Collision cross-section is an area of an imaginary sphere of radius $σ$ around the molecule within which the centre of another molecule cannot penetrate.
The volume swept by a single molecule in unit time is

$V = (π σ^{2}) \overline{u}$ where $\overline{u}$ is the average speed
If $N^{}$ is the number of molecules per unit volume, then the number of molecules within the volume V is
$N = V N^{} = (π σ^{2} \overline{u}) N^{}$
Hence, the number of collision made by a single molecule in unit time will be
$Z = N = (π σ^{2} \overline{u}) N^{}$
In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is $2 \overline{u}$ as shown below.

Number of collision made by a single molecule with other molecule per unit time is given by
$Z_{1} = π σ^{2} (\overline{u}_{rel}) N^{} = 2 π σ^{2} \overline{2} N^{}$
$T h e \to t a l ν m b e r o f b i m o \leq c \underline{a} r c o l l i s i o n s$ Z_(1) $p e r u n i t v o l u m e p e r u n i t t i m e i s g i v e n b y$ Z_(1)=(1)/(2)(Z_(1)N^())or Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^())N^()=(1)/(sqrt(2))pisigma^(2)overline(u)N^(2) $I f t h e c o l l s i o n \in v o l v e t w o u n l i k e m o \leq c \underline{e} s t h e n t h e ν m b e r o f c o l l i s i o n s$ Z_(12) $p e r u n i t v o l u m e p e r u n i t t i m e i s g i v e n a s$ Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2) $w h e r e$ N_(1) $and$ N_(2) $a r e t h e ν m b e r o f m o \leq c \underline{e} s p e r u n i t v o l u m e o f t h e t w o t y p e s o f m o \leq c \underline{e} s,$ sigma_(12) $i s t h e a v e r a \geq d i a m e t e r o f t h e t w o m o \leq c \underline{e} s and$ mu $i s t h e r e d u c e d m a s s . T h e m e a n e e p a t h i s t h e a v e r a \geq d i s tan c e t r a v e l \leq d b y a m o \leq c \underline{e} b e t w e e n t w o ≻ e s s i v e c o l l i s i o n s . W e c a n exp r e s s i t a s f o l l o w s :$ lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1))or" "lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^())implies(1)/(sqrt(2)pisigma^(2)overline(u)N^()) $T h r e e i d e a l g a s s a m p \leq s \in s e p a r a t e e q u a l v o l u m e c o n t a \in e r s a r e t a k e n and f o l l o w \in g d a t a i s g i v e n :$ $Gas A Gas B Gas C Pressure 1 a t m 2 a t m 4 a t m Temperature 1600 K 200 K 400 K Mean free paths 0.16 n m 0.16 n m 0.04 n m Mol.wt. 204080$
Calculate number of collision by one molecule per sec $(Z_{1})$ .

Question

Collision cross-section is an area of an imaginary sphere of radius

σ

around the molecule within which the centre of another molecule cannot penetrate.
The volume swept by a single molecule in unit time is

V = (π σ^{2}) \overline{u}

where

\overline{u}

is the average speed
If

N^{*}

is the number of molecules per unit volume, then the number of molecules within the volume V is

N = V N^{*} = (π σ^{2} \overline{u}) N^{*}

Hence, the number of collision made by a single molecule in unit time will be

Z = N = (π σ^{2} \overline{u}) N^{*}

In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is

2 \overline{u}

as shown below.

Number of collision made by a single molecule with other molecule per unit time is given by

Z_{1} = π σ^{2} (\overline{u}_{rel}) N^{*} = 2 π σ^{2} \overline{2} N^{*}

T h e \to t a l ν m b e r o f b i m o \leq c \underline{a} r c o l l i s i o n s

Z_(1)

p e r u n i t v o l u m e p e r u n i t t i m e i s g i v e n b y

Z_(1)=(1)/(2)(Z_(1)N^(**))or Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^(**))N^(**)=(1)/(sqrt(2))pisigma^(2)overline(u)N^(**2)

I f t h e c o l l s i o n \in v o l v e t w o u n l i k e m o \leq c \underline{e} s t h e n t h e ν m b e r o f c o l l i s i o n s

Z_(12)

p e r u n i t v o l u m e p e r u n i t t i m e i s g i v e n a s

Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2)

w h e r e

N_(1)

and

N_(2)

a r e t h e ν m b e r o f m o \leq c \underline{e} s p e r u n i t v o l u m e o f t h e t w o t y p e s o f m o \leq c \underline{e} s,

sigma_(12)

i s t h e a v e r a \geq d i a m e t e r o f t h e t w o m o \leq c \underline{e} s and

mu

i s t h e r e d u c e d m a s s . T h e m e a n e e p a t h i s t h e a v e r a \geq d i s tan c e t r a v e l \leq d b y a m o \leq c \underline{e} b e t w e e n t w o ≻ e s s i v e c o l l i s i o n s . W e c a n exp r e s s i t a s f o l l o w s :

lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1))

or" "lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^(**))implies(1)/(sqrt(2)pisigma^(2)overline(u)N^(**))

T h r e e i d e a l g a s s a m p \leq s \in s e p a r a t e e q u a l v o l u m e c o n t a \in e r s a r e t a k e n and f o l l o w \in g d a t a i s g i v e n :

Gas A Gas B Gas C Pressure 1 a t m 2 a t m 4 a t m Temperature 1600 K 200 K 400 K Mean free paths 0.16 n m 0.16 n m 0.04 n m Mol.wt. 204080

Calculate number of collision by one molecule per sec

(Z_{1})

.

Filo · Accepted Answer

Solution:

Question Text	Collision cross-section is an area of an imaginary sphere of radius $σ$ around the molecule within which the centre of another molecule cannot penetrate. The volume swept by a single molecule in unit time is $V = (π σ^{2}) \overline{u}$ where $\overline{u}$ is the average speed If $N^{}$ is the number of molecules per unit volume, then the number of molecules within the volume V is $N = V N^{} = (π σ^{2} \overline{u}) N^{}$ Hence, the number of collision made by a single molecule in unit time will be $Z = N = (π σ^{2} \overline{u}) N^{}$ In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is $2 \overline{u}$ as shown below. Number of collision made by a single molecule with other molecule per unit time is given by $Z_{1} = π σ^{2} (\overline{u}_{rel}) N^{} = 2 π σ^{2} \overline{2} N^{}$ $T h e \to t a l ν m b e r o f b i m o \leq c \underline{a} r c o l l i s i o n s$ Z_(1) $p e r u n i t v o l u m e p e r u n i t t i m e i s g i v e n b y$ Z_(1)=(1)/(2)(Z_(1)N^())or Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^())N^()=(1)/(sqrt(2))pisigma^(2)overline(u)N^(2) $I f t h e c o l l s i o n \in v o l v e t w o u n l i k e m o \leq c \underline{e} s t h e n t h e ν m b e r o f c o l l i s i o n s$ Z_(12) $p e r u n i t v o l u m e p e r u n i t t i m e i s g i v e n a s$ Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2) $w h e r e$ N_(1) $and$ N_(2) $a r e t h e ν m b e r o f m o \leq c \underline{e} s p e r u n i t v o l u m e o f t h e t w o t y p e s o f m o \leq c \underline{e} s,$ sigma_(12) $i s t h e a v e r a \geq d i a m e t e r o f t h e t w o m o \leq c \underline{e} s and$ mu $i s t h e r e d u c e d m a s s . T h e m e a n e e p a t h i s t h e a v e r a \geq d i s tan c e t r a v e l \leq d b y a m o \leq c \underline{e} b e t w e e n t w o ≻ e s s i v e c o l l i s i o n s . W e c a n exp r e s s i t a s f o l l o w s :$ lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1))or" "lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^())implies(1)/(sqrt(2)pisigma^(2)overline(u)N^()) $T h r e e i d e a l g a s s a m p \leq s \in s e p a r a t e e q u a l v o l u m e c o n t a \in e r s a r e t a k e n and f o l l o w \in g d a t a i s g i v e n :$ $Gas A Gas B Gas C Pressure 1 a t m 2 a t m 4 a t m Temperature 1600 K 200 K 400 K Mean free paths 0.16 n m 0.16 n m 0.04 n m Mol.wt. 204080$ Calculate number of collision by one molecule per sec $(Z_{1})$ .
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