topic 4 kft 131

5
16 TOPIC 4 - THE COLLISION THEORY Example of solved problems 1. For molecular oxygen at 25 o C, (a) calculate the collision frequency z 11 and (b) the collision density Z 11 at a pressure of 1 atm. Given: d oxy. = 0.361 nm. Answer: The collision diameter of O 2 is 0.361 nm or 0.361×10 -9 m (a) Collision frequency z 11 = (πd 2 ) × rel C ×ρ = (πd 2 ) ×<v 11 > ×ρ Collision cross-section= σ = πd 2 = π(0.361×10 -9 m) 2 = 4.094×10 -19 m 2 Relative mean speed, rel C = <v 11 > > < = 1 2 v <v 1 > 2 / 1 1 3 1 10 32 298 314 . 8 8 × × × × = - - - mol kg K s m kg π = 444 ms -1 <v 11 >= 2 2 1 >= < v (444 ms -1 ) = 627.9 ms -1 Molecule density = ρ = ) 298 )( 08206 . 0 ( ) 10 )( 10 022 . 6 )( 1 ( 1 1 3 3 1 23 K mol K L atm m L mol atm RT PN V N A - - - - × = = = 2.46×10 25 m -3 So z 11 = (πd 2 ) <v 11 > ρ=(4.094×10 -19 m 2 )( 627.9 ms -1 )( 2.46×10 25 m -3 ) = 6.33×10 9 s -1 (b) Collision density Z 11 = × = × 2 1 2 1 11 ρ z (πd 2 ) × rel C ×ρ = 7.78×10 34 m -3 s -1 = 1 23 1 3 3 1 3 10 022 . 6 ) 10 )( 1034 78 . 7 ( - - - - - × × mol L m s m = 1.29×10 8 mol L -1 s -1 2. Large vacuum chambers have been built for testing space vehicles at 10 -7 Pa. Calculate the number of molecular impacts per square meter of wall per second for molecular CO 2 (d = 0.40 nm) at 35 o C? Answer: Use formula of collision with wall, J N = 4 > < v ρ Calculate one item after another: i. Molecule density: 3 13 1 1 3 1 23 7 10 34 . 2 15 . 308 3145 . 8 10 02 . 6 10 - - - - - × = × × × = = = m K mol K m Pa mol Pa RT PN V N A ρ

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Page 1: Topic 4 kft 131

16

TOPIC 4 - THE COLLISION THEORY

Example of solved problems

1. For molecular oxygen at 25 oC, (a) calculate the collision frequency z11 and (b) the collision density Z11 at a pressure of 1 atm. Given: doxy. = 0.361 nm.

Answer:

The collision diameter of O2 is 0.361 nm or 0.361×10-9 m

(a) Collision frequency z11 = (πd2) × relC ×ρ = (πd2) ×<v11> ×ρ

Collision cross-section= σ = πd2 = π(0.361×10-9 m)2 = 4.094×10-19 m2

Relative mean speed, relC = <v11> ><= 12 v

<v1>2/1

13

1

1032

298314.88

××

××=

−−

molkg

Ksmkg

π= 444 ms-1

<v11>= 22 1 >=< v (444 ms-1) = 627.9 ms-1

Molecule density = ρ =)298)(08206.0(

)10)(10022.6)(1(11

33123

KmolKLatm

mLmolatm

RT

PN

V

N A

−−

−−×==

= 2.46×1025m-3

So z11 = (πd2) <v11> ρ=(4.094×10-19 m2)( 627.9 ms-1)( 2.46×1025m-3)

= 6.33×109

s-1

(b) Collision density Z11 = ×=×21

21

11 ρz (πd2) × relC ×ρ = 7.78×1034

m-3

s-1

= 123

13313

10022.6

)10)(103478.7(−

−−−−

×

×

mol

Lmsm= 1.29×10

8mol L

-1s

-1

2. Large vacuum chambers have been built for testing space vehicles at 10-7 Pa. Calculate the number of molecular impacts per square meter of wall per second for molecular CO2 (d = 0.40 nm) at 35 oC?

Answer:

Use formula of collision with wall, JN =4

>< vρ

Calculate one item after another:

i. Molecule density:

313113

1237

1034.215.3083145.8

1002.610 −

−−

−−

×=×

××=== m

KmolKmPa

molPa

RT

PN

V

N Aρ

Page 2: Topic 4 kft 131

17

ii. Mean speed :

1

2/1

13

1112/1

07.3851044

308314.888 −

−−

−−−

=

××

××=

=⟩⟨ ms

molkg

KmolKsmkg

M

RTv

ππ

iii. Put together:

4)07.385)(1034.2(

4

1313 −−×=

><==

msmvJZw N

ρ= 2.25×10

15m

-2s

-1

3. A 10 mL container with a hole1µm in diameter is filled with hydrogen. This container is placed in an evacuated chamber at 10 oC. How long will it take for 80% of the hydrogen to effuse out?

Answer:

hhhh13.1513.1513.1513.15minminminmin789789789789 ≈≈=

×××=

×

××

×××

××

×

==><

=

==><

=−

=><><

=−

=><

×=><

=−=

−−

−−

∫∫

st

tmsm

tm

m

tV

vA

N

N

NAdtV

vA

N

dN

vdtV

vA

N

dN

Vv

V

Nv

dt

dN

AJ

t

o

tNt

N

N

47355

88.173010963.16094.1

102

2833145.88

10104

)2101

(

20100

ln4

ln

moleculesofno.,area4

speedmean4

volume44

1

18

2/1

336

26

00

π

π

ρ

4. An effusion cell has a circular hole of diameter 2.50 mm. If the molar mass of

the solid in the cell is 260 g mol-1 and its vapor pressure is 0.835 Pa at 400 K, by how much will the mass of the solid decrease in a period of 2 h?

Answer:

( )

gggg0.01040.01040.01040.0104=×=

×××××

××××××××

=∆

∆=

∆=∆

=

=

><=

∆==

−−

−−−

−−−

kg

Pa

smkg

molkgsmkg

smmolkgPa

w

RTM

tPMA

RTM

tmAPNw

RTM

PN

M

RT

RT

PNv

tmA

wZwJ

ooA

AA

o

N

4

21

2/11312

2413

2/12/1

2/1

2/1

10041.1

)102604003145.82(

60602)1050.22

1(10260835.0

2)2(

)2(

8

4

1

4

π

π

ππ

ππ

ρ

Page 3: Topic 4 kft 131

18

Formula for collisions Same molecule Different molecules 1 . Collision frequency, or

collision per second, or collision of a single molecule.

z11 = (πd2) × relC ×ρ

z12 = 2)12(2

12 )( ρπ ×× relCd

2. Collision density or total number of coll. or -coll. per volume per second.

Z11 = 1121

z ρ

×=21πd

2× relC ×ρ2

Z12 = 21)12(2

12 )( ρρπ ×× relCd

3. Collision flux or collision with wall, or with surface, or coll. through a hole, or coll. per area s-1 or rate of effusion.

Zw=JN =4

><=

v

dt

dN w ρ=

tMA

wN

o

A

Exersice 4a

1. What is the mean speed of nitrogen molecule relative to another nitrogen molecule at 290 K?(662 m s-1)

2. What is the average relative speed of hydrogen molecules with respect to

oxygen molecules at 300 K? (1837 m s-1) 3. Calculate the average number of molecules per cubic centimeter for

nitrogen gas at 1 bar pressure and 25 oC? (2.43×1019) 4. The pressure in interplanetary space is estimated to be of the order of 10-14

Pa. Calculate the collision frequency for an atom present in the space. Assume that only hydrogen atoms (d = 0.2 nm) are present and that the temperature is 1000 K and.(5.9×10-10s-1)

5. How many collisions does a single Ar atom (d = 0.34 nm) make in 1.0 s when the temperature is 25 oC and the pressure is 1.0 µatm? Given: Ar: MR = 39.95 (4.98×103s-1)

6. How many collisions per second does an N2 (d = 0.43 nm) molecule make

at an altitude of 15 km where the temperature is 217 K and the pressure 0.050 kPa. (4.1×106 s-1)

7. Calculate the collision frequency for hydrogen molecule in hydrogen gas

at 1.0 bar pressure and 20 oC? Given: dH2 = 2.47 Ǻ. (1.2×107s-1 )

Page 4: Topic 4 kft 131

19

8.

Calculate the collision density for molecular chlorine at 35 oC and 770 mmHg. The collision diameter is 5.44 ×10-6cm. (1.14×1035 s-1m3)

9. What is the average time between collisions of an oxygen molecule in

oxygen gas at 10-12 bar and 50 oC (d = 3.60 Ǻ)? 10. Calculate the mean free path for chlorine gas (d= 0.544 nm) at 0.1 Pa and

25 oC? (0.0312 m)

11. How many molecules of O2 strike the wall per unit area per unit time at 600 mmHg at 25 oC? (dO2 =3.60Å )( 2.1×1027 m-2s-1)

12. Large vacuum chambers have been built for testing space vehicles at 10-6 Pa.

Calculate the number of molecules of nitrogen (d = 0.375 nm) impact per square meter of wall per second at 28 oC. (2.6×1021 m-2s-1)

13. A solid surface with dimensions 2.5 mm×3.0 mm is exposed to He gas at 90 Pa

and 500 K. How many collision do the He atoms make with this surface in 15s?(He: MR= 4.00) (5.9 ×1020 m-2s-1)

Exercise 4b 1. A box contains H2 (molecule b) and He (molecule c) at a total pressure of 1.4

atm. If the mixture contains 18% H2 by weight, calculate the ratio of zb(b)/zc(c)? Given db =0.247 nm and dc= 2.20 nm.(6.97×10-3)

2. A gas mixture contains H2 at 2/3 atm and O2 at 1/3 atm at 27 oC. Calculate the

number of collision per second of H2 with other H2, O2 with other O2, and H2 with O2 (dH2 = 0.272 nm, dO2 = 0.361 nm) (9.5×109 s-1, 2.1×109 s-1 , 4.7×109) .

3. At 30 km above the Earth’s surface (roughly in the middle of the stratosphere) , the temperature is roughly 1000 K and the gas density is 3.74×1023 molecules/m3. Assuming N2 (d = 0.38 nm) is representative of the stratosphere, determine z11; undergoes in this region of the stratosphere in 1 s. (2.09×108).

4. An equal number of moles of H2 (d1 =0.272 nm) and Cl2 (d2=0.544 nm) are mixed and held at 298 K and the total pressure of 1 kPa. (a) Calculate the collision frequencies z12 and z21, where hydrogen is component 1 and chlorine is component 2. (b) Calculate the collision density, Z12. (1.14×108s-1, 1.14×108s-1, 2.30×104 mol L-1s-1)

5. For O2 (d = 0.361×10-9m) at 10-3 bar at 25 oC, (a) What is the collision

density Z11 ? (b) What is the average time between collisions of a single molecule?(1.26×102 mol L-1s-1 ,1.6×10-7 s).

Page 5: Topic 4 kft 131

20

6. The average surface temperature of Mars is 220 K, and the surface pressure is 4.7 torr. The Martian atmosphere is mainly CO2 and N2 with smaller amounts of Ar, O2, CO, H2O, and Ne. Considering only the two main components, we approximate the Martian atmospheric composition as xCO2= 0.97 and xN2= 0.03. The collision diameters are dCO2=4.6 Ǻ and dN2= 3.7 Ǻ. For gas at 220 K at the Martian surface, calculate (a) the collision rate for one particular CO2 molecule with other CO2 molecules; (b) the collision rate for one particular N2 molecule with CO2 molecules;(c) the number of collisions per second made by one particular N2 molecule;(d) the number of CO2-N2 collisions per second in 1.0 cm3.

7. A container of volume 1×10-5 cm3 holds three molecules of gas b, which we label b1, b2, and b3. In 1 s, there are two b1-b2 collisions, two b1-b3 collisions, and two b2-b3 collsions. Find zb(b) and Zbb. The vapor pressure of water at 25 oC is 3160 Pa. If every water molecule that strikes the surface of liquid water sticks, what is the rate of evaporation of molecules from a square meter of surface? (1.2×1026 m-2s-1)

8. A 5- mL container with a hole10µm in diameter is filled with hydrogen. This container is placed in an evacuated chamber at 0 oC. How long will it take for 90% of the hydrogen to effuse out? (347 min)

9. Exactly 1 dm3 of nitrogen, under a pressure of 1 bar, will it takes 5.80 minutes to effuse through an orifice. How long will it take for Helium to effuse under the same conditions? (2.19 min)

10. The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At 25 oC and assuming that air consists of N2 molecules with a collision diameter of 395 pm, calculate the mean free path of the gas. (4.46×104 m)

11. Dry air had a CO2 mole fraction of 0.004. Calculate the total mass of CO2 that strikes 0.5 cm2 of one side of a green leaf in 10 s in dry air at 39 o C and 1 atm? (0.332 g )

12. A certain sample of a pure oxygen ( d = 3.61 Ǻ) has mean speed, <v> = 450 m s-1 and the average time between two successive collisions of a given molecule with other molecules is 4.0×10-10 s. Find the mean free path and molecule density in this gas. (1.8×10-7 m, 9.6×1024 m-3)