) 1/2 ( J / K 298K. CHEM 4641 Physical Chemistry I Chapter 16 (long-book chapter 33) Recommended Problems

Transcription

1 CHEM 4641 hysical Cheistry I Chapter 16 (long-book chapter 33) Recoended robles Chapter 16 (33) roble 1 Maxwell speed distribution in two diensions The velocity distribution is siply the Maxwell-Boltzann distribution, N e -E/(kT), where E (1/)v. This ust be noralized in two diensions v x v y N e + 1 N [ e v x kt dv x] v x (v x +v y ) kt dv y dv x The integral is in Appendix B. 1 N [ ( kt π 4 ) 1/ ] ( π kt ) so N π kt To pass fro the velocity distribution to the speed distribution change to polar coordinates (v,θ), then integrate out the direction angle. Note that dv x dv y v dθ. π (v) (v x,v y )v d θ θ (v) v kt v e kt v π kt v e kt π dθ Chapter 16 (33) roble 5 Average speed and kinetic energy of O and CCl 4 at 98K v 8kT E π k 1 v 3RT 3 RT O and CCl 4 have the sae average kinetic energy. Their average speeds differ. v O ( J / K 98K π kg ) 1/ /s v CCl4 v O ( O CCl4 ) 1/ 444/ s /s

2 CHEM 4641 hysical Cheistry I Chapter 16 (33) Recoended robles, continued Chapter 16 (33) roble 7 Travel ties of N and H at 98 K and 1 at, oving 1 in any direction A siple approxiate way to solve this proble is to change the question slightly. On ay ask how long a olecule, if traveling at its average speed, takes to ove 1. Then t R / <v>, where R is the distance traveled. (R 1 ) a. hydrogen to travel 1, t 1. / <v> 1. X [π / (8 k T)] 1/. au X 1.665X1-7 kg/au 3.347X1-7 kg t.565 s b. nitrogen 8. au X 1.665X1-7 kg/au 4.653X1-6 kg t.565 X (8. /.) 1/.1 s nitrogen requires an additional.15 s for an average displaceent of 1 c. The fraction of nitrogen olecules requiring ore tie will siply be the fraction of nitrogen olecules traveling at speed less than the average speed. That is, (v< ) (v< ) 4 π ( (v)d v π k T ) 3/ v e v /( k T ) d v Let u v /( kt ). uv /( k T ) The upper liit becoes /(k T ) 8 k T /(π ) /( kt ) / π. In the integrand, v ( k T /)u and d v k T / du (v< ) 4 π ( π k T ) 3/ kt ( k T ) 1/ / π u e u d u (v< ) π 4 / π u e u d u Nuerical integration gives.533. That is the fraction of olecules having speeds less than the average speed.

3 CHEM 4641 hysical Cheistry I Chapter 16 (33) Recoended robles, continued Chapter 16 (33) roble 7, continued The question as asked, average tie for a particular distance, requires the distribution of olecular ties. The relationship between t and v is siple: t R/v, where R1. is the distance traveled. <t> R <1/v> (which will not equal R/<v>) One can calculate t fro 1 /v 1 t R v (v)d v t R 4 π ( π kt ) 3/ 1 v v e v /( kt ) d v t R 4 π ( π kt ) 3/ v e v /( kt ) d v Let u v /( k T ) Then vdv(1/)d(v )( kt /)u du t R 4 π ( π kt ) 3/ k T e u d u The integral e u du π 4 is given in Appendix A. t R 4 π ( kt ) 1/ 1 π π 3 / 4 R π 4 ( π 8 k T ) 1/ π 4 R Average tie is longer than that calculated by the approxiate ethod. The fraction of longer-tie olecules can be calculated using the sae integral as in the approxiate solution, except that the upper liit is R/<t> instead of. Chapter 16 (33) roble 11 robabilities of velocity ranges in one diension Given: ( v x v o ) erf( v o /(kt)) where erf ( z) π z e x d x. Let v k T ( v x kt ) erf (1).84 ( v x k T ) 1 ( v x k T ) 1 erf (1)

4 CHEM 4641 hysical Cheistry I Chapter 16 (33) Recoended robles, continued Chapter 16 (33) roble 1 speed of sound at T1K. v sound ( γ RT /M) 1/ where γ C /C V C /(C -R) For a onatoic ideal gas C (5/) R.79 J/(ol.K) so γ 5 / a) gas M (kg/ol) v sound (/s) Ne Ar Kr b) v sound (Ar, 1K) v sound (Kr, T) 1 / T / 83.8 T 1K X (83.8 / 39.95) 98 K Chapter 16 roble 19 Translational kinetic energy distribution for an ideal gas (v)4 π ( π k T ) 3 / v e v /( k T ) (E tr ) (v) d v d E tr E tr v v E 1/ tr d v 1 / ( E d E tr ) tr (E tr ) 4 π ( π kt ) 3/ E tr e E tr /(kt ) 1 / ( E tr ) (E tr ) π (π k T ) 3/ E 1/ E tr e tr /(kt ) Chapter 16 (33) roble 3. Collision rate with walls of a cube Wall area 6 sides X area of one side 6 c 6X1-4. Collisions per area per second ( π kt ) 1/ 1135 a (π au kg/au J / K 98 K ) 1/ s 1 Collisions/second X area 1.5X1 4 /s

5 CHEM 4641 hysical Cheistry I Chapter 16 (33) Recoended robles, continued Chapter 16 (33) roble 6 rate of collisions of oxygen with a surface kt. kg 3.g/ol au kg 1135 a (a) Z c π kg J /K 98K Z s c s 1 c ultiply by 6 to get collisions per inute: c in 1 (b) 1 1 Torr 1 1 ( ) a a At that pressure, c s J / J s / Chapter 16 (33) roble 8 collision frequency and ean free path in nitrogen N gas. 8. au kg. (a) 1 at. T 98K. The single-particle collision frequency kt σ v. v 8RT 475 /s π M σ.43 n (Table 16.1 or 33.1) kt 1135 a J / K 98K ( ) ( ) (475 /s) s 1 The answer does not disagree with 1 1 reported in the CRC Handbook. (b) It is convenient to note that collision frequency is proportional to pressure and inversely proportional to the square root of teperature s s 1 98K 1135 K a.38 at (c) At TK, v 475 / s 48 /s 98 λ v 48 /s n 1 s

6 CHEM 4641 hysical Cheistry I Chapter 16 (33) Recoended robles, continued Chapter 16 (33) roble 3 Collision frequency of carbon dioxide a) 1at. T98K. Fro Table 16.1 (or 33.1) in this chapter, σ.5 n 5.X1-19. The total collision frequency (1/) (N/V) σ <v rel > -1/ (/kt) σ [8RT/(πM)] 1/ /(kt) 1135 a / ( 1.386X1-3 J/K X 98K).463X1 5-3 [8RT/(πM)] 1/ [ 8X J/(ol K) X 98K / (π X 44.1X1-3 kg/ol)] 1/ [ (/s) ] 1/ 379 /s.771 X (.463X1 5-3 ) ( 5.X1-19 ) X 379 /s 8.45X s -1 b) At what T does.1 X 8.45X s -1? Suppose pressure is unchanged. is proportional to T -3/ at constant pressure. 8.45X s -1 X (98) 3/.1 X 8.45X s -1 X (T) 3/ 1 X (98) 3/ T 3/ T 1 /3 X 98K 138 K Chapter 16 (33) roble 3 ean free path of Argon at 98K and various pressures. σ.36 n. λ kt σ J /K 98 K 8.8 X 1 3 a a).5 at. λ 8.8X1-3 a /(.5 X 1135 a) 1.6X n b).5 at. λ 8.8X1-3 a /(.5 X 1135 a) 1.6X μ c) 5X1-6 at. λ 8.8X1-3 a /(5X1-6 X 1135 a) Chapter 16 (33) roble 33 ean free path of Ne, Kr and CH 4 at 5K and 1 at. λ kt σ J / K 5 K σ 1135 J X 1 6 σ a) Ne. σ.4 X λ.1 X 1-7 b) Kr. λ Kr λ Ne ( σ Ne /σ Kr ).X1-7 (.4 /.5) 9.3X1-8 c) CH 4. λ CH4 λ Ne ( σ Ne /σ CH4 ).X1-7 (.4 /.48) 1.1X1-7

7 CHEM 4641 hysical Cheistry I Chapter 16 (33) Recoended robles, continued Chapter 16 (33) roble 34. ressure required to obtain a desired ean free path. The basic equation is λ kt σ. It is easily rearranged to give as a function of T. k T λ σ The collision cross section σ.4 n 4.X1-19. λ. c.. k 1.386X1-3 J/K. T 5. K..61 a 4.6X1-4 torr Chapter 16 (33) roble 35 Copare v p,, and v rs for the 1-d Maxwell speed distribution The one-diensional speed distribution is (v) ( π k T ) 1/ e v /(kt ), where v is the speed and runs fro fro to. Noralization is (v)d v 1. Most probable speed v p, because (v) is greatest at v. Average speed v (v)d v ( πk T ) 1/ v e v /(k T ) d v Let u v /(kt). ( π k T ) 1/ k T e u d u The integral equals 1. k T π Root-ean-square speed v rs v rs ( v rs π k T ) 1/ ( π k T ) 1/ k T 4 v rs k T v (v)d v. v e v /(kt ) d v The integral is in Appendix A. π kt k T Just as in three diensions, v p < < v rs.