Lesson 12 Physics 168 1
Temperature and Kinetic Theory of Gases 2
Atomic Theory of Matter On microscopic scale, arrangements of molecules in solids, liquids, and gases are quite different 3
Temperature and Thermometers Temperature is a measure of how hot or cold something is Most materials expand when heated 4
Temperature and Thermometers (cont d) Temperature is generally measured using either Fahrenheit or Celsius scale Freezing point of water is 0 C = 32 F boiling point is 100 C = 212 F T C = 5 9 (T F 32 ) T F = 9 5 T C + 32 5
Thermal Expansion Linear expansion occurs when an object is heated L = L 0 (1 + T ) coefficient of linear expansion 6
Volume expansion is similar except that it is relevant for liquids and gases as well as solids: V = V 0 T coefficient of volume expansion For uniform solids 3 7
Equation of state and Boyle s law Relationship between volume, pressure, temperature, and mass of a gas is called an equation of state We will deal here with gases that are not too dense Boyle s Law: volume of a given amount of gas is inversely proportional to pressure as long as temperature is constant V / 1 P 8
Gas Laws and Absolute Temperature Volume is linearly proportional to temperature, as long as temperature is somewhat above condensation point and pressure is constant: V / T Extrapolating volume becomes zero at 273.15 C this temperature is called absolute zero 9
Gas Laws and Absolute Temperature (cont d) Concept of absolute zero allows us to define a third temperature scale absolute (or Kelvin) scale This scale starts with 0 K at absolute zero but otherwise is same as Celsius scale Freezing point of water is 273.15 K and boiling point is 373.15 K When volume is constant pressure is directly proportional to temperature P / T 10
Temperature of various places and phenomena 11
Ideal Gas Law We can combine three relations just derived into a single relation: PV / T What about amount of gas present? If temperature and pressure are constant volume is proportional to amount of gas: PV / mt 12
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microworld Micro World atoms & molecules macroworld Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) Atomic mass mass of an atom in atomic mass units (u) By definition: 1 atom 12 C weighs 12 amu u On this scale u u 1 H = 1.008 amu 16 O = 16.00 amu 14
Mole (mol): A unit to count numbers of particles Dozen = 12 Pair = 2 Mole amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C 1 mol = N A = 6.022 x 10 23 N A Avogadro s number 15
Molar mass is the mass of 1 mole of eggs shoes marbles atoms in grams 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g u 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li u For any element (u) atomic mass (amu) = molar mass (grams) 1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 10 23 amu u M = molar mass in g/mol M = molar mass in g/mol M N A = Avogadro s number 16
Ideal Gas Law (cont d) A mole (mol) is defined as number of grams of a substance that is numerically equal to molecular mass of substance 1 mol H 2 1 mol N 2 has a mass of has a mass of 2g 28 g 1 mol CO 2 has a mass of 44 g Number of moles in a certain mass of material: n (mol) = mass (grams) Molecular mass (g/mol) 17
Ideal Gas Law (cont d) We can now write ideal gas law PV = nrt number of moles universal gas constant R =8.315 J/(mol K) =0.0821 (L atm)/(mol K) = 1.99 calories/(mol K) 18
Ideal Gas Law in Terms of Molecules: Avogadro s Number Since gas constant is universal number of molecules in one mole is same for all gases That number is called Avogadro s number: N A =6.02 10 23 Number of molecules in a gas is number of moles times Avogadro s number: N = nn A 19
Ideal Gas Law in Terms of Molecules: Avogadro s Number Therefore we can write: PV = NkT Boltzmann s constant k K = R N A = 8.315 J/mol K 6.02 10 23 / mol =1.38 10 23 J/K 20
Kinetic Theory and Molecular Interpretation of Temperature Assumptions of kinetic theory: Large number of molecules moving in random directions with a variety of speeds Molecules are far apart on average Molecules obey laws of classical mechanics and interact only when colliding Collisions are perfectly elastic 21
Kinetic Theory and Molecular Interpretation of Temperature Force exerted on the wall by collision of one molecule is F = (mv) t = 2 mv x 2l/v x = mv2 x l Then force due to all N molecules colliding with that wall is F = m l N v2 x 22
Kinetic Theory and Molecular Interpretation of Temperature Averages of squares of speeds in all three directions are equal: F = m l N v2 3 So pressure is P = F A = 1 3 Nm v 2 Al = 1 3 Nm v 2 V Rewriting PV = 2 3 N (1 2 m v2 ) so hkei = KE = 1 2 m v2 = 3 2 kt Average translational kinetic energy of molecules in an ideal gas is directly proportional to temperature of gas 23
Kinetic Theory and Molecular Interpretation of Temperature We can invert this to find average speed of molecules in a gas as a function of temperature v rms = p v 2 = r 3kT m 24
Mean Free Path Average speed of molecules in a gas at normal pressure is several hundred meters per second yet if somebody across room from you opens a perfume bottle you do not detect odor for several minutes Reason for time delay perfume molecules do not travel directly towards you but instead travel a zigzag path due to collisions with air molecules Average distance travelled by a molecule between collisions is called its mean free path 25
Mean Free Path (cont d) Consider one gas molecule of radius moving with speed through a region of stationary molecules Moving molecule will collide with another molecule of radius if centers of two molecules come within a distance r v r d = r 1 + r 2 from each other If all molecules are of same type d = molecular diameter 26
After a time Mean Free Path (cont d) As molecule moves it will collide with any molecule t whose center molecule moves a distance is in a circle of radius vt in cylindrical volume d and collides with every molecule d 2 vt Number of molecules in this volume is n V d 2 vt Number density number of molecules per unit volume n V = N/V After each collision direction of molecule changes so path actually zigs and zags Total path length divided by number of collisions is mean free path = vt n V d 2 vt = 1 n V d 2 27
Mean Free Path (cont d) Our previous calculation of mean free path assumes that all but one of gas molecules are stationary which is not a realistic situation When motion of all molecules is taken into account mean free path becomes = 1 p 2 nv d 2 Average time between collisions is called collision time Reciprocal of collision time is equal to average number of collisions per second Then ifv av is average speed average distance travelled between collisions is = v av 28
What is heat? Subjective sensation (familiar to everyone): when you touch something hot heat goes into your hand and burns it We talk of tansfer of heat e.g. + from one object (the radiator) to another (your hand) Fact that radiator is hot expressed as radiator is @ high temperature Heat and work are mutually convertible: you can do work and get heat or use heat and do work (a car engine or steam engine) The catch is + when you go from heat to useful work you never achieve 100% conversion there is always some heat wasted Thermal energy + random motion of molecules in matter sample which constitutes form of kinetic energy Temperature + measure of average kinetic energy/molecule 29
So...what is heat? Heat is also kinetic energy + moving energy It is kinetic energyof zillions of atoms in material vibrating (in solid) or zapping around (in gas) in a random way going nowehere in particular Heat is random energy of atoms and molecules This is very different from organized energy It is like throwing bag of hot popcorn (organized energy) vs. letting them pop around inside popper (random energy) Popcorn popping very quickly has a lot of heat energy 30
So...what is heat? Total heat energy of material depends on two things: 1 how many atomos there are 2 how energetic each atom is (on average) Heat energy Q can be expressed as product of number of atoms N and average randon energy of single atom h#i Q = N h#i Temperature + direct measurement of second factor: average energy of an atom High temperature means peppy atoms Low temperature means sluggish atoms When atoms are so sluggish that they don t move you ve hit bottom in temperature scale + absoulte zero which is T = 273 C + very cold! 31
So...what is heat? When you push your hand across table (mechanical energy) you set molecules at surface of table into random vibration and create heat energy (friction in this case) 32
So...what is heat? You touch hot poker + madly vibrrating atoms on surface of poker bombard molecules of your skin these start vibrating + causing all kind of neurochemical reactions which tell your brain: It s hot, let go! 33
Laws of Thermodynamics Via work or via heat, every energy transfer happen can! 34
THERMODYNAMICS (developed in 19th century) Phenomenological theory to describe equilibrium properties of macroscopic systems based on few macroscopically measurable quantities 35
A thermodynamic system A system is a closed environment in which heat transfer can take place -For example, gas, walls, and cylinder of an automobile engine- Work done on gas or work done by gas 36
Internal energy Internal energy of system U of a system is total of all kinds of energy possessed by particles that make up system Usually internal energy consists of sum of potential and kinetic energies of working gas molecules 37
Two ways to increase internal energy U + U WORK DONE ON A GAS Positive HEAT PUT INTO A SYSTEM Positive 38
Two ways to decrease internal energy U W out hot U DECREASE hot Q out WORK DONE BY EXPANDING GAS is Positive W HEAT LEAVES A SYSTEM is Negative Q 39
Thermodynamic state STATE of a thermodynamic system is determined by four factors: Absolute Pressure P in Pascals Temperature T in Kelvins Volume V in cubic meters Number of moles, n, of working gas 40
Thermodynamic process Increase in Internal Energy U W out Q in Initial State: P 1 V 1 T 1 n 1 Heat input Work by gas Final State: P 2 V 2 T 2 n 2 41
Reverse process Decrease in Internal Energy W in U Q out Initial State: Work on gas Final State: P 1 V 1 T 1 n 1 Loss of heat P 2 V 2 T 2 n 2 42
Thermal Equilibrium: Zeroth Law of Thermodynamics Two objects placed in thermal contact will eventually come to same temperature When they do we say they are in thermal equilibrium In the process of reaching thermal equilibrium heat Q is transferred from one body to the other Zeroth law of thermodynamics says that if 2 objects are each in equilibrium with a 3 object they are also in thermal equilibrium with each other 43
First Law Thermodynamics Energy of universe is constant Energy can be neither created nor destroyed so while energy can be converted to another form total energy remains constant Energy In Any Isolated System Energy Out This is merely a statement of conservation of energy 44
First law of Thermodynamics First law of thermodynamics is application of conservation of energy principle to heat and thermodynamic processes Change in internal energy of a system is equal to heat added to system minus work done by system U = Q W Change in internal energy heat added to system work done by system Energy can t be created or destroyed it can only change form 45
First law of thermodynamics Net heat put into a system is equal to change in internal energy of system plus work done BY system Q = U + W = (final initial) Conversely, work done ON a system is equal to change in internal energy plus heat lost in process 46
Sign conventions for first law +W out Heat Q INPUT is positive Work BY a gas is positive +Q in + U Heat OUT is negative Work ON a gas is negative W in U Q = U + W = (final initial) Q out 47
Application of first law of thermodynamics Example 1 In figure, gas absorbs 400 J of heat and at same time does 120 J of work What is change in internal energy of system? on piston W out = 120 J Apply First Law: Q = U + W Q in = 400 J 48
Application of first law of thermodynamics (cont d) Q is positive +400 J Heat In W is positive Q = U + W U = Q W +120 J Work Out Q in = 400 J W out = 120 J U = Q W = (+400 J) (+120 J) = +280 J U = +280 J 49
Application of first law of thermodynamics (cont d) Energy is conserved 400 J of input thermal energy is used to perform is used to perform 120 J of external work, W out = 120 J increasing internal energy of system by 280 J Q in = 400 J Increase in internal energy is U = +280 J 50
It is common experience that a cup of hot coffee left in a cooler room eventually cools off This process satisfies first law of thermodynamics since amount of energy lost by coffee is equal to amount gained by surrounding air Now let us consider reverse process-hot coffee getting even hotter in a cooler room as a result of heat transfer from room air We all know that this process never takes place Yet, doing so would not violate first law as long as amount of energy lost by air is equal to amount gained by coffee It is clear from above that processes proceed in a certain direction and not in reverse direction First law places no restriction on direction of a process, but satisfying first law does not ensure that that process will actually occur This inadequacy of first law to identify whether a process can take place is remedied by introducing another general principle, second law of thermodynamic 51
Second Law of Thermodynamics Kelvin statement: No system can absorb heat from a single reservoir and convert it entirely to work without additional net changes in the system or its surroundings Clausius statement: A process whose only net result is to absorb heat from a cold reservoir and release the same amount of heat to a hot reservoir is impossible 52
Entropy A measure of disorder of a system Systems tend to change from a state of low entropy to a state of higher entropy If left to themselves, systems tend to increase their entropy 53
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Four thermodynamic processes Isochoric Process V = 0, W = 0 Isobaric Process P = 0 Isothermal Process T = 0, U = 0 Adiabatic Process Q = 0 Q = U + W 56
CONSTANT VOLUME 0 Q = U + W Isochoric process V = 0, W = 0 so that Q = U Q in Q out No Work Done + U U HEAT IN = INCREASE IN INTERNAL ENERGY HEAT OUT = DECREASE IN INTERNAL ENERGY 57
No Change in volume Isochoric example P 2 B P 1 A P A T A = P B T B 400 J V 1 =V 2 Heat input increases with constant P 400 J heat input increases internal energy V by and zero work is done 400 J 58
CONSTANT PRESSURE Isobaric process P = 0 Q = U + W But W = P V Q in Q out Work Out Work In + U U HEAT IN HEAT OUT =W out + =W out + INCREASE IN INTERNAL ENERGY DECREASE IN INTERNAL ENERGY 59
Isobaric example P A B V A T A = V B T B 400 J Heat input increases with constant P V V 1 V 2 400 J heat does 120 J of work increases internal energy by 280 J 60
Isobaric work P A B V A T A = V B T B 400 J V 1 V 2 Heat input increases V with constant P Work = Area under PV curve Work = P V 61
Isothermal process CONSTANT TEMPERATURE T = 0, U = 0 Q = U + W and Q = W Q in Q out Work Out Work In U = 0 U = 0 NEAT HEAT INPUT = WORK OUTPUT WORK INPUT = NET HEAT OUT 62
Isothermal example Constant T P A A B P B U = T = 0 P A V A =P B V B V 2 V 1 Slow compression at constant temperature No change in U 63
Isothermal expansion Constant T P A A P B B P A V A =P B V B V A V B T A =T B U = T = 0 400 J as of energy is absorbed by gas 400 J T = U = 0 of work is done on gas Isothermal Work W=nRTln V B V A 64
Adiabatic process NO HEAT EXCHANGE Q = 0 Q = U + W; W = U or U = W W = U U = W Work Out Work In U + U Q = 0 Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy 65
Adiabatic example P A A P B B V 1 V 2 Insulated Walls Q = 0 Expanding gas does work with zero heat loss Work = U 66
Adiabatic expansion P A A P A V A T A = P BV B T B P B B Q = 0 V 1 V 2 400 J of WORK is done, DECREASING internal energy by Net heat exchange is ZERO Q = 0 400 J P A V A =P B V B 67
REMEMBER, FOR ANY PROCESS INVOLVING AN IDEAL GAS PV = n R T P A V A T A = P BV B T B Q = U + W U = nc V T 68
A 2 L Example problem sample of Oxygen gas has an initial temperature and pressure of 200 K and 1 atm Gas undergoes four processes: AB : Heated at constant V to 400 K BC : Heated at constant P to 800 K CD : Cooled at constant V back to 1 atm DA : Cooled at constant P back to 200 K 69
How many moles of Consider point PV-DIAGRAM FOR PROBLEM A PV = n R T O 2 are present? P B B 400 K 800 K 1 atm A 200 K 2L n= PV RT = (101.300 Pa)(0.002 m3 ) (8.314 J/mol K)(200 K) =0.122 mol 70
Process AB Isochoric B What is pressure at point? P A T A = P B T B P B B 400 K 800 K 1 atm A 200 K 1 atm 200 K = P B 400 K 2L P B = 2 atm or 203 kpa 71
Process AB Analyze first law for ishochoric process Q = U + W AB W = 0 P B B 400 K 800 K Q = U = n C V T 1 atm A 200 K 2L U = (0.122 mol)(21.1 J/mol K)(400 K 200 K) Q = +514 J U = +514 J W = 0 72
What is Volume at point Process BC Isobaric C( & D)? V B T B = V C T C P B B 400 K 800 K C 2L 400 K = V C 800 K 1 atm 200 K D 2L 4L V C =V D =4L 73
Process BC Finding is ISOBARIC U for process BC P = 0 2 atm B 400 K 800 K C U = n C V T 1 atm 200 K 2L 4L U = (0.122 mol)(21.1 J/mol K)(800 K 400 K) U = +1028 J 74
Finding Work depends on change in V W for process BC P = 0 2 atm B 400 K 800 K C 1 atm 200 K Work = P V 2L 4L W = (2 atm)(4 L 2 L) = 4 atm L = 405 J W = +405 J 75
Finding Analyze first law for BC Q for process BC Q = U + W Q = +1028 J + 405 J Q = +1433 J 2 atm 1 atm B 400 K 800 K C 200 K 2L 4L Q = 1433 J U = 1028 J W = +405 J 76
Process CD Isochoric D What is temperature at point? P C T C = P D T D 2 atm 1 atm B A 400 K 800 K C 200 K D 2 atm 800 K = 1 atm T D 2L T D = 400 K 77
Process CD Analyze first law for ISOCHORIC process CD Q = U + W W = 0 Q = U = n C V T 2 atm 1 atm B A 400 K 800 K C 200 K 400 K D 2L U = (0.122 mol)(21.1 J/mol K)(400 K 800 K) Q = 1028 J U = 1028 J W = 0 78
Process DA Finding U is ISOBARIC for process DA P = 0 U = n C V T 2 atm 1 atm A 400 K 800 K 2L 200 K 4L 400 K D U = (0.122 mol)(21.1 J/mol K)(400 K 200 K) U = 514 J 79
Finding Wfor process DA Work depends on change in V P = 0 2 atm 400 K 800 K Work = P V 1 atm A 200 K 400 K D 2L 4L W = (1 atm)(2 L 4 L) = 2 atm L = 203 J W = 203 J 80
Finding Analyze first law for DA Q for process DA Q = U + W 2 atm 400 K 800 K Q = 514 J 203 J Q = 717 J 1 atm A 2L 200 K 4L 400 K D Q = 717 J U = 514 J W = 203 J 81
PROBLEM SUMMARY For all processes Q = U + W PROCESS AB BC CD DA TOTALS Q U W 514 J 514 J 1433 J 1028 J 405 J 1028 J 1028 J 717 J 514 J 203 J 202 J 0 0 0 202 J 82
NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREA 2 atm 1 atm +404 J B C B 2 atm 202 J Neg 1 atm C 2L 4L 2L 4L 2 atm 1 atm B C Area = (1 atm)(2 L) Net work = 2 atm L = 202 J 2L 4L 83