v rms 3kT m What is m? m is the mass of one oxygen molecule. 23 3( / ) (32 u)(1.66 x10 kg / u) x J K K 466 m/ s Is this fast? YES! Speed of sound: 343m/s!

17 A cylinder contains a mixture of helium and argon gas in equilibrium at 150 C. (a) What is the average kinetic energy for each type of gas molecule? (b) What is the root-mean-square speed of each type of molecule?

18 More Kinetic Theory Problems A gas molecule with a molecular mass of 32.0 u has a speed of 325 m/s. What is the temperature of the gas molecule? A) 72.0 K B) 136 K C) 305 K D) 459 K E) A temperature cannot be assigned to a single molecule. Temperature ~ Average KE of all particles

19 Equipartition of Energy Each translational degree of freedom contributes an equal amount to the energy of the gas In general, a degree of freedom refers to an independent means by which a molecule can possess energy Each degree of freedom contributes ½k B T to the energy of a system, where possible degrees of freedom are those associated with translation, rotation and vibration of molecules With complex molecules, other contributions to internal energy must be taken into account One possible energy is the translational motion of the center of mass Rotational motion about the various axes also contributes There is kinetic energy and potential energy associated with the vibrations

20 Monatomic and Diatomic Gases The thermal energy of a monatomic gas of N atoms is A diatomic gas has more thermal energy than a monatomic gas at the same temperature because the molecules have rotational as well as translational kinetic energy.

21 Molar Specific Heat We define specific heats for two processes that frequently occur: Changes with constant pressure Changes with constant volume Using the number of moles, n, we can define molar specific heats for these processes Molar specific heats: Q = nc V DT for constant-volume processes Q = nc P DT for constant-pressure processes

22 Ideal Monatomic Gas Therefore, DE int = 3/2 nrt DE is a function of T only In general, the internal energy of an ideal gas is a function of T only The exact relationship depends on the type of gas At constant volume, Q = DE int = nc V DT This applies to all ideal gases, not just monatomic ones

23 Ratio of Molar Specific Heats We can also define the ratio of molar specific heats g CP 5 R / C V 3 R/ 2 Theoretical values of C V, C P, and g are in excellent agreement for monatomic gases But they are in serious disagreement with the values for more complex molecules Not surprising since the analysis was for monatomic gases

24 Agreement with Experiment Molar specific heat is a function of temperature At low temperatures, a diatomic gas acts like a monatomic gas C V = 3/2 R At about room temperature, the value increases to C V = 5/2 R This is consistent with adding rotational energy but not vibrational energy At high temperatures, the value increases to C V = 7/2 R This includes vibrational energy as well as rotational and translational

25 Sample Values of Molar Specific Heats

26 In a constant-volume process, 209 J of energy is transferred by heat to 1.00 mol of an ideal monatomic gas initially at 300 K. Find (a) the increase in internal energy of the gas, (b) the work done on it, and (c) its final temperature

27 Molar Specific Heats of Other Materials The internal energy of more complex gases must include contributions from the rotational and vibrational motions of the molecules In the cases of solids and liquids heated at constant pressure, very little work is done, since the thermal expansion is small, and C P and C V are approximately equal

28 Adiabatic Processes for an Ideal Gas An adiabatic process is one in which no energy is transferred by heat between a system and its surroundings (think styrofoam cup) Assume an ideal gas is in an equilibrium state and so PV = nrt is valid The pressure and volume of an ideal gas at any time during an adiabatic process are related by PV g = constant g = C P / C V is assumed to be constant All three variables in the ideal gas law (P, V, T ) can change during an adiabatic process

29 Adiabatic Process The PV diagram shows an adiabatic expansion of an ideal gas The temperature of the gas decreases T f < T i in this process For this process P i V i g = P f V f g and T i V i g-1 = T f V f g-1

30 A 2.00-mol sample of a diatomic ideal gas expands slowly and adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L. (a) What is the final pressure of the gas? (b) What are the initial and final temperatures? (c) Find Q, W, and DE int.

31 Cyclic Processes DEint Q W A cyclic process is one that starts and ends in the same state On a PV diagram, a cyclic process appears as a closed curve DE int = 0, Q = -W In a cyclic process, the net work done on the system per cycle equals the area enclosed by the path representing the process on a PV diagram

32 Isothermal Process At right is a PV diagram of an isothermal expansion The curve is a hyperbola The curve is called an isotherm The curve of the PV diagram indicates PV = constant The equation of a hyperbola Because it is an ideal gas and the process is quasi-static, PV = nrt and nrt W P dv dv nrt V V V V f f f V V V i i i W V i nrt ln V f dv V

33 DEint Q W PV nrt W V V f i P dv Isothermal Process An isothermal process is one that occurs at a constant temperature Since there is no change in temperature, DE int = 0 Therefore, Q = - W Any energy that enters the system by heat must leave the system by work

34 DEint Q W PV nrt W V V f i P dv Isobaric Processes An isobaric process is one that occurs at a constant pressure The values of the heat and the work are generally both nonzero The work done is W = P (V f V i ) where P is the constant pressure

35 DEint Q W PV nrt W V V f i P dv Isovolumetric Processes An isovolumetric process is one in which there is no change in the volume Since the volume does not change, W = 0 From the first law, DE int = Q If energy is added by heat to a system kept at constant volume, all of the transferred energy remains in the system as an increase in its internal energy

Special Case: Adiabatic Free Expansion This is an example of adiabatic free expansion The process is adiabatic because it takes place in an insulated container Because the gas expands into a vacuum,

36 Special Case: Adiabatic Free Expansion This is an example of adiabatic free expansion The process is adiabatic because it takes place in an insulated container Because the gas expands into a vacuum, it does not apply a force on a piston and W = 0 Since Q = 0 and W = 0, DE int = 0 and the initial and final states are the same and no change in temperature is expected. No change in temperature is expected

37 Thermo Processes DEint Q W Adiabatic No heat exchanged Q = 0 and DE int = W Isobaric Constant pressure W = P (V f V i ) and DE int = Q + W Isovolumetric Constant Volume W = 0 and DE int = Q Isothermal Constant temperature DE int = 0 and Q = -W W V i nrt ln V f

38 DEint Q W W V V f i P dv A gas is taken through the cyclic process as shown. (a) Find the net energy transferred to the system by heat during one complete cycle. (b) What If? If the cycle is reversed that is, the process follows the path ACBA what is the net energy input per cycle by heat?

39 DEint Q W PV nrt W V V f i P dv A sample of an ideal gas goes through the process as shown. From A to B, the process is adiabatic; from B to C, it is isobaric with 100 kj of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 150 kj of energy leaving the system by heat. Determine the difference in internal energy E(B) E(A).

40 Important Concepts

$Heat Engine DEint = 0 for the entire cycle A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work A heat engine$

41 Heat Engine DEint = 0 for the entire cycle A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work A heat engine carries some working substance through a cyclical process The working substance absorbs energy by heat from a high temperature energy reservoir (Q h ) Work is done by the engine (W eng ) Energy is expelled as heat to a lower temperature reservoir (Q c )

42 e W Q eng h DEint = 0 for the entire cycle

43 Thermal Efficiency of a Heat Engine DEint = 0 for the entire cycle Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature e W Q Q Q eng h c c 1 Q Q Q h h h We can think of the efficiency as the ratio of what you gain to what you give

44 Rank in order, from largest to smallest, the work W out performed by these four heat engines. e W Q Q Q eng h c c 1 Q Q Q h h h A. W b > W a > W c > W d B. W b > W a > W b > W c C. W b > W a > W b = W c D. W d > W a = W b > W c E. W d > W a > W b > W c

45 Rank in order, from largest to smallest, the work W out performed by these four heat engines. e W Q Q Q eng h c c 1 Q Q Q h h h A. W b > W a > W c > W d B. W b > W a > W b > W c C. W b > W a > W b = W c D. W d > W a = W b > W c E. W d > W a > W b > W c

46 Perfect Heat Engine No energy is expelled to the cold reservoir It takes in some amount of energy and does an equal amount of work e = 100% It is an impossible engine

47 W Q Q e W Q H H Could this heat engine be built? C Q T T and ec 1 Q T T c c c h h h A. Yes. B. No. C. It s impossible to tell without knowing what kind of cycle it uses.

48 W Q Q e W Q H H Could this heat engine be built? C Q T T and ec 1 Q T T c c c h h h A. Yes. B. No. C. It s impossible to tell without knowing what kind of cycle it uses.

49 Analyze this engine to determine (a) the net work done per cycle, (b) the engine s thermal efficiency and (c) the engine s power output if it runs at 600 rpm. Assume the gas is monatomic and follows the idealgas process above.

50 Gasoline Engine In a gasoline engine, six processes occur during each cycle For a given cycle, the piston moves up and down twice This represents a four-stroke cycle The processes in the cycle can be approximated by the Otto cycle

51 The Conventional Gasoline Engine

52 Gasoline Engine Intake Stroke During the intake stroke, the piston moves downward A gaseous mixture of air and fuel is drawn into the cylinder Energy enters the system as potential energy in the fuel

53 Gasoline Engine Compression Stroke The piston moves upward The air-fuel mixture is compressed adiabatically The temperature increases The work done on the gas is positive and equal to the negative area under the curve

54 Gasoline Engine Spark Combustion occurs when the spark plug fires This is not one of the strokes of the engine It occurs very quickly while the piston is at its highest position Conversion from potential energy of the fuel to internal energy

55 Gasoline Engine Power Stroke In the power stroke, the gas expands adiabatically This causes a temperature drop Work is done by the gas The work is equal to the area under the curve

56 Gasoline Engine Valve Opens An exhaust valve opens as the piston reaches its bottom position The pressure drops suddenly The volume is approximately constant So no work is done Energy begins to be expelled from the interior of the cylinder

57 Gasoline Engine Exhaust Stroke In the exhaust stroke, the piston moves upward while the exhaust valve remains open Residual gases are expelled to the atmosphere The volume decreases

58 The Otto cycle approximates the processes occurring in an internal combustion engine Otto Cycle

59 Otto Cycle Efficiency If the air-fuel mixture is assumed to be an ideal gas, then the efficiency of the Otto cycle is e 1 1 V V g g is the ratio of the molar specific heats V 1 / V 2 is called the compression ratio

60 Otto Cycle Efficiency, cont Typical values: Compression ratio of 8 g = 1.4 e = 56% Efficiencies of real engines are 15% to 20% Mainly due to friction, energy transfer by conduction, incomplete combustion of the airfuel mixture

62 Heat Pumps and Refrigerators Heat engines can run in reverse This is not a natural direction of energy transfer Must put some energy into a device to do this Devices that do this are called heat pumps or refrigerators Examples A refrigerator is a common type of heat pump An air conditioner is another example of a heat pump

64 Coefficient of Performance The effectiveness of a heat pump is described by a number called the coefficient of performance (COP) In heating mode, the COP is the ratio of the heat transferred in to the work required COP = energy transferred at high temp work done by heat pump Q h W

65 COP, Heating Mode COP is similar to efficiency Q h is typically higher than W Values of COP are generally greater than 1 It is possible for them to be less than 1 We would like the COP to be as high as possible

66 COP, Cooling Mode In cooling mode, you gain energy from a cold temperature reservoir Q c COP W A good refrigerator should have a high COP Typical values are 5 or 6

67 Carnot Engine Carnot Cycle A heat engine operating in an ideal, reversible cycle (now called a Carnot cycle) between two reservoirs is the most efficient engine possible. This sets an upper limit on the efficiencies of all other engines

68 Carnot Cycle, PV Diagram The work done by the engine is shown by the area enclosed by the curve, W eng The net work is equal to Q h Q c DE int = 0 for the entire cycle

69 Efficiency of a Carnot Engine Carnot showed that the efficiency of the engine depends on the temperatures of the reservoirs Qc Tc Tc and ec 1 Qh Th Th Temperatures must be in Kelvins All Carnot engines operating between the same two temperatures will have the same efficiency

70 Carnot s Theorem No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs All real engines are less efficient than a Carnot engine because they do not operate through a reversible cycle The efficiency of a real engine is further reduced by friction, energy losses through conduction, etc.

71 Notes About Carnot Efficiency Efficiency is 0 if T h = T c Efficiency is 100% only if T c = 0 K Such reservoirs are not available Efficiency is always less than 100% The efficiency increases as T c is lowered and as T h is raised In most practical cases, T c is near room temperature, 300 K So generally T h is raised to increase efficiency

72 Carnot Cycle in Reverse Theoretically, a Carnot-cycle heat engine can run in reverse This would constitute the most effective heat pump available This would determine the maximum possible COPs for a given combination of hot and cold reservoirs

73 Carnot Heat Pump COPs In heating mode: COP In cooling mode: C Qh Th W T T h c COP C Qc Tc W T T h c

26. A heat pump, shown in Figure P22.26, is essentially an air conditioner installed backward. It extracts energy from colder air outside and deposits it in a warmer room.

74 26. A heat pump, shown in Figure P22.26, is essentially an air conditioner installed backward. It extracts energy from colder air outside and deposits it in a warmer room. Suppose that the ratio of the actual energy entering the room to the work done by the device s motor is 10.0% of the theoretical maximum ratio. Determine the energy entering the room per joule of work done by the motor, given that the inside temperature is 20.0 C and the outside temperature is 5.00 C. Q W h Qh W Carnotefficiency Carnotcycle Q h Th 293 K W T T 293 K 268 K h c 1.17 joules of energy enter the room by heat for each joule of work done.

75 Can this refrigerator be built? W Q Q H COP Q c W C COP C Qc Tc W T T h c

Speed Distribution at CONSTANT Temperature is given by the Maxwell Boltzmann Speed Distribution

Speed Distribution at CONSTANT Temperature is given by the Maxwell Boltzmann Speed Distribution Temperature ~ Average KE of each particle Particles have different speeds Gas Particles are in constant RANDOM motion Average KE of each particle is: 3/2 kt Pressure is due to momentum transfer Speed Distribution