Class 11

Chemistry

Physical Chemistry

States of Matter

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

$V=(πσ_{2})u$ where $u$ is the average speed

If $N_{∗}$ is the number of molecules per unit volume, then the number of molecules within the volume V is

$N=VN_{∗}=(πσ_{2}u)N_{∗}$

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

$Z=N=(πσ_{2}u)N_{∗}$

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 $2 u$ as shown below.

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

$Z_{1}=πσ_{2}(u_{rel})N_{∗}=2 πσ_{2}2N_{∗}$

$The→talνmberofbimo≤ca rcollisions$Z_(1)$perunitvolumeperunittimeisgivenby$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)$Ifthecollsion∈volvetwounlikemo≤ce sthentheνmberofcollisions$Z_(12)$perunitvolumeperunittimeisgivenas$Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2)$where$N_(1)$and$N_(2)$aretheνmberofmo≤ce sperunitvolumeofthetwotypesofmo≤ce s,$sigma_(12)$istheavera≥diameterofthetwomo≤ce sand$mu$isthereducedmass.Themeaneepathistheavera≥distancetravel≤dbyamo≤ce betweentwo≻essivecollisions.Wecanexpressitasfollows:$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^(**))$Threeidealgassamp≤s∈separateequalvolumeconta∈ersaretakenandfollow∈gdataisgiven:$$Gas AGas BGas C Pressure1atm2atm4atm Temperature1600K200K400K Mean free paths0.16nm0.16nm0.04nm Mol.wt.204080 $

Calculate number of collision by one molecule per sec $(Z_{1})$.

- $4:1:4$
- $1:4:4$
- $4:3:2$
- $1:2:4$