Temperature - Santa Rosa Junior College

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 transferSpeed ‘Distribution’ atCONSTANT Temperatureis given by theMaxwell BoltzmannSpeed Distribution

Mean Free Path A single molecule follows azig-zag path through a gasas it collides with othermolecules. The average distancebetween the collisions iscalled the mean free path: (N/V) is the number density of the gas in m 3. r is the the radius of the molecules when modeled ashard spheres; for many common gases r 10 10 m. 2013 Pearson Education, Inc.Slide 18-20

QuickCheck 18.1The temperature of a rigid container of oxygen gas(O2) is lowered from 300 C to 0 C. As a result, themean free path of oxygen moleculesA. Increases.B. Is unchanged.C. Decreases. 2013 Pearson Education, Inc.Slide 18-21

QuickCheck 18.1The temperature of a rigid container of oxygen gas(O2) is lowered from 300 C to 0 C. As a result, themean free path of oxygen moleculesA. Increases.B. Is unchanged.C. Decreases. 2013 Pearson Education, Inc.λ depends only on N/V, not T.Slide 18-22

Pressure and Kinetic Energy Assume a container is a cube withedges d.Look at the motion of the molecule interms of its velocity components andmomentum and the average forcePressure is proportional to the numberof molecules per unit volume (N/V) andto the average translational kineticenergy of the molecules.This equation also relates themacroscopic quantity of pressure with amicroscopic quantity of the averagevalue of the square of the molecularspeedOne way to increase the pressure is toincrease the number of molecules perunit volumeThe pressure can also be increased byincreasing the speed (kinetic energy) ofthe molecules2 N 12 P mo v 3 V 2

Molecular Interpretation ofTemperature We can take the pressure as it relates to the kineticenergy and compare it to the pressure from theequation of state for an ideal gas2 N 1 2 NkBTPv nRT m 3 V 2 Temperature is a direct measure of the averagemolecular kinetic energy

Molecular Interpretation ofTemperature Simplifying the equation relatingtemperature and kinetic energy gives132mo v kBT22 This can be applied to each direction,1 2 1m v x kBT22– with similar expressions for vy and vz

Total Kinetic Energy The total kinetic energy is just N times the kineticenergy of each molecule 1 2 33K N mv NkBTnRTtot trans2 2 2 If we have a gas with only translational energy, this isthe internal energy of the gas This tells us that the internal energy of an ideal gasdepends only on the temperature

QuickCheck 18.4A rigid container holds both hydrogen gas (H2) andnitrogen gas (N2) at 100 C. Which statementdescribes the average translational kinetic energiesof the molecules?A. єavg of H2 єavg of N2.B. єavg of H2 єavg of N2.C. єavg of H2 єavg of N2. 2013 Pearson Education, Inc.Slide 18-37

QuickCheck 18.4A rigid container holds both hydrogen gas (H2) andnitrogen gas (N2) at 100 C. Which statementdescribes the average translational kinetic energiesof the molecules?A. єavg of H2 єavg of N2.B. єavg of H2 єavg of N2.C. єavg of H2 єavg of N2. 2013 Pearson Education, Inc.Slide 18-38

Kinetic Theory ProblemA 5.00-L vessel contains nitrogen gas at27.0 C and 3.00 atm. Find (a) the totaltranslational kinetic energy of the gasmolecules and (b) the average kinetic energyper molecule.

Hot QuestionSuppose you apply a flame to 1 liter of water for a certaintime and its temperature rises by 10 degrees C. If you applythe same flame for the same time to 2 liters of water, by howmuch will its temperature rise?a) 1 degreeb) 5 degreesc) 10 degreesd) zero degrees

Ludwig Boltzmann or Dean Gooch? 1844 – 1906 Austrian physicist Contributed to– Kinetic Theory of Gases– Electromagnetism– Thermodynamics Pioneer in statisticalmechanics

Distribution of Molecular Speeds The observed speed distribution of gasmolecules in thermal equilibrium isshown at rightNV is called the Maxwell-Boltzmannspeed distribution functionmo is the mass of a gas molecule, kB isBoltzmann’s constant and T is theabsolute temperature mo NV 4π N kTπ2B 3/22v e mv 2 / 2 kBT

Molecular Speeds andCollisions

Speed Summary Root mean square speed v rmskT3kBT 1.73 Bmomo The average speed is somewhat lower thanthe rms speed8k TkT v avgB 1.60 Bmoπ mo The most probable speed, vmp is the speed atwhich the distribution curve reaches a peak vrms vavg vmp v mp2kBTkT 1.41 Bmm

Some Example vrms ValuesAt a given temperature, lighter molecules move faster, onthe average, than heavier molecules

Speed Distribution The peak shifts to the rightas T increases– This shows that the averagespeed increases withincreasing temperature The asymmetric shapeoccurs because the lowestpossible speed is 0 and thehighest is infinity

The Kelvin Temperature ofan ideal gas is a measure ofthe average translationalkinetic energy per particle:3 / 2kT KE 1/ 2mv2rmsk 1.38 x 10-23 J/K Boltzmann’s ConstantRoot-mean-square speed:v rms v23kTm

Kinetic Theory ProblemCalculate the RMS speed of an oxygen moleculein the air if the temperature is 5.00 C.The mass of an oxygen molecule is 32.00 u(k 1.3 8x 10 -23 J/K, u 1.66 x 10 -27 kg)vrms3kT mWhat is m?m is the mass of oneoxygen molecule in kg.What is u?How do we get the mass in kg?

QuickCheck 18.3A rigid container holds both hydrogen gas (H2)and nitrogen gas (N2) at 100 C. Which statementdescribes their rms speeds?A. vrms of H2 vrms of N2.B. vrms of H2 vrms of N2.C. vrms of H2 vrms of N2. 2013 Pearson Education, Inc.Slide 18-34

QuickCheck 18.3A rigid container holds both hydrogen gas (H2)and nitrogen gas (N2) at 100 C. Which statementdescribes their rms speeds?A. vrms of H2 vrms of N2.B. vrms of H2 vrms of N2.C. vrms of H2 vrms of N2. 2013 Pearson Education, Inc.Slide 18-35

Kinetic Theory ProblemCalculate the RMS speed of an oxygen moleculein the air if the temperature is 5.00 C.The mass of an oxygen molecule is 32.00 u(k 1.3 8x 10 -23 J/K, u 1.66 x 10 -27 kg)vrms3kT mWhat is m?m is the mass of oneoxygen molecule. 233(1.38 x10 J / K )278 K (32u )(1.66 x10 27 kg / u ) 466m / sIs this fast?YES!Speed ofsound:343m/s!

A cylinder contains a mixture of heliumand argon gas in equilibrium at 150 C.(a) What is the average kinetic energyfor each type of gas molecule?(b) What is the root-mean-square speedof each type of molecule?

More Kinetic Theory ProblemsA gas molecule with a molecular mass of 32.0 u has aspeed of 325 m/s. What is the temperature of the gasmolecule?A) 72.0 KB) 136 KC) 305 KD) 459 KE) A temperature cannot be assigned to a singlemolecule.Temperature Average KE of all particles

Equipartition of Energy Each translational degree of freedom contributes anequal amount to the energy of the gas– In general, a degree of freedom refers to anindependent means by which a molecule canpossess energyEach degree of freedom contributes ½kBT to theenergy of a system, where possible degrees offreedom are those associated with translation,rotation and vibration of moleculesWith complex molecules, other contributions tointernal energy must be taken into accountOne possible energy is the translational motion ofthe center of massRotational motion about the various axes alsocontributesThere is kinetic energy and potential energyassociated with the vibrations

Monatomic and Diatomic GasesThe thermal energy of a monatomic gas of N atoms isA diatomic gas has more thermal energy than a monatomicgas at the same temperature because the molecules haverotational as well as translational kinetic energy.

Molar Specific Heat We define specific heats for two processesthat frequently occur:– Changes with constant pressure– Changes with constant volume Using the number of moles, n, we candefine molar specific heats for theseprocesses Molar specific heats:– Q nCV DT for constant-volume processes– Q nCP DT for constant-pressure processes

Ideal Monatomic Gas Therefore, Eint 3/2 nRT E is a function of T only In general, the internal energy of an idealgas is a function of T only– The exact relationship depends on the type ofgas At constant volume, Q Eint nCV T– This applies to all ideal gases, not justmonatomic ones

Ratio of Molar Specific Heats We can also define the ratio of molar specific heatsCP 5R / 2 γ 1.67CV 3R / 2 Theoretical values of CV , CP , and γ are in excellentagreement for monatomic gases But they are in serious disagreement with the valuesfor more complex molecules– Not surprising since the analysis was for monatomic gases

Agreement with Experiment Molar specific heat is a function oftemperatureAt low temperatures, a diatomic gasacts like a monatomic gas CV 3/2 RAt about room temperature, the valueincreases to CV 5/2 R– This is consistent with addingrotational energy but notvibrational energyAt high temperatures, the valueincreases to CV 7/2 R– This includes vibrational energyas well as rotational andtranslational

Sample Values of Molar SpecificHeats

In a constant-volume process, 209 J ofenergy is transferred by heat to 1.00 molof an ideal monatomic gas initially at300 K. Find (a) the increase in internalenergy of the gas, (b) the work done onit, and (c) its final temperature

Molar Specific Heats of OtherMaterials The internal energy of more complex gasesmust include contributions from therotational and vibrational motions of themolecules In the cases of solids and liquids heated atconstant pressure, very little work is done,since the thermal expansion is small, and CPand CV are approximately equal

Adiabatic Processes for anIdeal Gas An adiabatic process is one in which no energy istransferred by heat between a system and itssurroundings (think styrofoam cup) Assume an ideal gas is in an equilibrium state and soPV nRT is valid The pressure and volume of an ideal gas at any timeduring an adiabatic process are related byPV γ constantγ CP / CV is assumed to be constantAll three variables in the ideal gas law (P, V, T ) canchange during an adiabatic process

Special Case: Adiabatic FreeExpansion This is an example of adiabatic freeexpansion The process is adiabatic because ittakes place in an insulated container Because the gas expands into avacuum, it does not apply a force ona piston and W 0 Since Q 0 and W 0, Eint 0 andthe initial and final states are thesame and no change in temperature isexpected.– No change in temperature is expected

Reading Question 18.2What additional kind of energy makes CV largerfor a diatomic than for a monatomic gas?A.B.C.D.E.Charismatic energy.Translational energy.Heat energy.Rotational energy.Solar energy.Slide 18-12

Reading Question 18.2What additional kind of energy makes CV largerfor a diatomic than for a monatomic gas?A.B.C.D.E.Charismatic energy.Translational energy.Heat energy.Rotational energy.Solar energy.Slide 18-13

Reading Question 18.3The second law of thermodynamics says thatA. The entropy of an isolated system neverdecreases.B. Heat never flows spontaneously fromcold to hot.C. The total thermal energy of an isolatedsystem is constant.D. Both A and B.E. Both A and C.Slide 18-15

Reading Question 18.4In general,A. Both microscopic and macroscopic processesare reversible.B. Both microscopic and macroscopic processesare irreversible.C. Microscopic processes are reversible andmacroscopic processes are irreversible.D. Microscopic processes are irreversible andmacroscopic processes are reversible.Slide 18-16

QuickCheck 18.6Systems A and B are both monatomic gases. At this instant,A.TA TB.B.TA TB.C.TA TB.D.There’s not enough information to compare theirtemperatures.Slide 18-51

QuickCheck 18.6Systems A and B are both monatomic gases. At this instant,A.TA TB.B.TA TB.C.TA TB.D.There’s not enough information to compare theirtemperatures.A has the larger average energy per atom.Slide 18-52

Adiabatic Process The PV diagram showsan adiabatic expansionof an ideal gas The temperature of thegas decreases– Tf Ti in this process For this processPi Viγ Pf Vfγ andTi Viγ-1 Tf Vfγ-1