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PPT – Collision Theory PowerPoint presentation | free to download - id: 51cd1-YTM4Z

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Collision Theory Molecules and atoms exchange

energy, change direction, react, etc. through

collisions. The collision number or frequency

which is the number of collisions per unit volume

per second is thus an important

quantity. Consider a molecule A whizzing (keep it

clean) through space. Along its line of travel

molecule A will collide with another molecule B

(which can be the same as A), if at closest

approach the distance between A and B would be

less than the sum of their respective radii, rA

rB

If A is traveling with a speed cA, then in one

second A would collide with any stationary

molecule B whose center is within a cylinder of

volume collision volume p ( rA rB )2 cA

(1 sec) Since the B molecules are not stationary,

but are also moving, the appropriate speed to use

is the relative mean speed of A and B molecules

as given by the expression ltcABgt 8 k T / p

u 1/2 where u is the reduced mass, defined

by u mA mB / (mA mB)

Multiplying this collision volume by the number

density of B molecules will give the total number

of collisions between a single A molecule and all

the B molecules present in the volume collision

volume (NB / V) p ( rA rB )2 ltcABgt (1

sec) (NB / V) Dividing this expression by the

volume and by the time interval of 1 second

will give zAB, the collision number for a single

A molecule, which is the total number of

collisions per unit volume per second between a

single A molecule and all the B molecules.

zAB collision volume (NB / V) (1 / V) (1 / 1

sec) p ( rA rB )2 ltcABgt (NB / V) (1 /

V) If the collision number for a single A

molecule is multiplied by all the A molecules

present we get the ZAB, the collision number for

A molecules colliding with B molecules, which is

the total number of collisions per unit volume

per second that all A molecules make with all B

molecules ZAB zAB NA p ( rA rB

)2 ltcABgt (NB / V) (NA / V) No2 p ( rA rB

)2 ltcABgt A B No2 p ( rA rB )2 8 k T

/ p u 1/2 A B Can you justify the steps in

this derivation? Note that in the above

expressions the lower case z in zAB implies the

number of collisions a single A molecule makes

with all B molecules, while the upper case Z in

ZAB implies the number of collisions all A

molecules makes with all B molecules, per unit

volume per second in both cases.

These expressions can be reduced even further, if

only A molecules are present. For zA, the

collision number for a single A molecule, which

is the total number of collisions per unit volume

per second between a single A molecule and all

the other A molecules we have zA p

(rArA)2 (8kT/(p (mAmA / (mAmA))))1/2 (NA/V)

(1/V) 25/2 p rA2 ( 8 k T / (p mA))1/2 (NA /

V) (1 / V) 25/2 p rA2 ltcAgt (NA / V) (1 /

V) In developing a similar expression for ZAA,

the collision number for A molecules colliding

with A molecules, which is the total number of

collisions per unit volume per second that all A

molecules make with all other A molecules, we

have to divide by 2 to avoid counting the

collisions twice. Derive an expression for ZAA.

The mean free path, l, is defined as the average

distance a molecule travels between collisions

and is calculated for molecules of a given type,

e.g., A, by dividing the molecules mean speed,

ltcAgt, or the distance the molecule would on

average travel in 1 sec by the collision number

for a single molecule times the volume of the

container, zA V, or the number of collisions that

a single A molecule would make on average with

other A molecules in 1 sec lA ltcAgt

/ (zA V) ltcAgt / ( 25/2 p rA2 ltcAgt ( NA / V )

) k T / ( 25/2 p rA2 P ) Could you justify

the steps in this derivation? In a closed

constant volume container how does the mean free

path vary with temperature? What is the mean

free path of O2 (g) molecules at 25.0 oC and

0.200 atm? The van der Waals radius for O2 (g)

is 0.188 nm.

Another useful quantity is Zsurface, the

collision number for molecules with a surface,

which is equal to the number of collisions of

molecules with the surface per unit area per

second. Consider a molecule in a rectangular box

traveling toward the right wall of area A with

component of speed in the x direction of cx

If this molecule is within a distance of cx times

1 sec of the right wall it will strike the wall

within 1 sec. Since in the collection of

molecules there is a distribution of speeds, on

average all molecules with a component of

velocity in the plus x direction will strike the

wall within 1 sec if they are within the

volume A (1 sec) ltcxgt A (1 sec) 0 ? oo cx P

(cx) A (1 sec) 0 ? oocx (m / 2 p

k T)1/2 e - m cx2 / 2 k T dcx A (1

sec) ( k T / 2 p m )1/2 In this derivation P

(cx) is the one-dimensional distribution of

molecular speeds or the probability that a

molecule has a speed between cx and cx dcx.

If the concentration of molecules is N / V, then

the number of molecules colliding with the

surface per unit area per second is Zsurface

(N / V) ( k T / 2 p m )1/2 P / ( 2 p m k T

)1/2 What happens to Zsurface as the

temperature of a gas in a closed constant volume

container is increased?

Calculate the average amount of time in seconds

that a water molecule exists on the liquid

surface of water in equilibrium with water vapor

at 25.0 oC. At 25.0 oC the equilibrium vapor

pressure of water is 23.76 mm Hg. Assume that a

water molecule at the liquid surface occupies an

area of 0.1000 nm2. Further assume that every

molecule of water vapor that collides with the

liquid surface enters the liquid phase.

Kinetic molecular theory can be used to develop a

molecular interpretation of pressure. Starting

with the ideal gas law P n R T / V ( n /

V ) ( R T )1/2 ( R T )1/2 and substituting a

result for the average molar energy of a gas,

ltEgt 3 /2 R T, derived from the assumptions of

kinetic molecular theory we get P ( n / V )

( R T )1/2 (2/3)1/2 ltEgt1/2 Now writing the

average molar energy in terms of the average

kinetic energy per molecule, ltEgt1/2 lt No (m c2

/ 2) gt1/2 gives P ( n / V ) ( R T )1/2

(2/3)1/2 lt No (m c2 / 2) gt1/2 multiplying by

(m/m)1/2 yields P ( n / V ) ( No R T / 3

m)1/2 lt (mc)2 gt1/2 Since lt(mc)2gt1/2 is the

root mean squared momentum, ltp2gt1/2, of the gas

molecules, we can use n N / No and R No k and

recognize that the pressure is P

( (N / No) / V ) ( No (No k) T / 3 m)1/2

lt(p)2gt1/2 (2 p / 3)1/2 (N / V) (k T / (2

p m))1/2 lt(p)2gt1/2 (2 p / 3)1/2 Zsurface

lt(p)2gt1/2 proportional to the product of the

average number of collisions per second with a

unit area of the container surface and the root

mean squared momentum of those collisions.

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