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- ch18_2_young_freedman.pdf

Noor E.

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Pressure of an Ideal Gas Rd di i (i i) i5. Random direct on (isotrop c assumpt on: (v x 2 ) av = (v y 2 ) av = (v z 2 ) av (x,y,z are the same) Since v 2 = v x 2 + v y 2 + v z 2 , we have (v 2 ) av = 3(v x 2 ) av This gives, 2 () 3 av tot vmN F L ? 6. Finally, the pressure on the wall is: 22 () ()1 tot av av Fv NmvmN P ?? ? 33AAL V Pressure of an Ideal Gas Rewriting, we have ? ? ? ? 2 21 2 PV N N KE ?? ?? (avg KE per molecule) ? This tells us that P inside the container: 32 3 av av mv NK? ? ?? container: ? is proportional to the # of molecules N ? is proportional to the a g KE of molec les to the v . ules (These are microscopic properties of the gas.) Molecular Interpretation of Temperature From before: 2 3 av PV N KE ?? ?? ?? ? From Idea Gas Law: PV=NkT (Recall k ? Boltzmann Constant) Fbthfth tbt dthFor both o these o e rue, we nee o ave: 3 (per molecule) 2 KE kT ?? ?? ?? ? (Note: K tr in your book is the total KE for all molecules in volume V )av all in . ?Temperature is a direct measure of the average translational KE of the molecules in an ideal gas.? Distribution of Molecular Speeds ihi id l? Within an idea gas, molecules moves with a diversity of speedsyp ? A mathematical rigorous way to describe this statistically is th di t ib ti f tiru a s ribution unc on f(v). ? Properties: ? f(v) ~ 0, v ~ 0 ? f(v) ? 0, v ? large ? f(v) largest at mid-range Maxwell-Boltzmann Distribution ? f(v)dv gives the probability of finding molecules with speed in [ d ] ?? 2 32 22 4 2 / m/kT m fe kT ? ?? ? ? ? ?? ? ?? ?? range v,v+ v . ? Averages with respect to the distribution of molecular speeds of molecular speeds can be calculated using f(v): ? 1. (avg of v) 2 (f 2 ) 0 () av vvfvdv? ? 22 () ()fd ? ? . (avg of v 0 av vvf v dv? Different Types of Averages d( l ) 8kT ? Average speed (mean va ue : ? Root Mean Square (RMS) speed: av v m? ? 2 3kT vv ?? ???? speed: Note: v av does not equal v rms ! ? Most Probable Speed - the maximum value of the distribution rms av m ?? Speed maximum value distribution function f(v): 2 mp kT v m ? Similar to ?mean? and ?median?, there are multiple ways to statistically describe the ?average? values for a distribution of molecules moving at different speeds! moving speeds! D drinking Mean Free Path for Gas Molecules For simplicity, let assume that the darker shaded molecule is the only one which is moving and to the right with speed v. r v 2r 2 4 r? Then within a time interval dt, the number of molecules that it might collide will be given by: spatial volume ?? Effective collision area = 2 (2 )r? Assumptions: llihfii di ( ? ? ? ? ? ? 2 volume number density indicated (blue) 4 (we take ) rms dn rvdtNV vv? ? ?? ?? ? ?? ?Molecule w t fin te radius r note: point-like particles do not collide) ? The number density (# per unit vol) is given by N/V. () Thus, the number of collisions/unit time is, 2 4dN gve by ? Molecules move at an average speed v (we will this average to be v rms ). dnrvN dt V ? ? Mean Free Path for Gas Molecules To take into account that other molecules besides the red one are also moving, the estimated collision rate should be higher and it can be shown that this mean collision rate will be larger by . So, 2 2 42dn r vN dt V ? ? The average time between collisions (mean free time) is then the reciprocal of this note average is then the value, 2 42 mean V t rvN? ? And, the mean free path will be, 2 42 mean V vt rN ? ? ?? or 2 42 kT rP ? ? ? (we used )PV NkT? Derivation of the Factor 2 Key Idea: If all the molecules are moving, the relevant quantity for consideration should be the relative velocity between a pair of molecules. v 2 v 1 1 v 2 v 21 ?vv Again, we take the average of this relative velocity to be the root-mean-square value: ???? 2 21 21 22 21 11 2 rel v ?? ? ?? ? vvvv vv vv vv ? ??? Note: are in 12 andvv random relative directions ! 22 21 11 2?? ?vv vv vv??? 21 0?vv? Derivation of the Factor 2 This gives, 2 22 21 2 rel v ??vv vv?? 11 22 2 2vv v ? ? vv? 21 ??? Again, random direction assumption implies these two averages to be the same ! these averages to be the Taking the square root of the above equation, ?? 22 22 rel rel rms rms vvvv? ?? 2 4dN 2 4 2dN dnrN dt V v? ? dnrN dt V v? ?should be written as Thus, back v rms and ? Example: 18.6 & 18.8 18.6: a) What is the average translational KE of an ideal gas molecule at 27 o C? Kinetic Theory ? ? ? ? ? 23 21 33 138 10 / 27315 27 621 10KE kT J K K J ? ? ?? ? ? ?? 1.38 10 73.15 27 .21 10 22 per molecule b) What is the root-mean-square speed of O 2 at this T? rms v 2 23 3 3(1.38 10 / )(300 ) 484 rms O kT v m JK K ms ? ? ? 2 26 5.31 10 O mkg ? ?? (molecular mass) 26 5.31 10 kg ? ?? ? 18.8: a) Evaluate the mean free path ??of an air molecule at 27 o C. ? ?? ? 23 D perfume ???? 8 2 2 10 5 1.38 10 / 300 5.8 10 42 422.010 1.0110 JK K kT m rp mPa ? ? ? ? ? ? ? ?? ?? ?? ? ? 10 210rm ? ?? Heat Capacities of Gases (at constant V) Using the Kinetic-Molecular model, one can calculate heat capacity for an Ideal Gas! ? For point-like molecules (monoatomic gases), molecular energy consists only of translational kinetic energy K tr ? We just learned that K tr is directly proportional to T. ? Wh ifiit i l t fh tdQWhen an infinites ma amoun o ea enters the gas, dT increases, and dK tr increases accordingly, 33 () 22 tr dK NkdT or nRdT?? Heat Capacities of Gases F d fi iti f l h t? rom e n on o mo ar ea capacity, we also know: dQ nC dT? ? From energy conservation, requiring dQ=dK (dW=0) gives v tr , 3 2 v nC dT nRdT? ? Monoatomic ~ Ideal Gas ( 12.47 / ) 3 2 v J mol KCR? ?? (matches well with prediction) Heat Capacity (diatomic) A diatomic molecule can absorb energy in its translation, and also in its rotation and in the vibrations of its molecular structure. Equipartition of Energy Thi i i l t t th t h dffd(? ts pr nc p e s a es a eac degree of ree om separa e mechanisms in storing energy?) will contribute (½ kT) to the total average energy per molecule. ? Monoatomic: 3 translational dofs ? 3 (½ kT) This give K =3/2NkT (same as before) tot = 3/2 as before). ? Diatomic (without vibration): 3 trans dofs + 2 rotational dofs This give K tot = 5/2 NkT or = 5/2 nRT. Again, consider an infinitesimal energy change, we have and this gives C 5/2 R 5 CdT RdT , this gives v = . 2 v n nRdT? Heat Capacities (real gases) ? At low T, only the 3 translational dofs are activated ? At higher T, additional rotational dofs are are activated ? At higher T still , vibrational dofs might also get activated Heat capacity for a H 2 gas Heat Capacities of Ideal Solids At t d? oms are connec e together by springs ? Assume harmonic motions for these springs (Hooke?s law) ? For each spatial direction, direction, there are two dofs (vibrational KE, vibrational PE) and we have 3 dims and dims ? For the entire solid, we have 3/2 kT (KE) + 3/2 kT (PE) ? E = 3kT tot C v = 3R Dulong Petit Prediction Phases of Matter (reading phase diagrams) Administrator Microsoft PowerPoint - ch18_young_freedman.ppt [Compatibility Mode]

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