Page Physics 207 – Lecture 25Physics 207: Lecture 25, Pg 1Lecture 25Goals:Goals:••Chapters 18, microChapters 18, micro--macro connectionmacro connection• Third test on Thursday at 7:15 pm.Physics 207: Lecture 25, Pg 2(m/s)Percentage ofmolecules510150-100100-200200-300300-400400-500500-600600-700700-800800-900900-10001000-11001100-1200Nitrogen moleculesnear room temperaturePage Physics 207 – Lecture 25Physics 207: Lecture 25, Pg 3lWhat is the typical size of an atom or a small molecule?Atomic scaleA) 10-6m B) 10-10 m C) 10-15 mr ≈1 angstrom=10-10 mrPhysics 207: Lecture 25, Pg 4Mean free pathlAverage distance particle moves between collisions:λ=14 2πN /V( )r2N/V: particles per unit volumelThe mean free path at atmospheric pressure is:λ=68 nmPage Physics 207 – Lecture 25Physics 207: Lecture 25, Pg 5Pressure of a gasvxvxmPhysics 207: Lecture 25, Pg 6Consider a gas with all molecules traveling at a speed vxhitting a wall. l If (N/V) increases by a factor of 2, the pressure would:A) decrease B) increase x2 C) increase x4l If m increases by a factor of 2, the pressure would:A) decrease B) increase x2 C) increase x4l If vxincreases by a factor of 2, the pressure would:A) decrease B) increase x2 C) increase x4vxPage Physics 207 – Lecture 25Physics 207: Lecture 25, Pg 7P=(N/V)mvx2lBecause we have a distribution of speeds:P=(N/V)m(vx2)avglFor a uniform, isotropic system:(vx2)avg= (vy2)avg= (vz2)avglRoot-mean-square speed:(v2)avg=(vx2)avg+(vy2)avg+(vz2)avg=Vrms2Physics 207: Lecture 25, Pg 8Microscopic calculation of pressureP=(N/V)m(vx2)avg=(1/3) (N/V)mvrms2PV = (1/3) Nmvrms2Page Physics 207 – Lecture 25Physics 207: Lecture 25, Pg 9Micro-macro connectionPV = (1/3) Nmvrms2PV = NkBT (ideal gas law)kBT =(1/3) mvrms2lThe average translational kinetic energy is:avg=(1/2) mvrms2avg=(3/2) kBTPhysics 207: Lecture 25, Pg 10l The average kinetic energy of the molecules of an ideal gas at 10°C has the value K1. At what temperature T1(in degrees Celsius) will the average kinetic energy of the same gas be twice this value, 2K1?(A) T1= 20°C(B) T1= 293°C(C) T1= 100°Cl Suppose that at some temperature we have oxygen molecules moving around at an average speed of 500 m/s. What would be the average speed of hydrogen molecules at the same temperature? (A) 100 m/s(B) 250 m/s(C) 500 m/s(D) 1000 m/s(E) 2000 m/sPage Physics 207 – Lecture 25Physics 207: Lecture 25, Pg 11Equipartition theoreml Things are more complicated when energy can be stored in other degrees of freedom of the system. monatomic gas: translationsolids: translation+potential energydiatomic molecules: translation+vibrations+rotationsPhysics 207: Lecture 25, Pg 12Equipartition theoreml The thermal energy is equally divided among all possible energy modes (degrees of freedom). The average thermal energy is (1/2)kBT for each degree of freedom. avg=(3/2) kBT (monatomic gas)avg=(6/2) kBT (solids)avg=(5/2) kBT (diatomic molecules)l Note that if we have N particles:Eth=(3/2)N kBT =(3/2)nRT (monatomic gas)Eth=(6/2)N kBT =(6/2)nRT (solids)Eth=(5/2)N kBT =(5/2)nRT (diatomic molecules)Page Physics 207 – Lecture 25Physics 207: Lecture 25, Pg 13Specific heatl Molar specific heats can be directly inferred from the thermal energy.Eth=(6/2)N kBT =(6/2)nRT (solid)∆Eth=(6/2)nR∆T=nC∆TC=3R (solid) l The specific heat for a diatomic gas will be larger than the specific heat of a monatomic gas:Cdiatomic=Cmonatomic+RPhysics 207: Lecture 25, Pg 14Entropyl A perfume bottle breaks in the corner of a room. After some time, what would you expect?A)B)Page Physics 207 – Lecture 25Physics 207: Lecture 25, Pg 15very unlikelyprobability=(1/2)Nl The probability for each particle to be on the left half is ½.Physics 207: Lecture 25, Pg 16Second Law of thermodynamicsl The entropy of an isolated system never decreases. It can only increase, or in equilibrium, remain constant.l The laws of probability dictate that a system will evolve towards the most probable and most random macroscopic statel Thermal energy is spontaneously transferred from a hotter system to a colder
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