Prakash Gupta Heat and Thermodynamics

Transport Phenomenon

1 Deﬁne Collision Cross-Section and obtain a relation between the collision

frequency and mean free path of gas.

Let us consider a molecule moving in an assembly of other molecules.

Let the diameter of each molecule be σ. Thenthe moving molecule

will suﬀer collision with all these molecules whose centres lie in

a cylinder of cross-section πσ2. This area is called the Collision

Cross-section. It is the eﬀective area which determines the proba-

bility that a molecule of diameter will collide an another molecule

of the same diameter.

Prakash Gupta Heat and Thermodynamics

1.1 Mean free path

Molecules move in a straight line with constant speeds between

two successive collisions. Thus, the path of a single molecule is

a series of short zigzag path of diﬀerent lengths. These paths of

diﬀerent lengths are called the free paths of the molecules and

their mean is called mean free path.

The mean free path is deﬁned as the average distance transversed

by a molecule between two successive collisions.

If λ1, λ2, λ3, .. are the successive paths traversed in the total time

t, then we must have

λ1+λ2+λ3+...... = avg. speed ×t

Prakash Gupta Heat and Thermodynamics

If Nbe the total no. of collisions

∴Mean free path = λ1+λ2+λ3+...... +λN

=avg. speed ×t

1.2 Relation between Collision frequency & mean free path

Let us consider a gas containing n molecules per unit volume.

Consider one molecule in motion while all other molecules are at

rest. If σbe the diameter of each molecules, then the moving

molecule will collide with all those molecules, if their centre lie

within a distance σfrom the centre of moving molecules.

Prakash Gupta Heat and Thermodynamics

If v be the ve-

locity of moving molecule, the no. of molecular collision per

second is

N= (πσ2v)n

Prakash Gupta Heat and Thermodynamics

where nis the no. of molecules per unit volume.

∴Mean free path = distance ravelled in 1 second

total no. collision in 1 second

πσ2vn

∴λ=1

πσ2n

Again, no. of collisions in one second is πσ2vn. Therefore average

time between two successive collision is

τ=1

πσ2vn

Prakash Gupta Heat and Thermodynamics

So, the collision frequency is

f=1

τ=πσ2vn

or, f=1

λv

∴f=v

This is the required relation between collision frequency and mean

free path of a gas.

Prakash Gupta Heat and Thermodynamics

2 Q. How is mean free path of a gas related with absolute temperature

and pressure. Explain

We have, mean free path given by

λ=1

πσ2n(1)

where σis the diameter of molecule and nis the no. of molecules

per unit volume.

For random collision, Maxwell applied distribution law of molec-

ular speed to get

λ=1

√2πσ2n(2)

Prakash Gupta Heat and Thermodynamics

If m be the mass of each molecule, then mn =ρ, where rho is

the density of the gas. Thus,

λ=m

√2(πσ2n)(3)

According to kinetic theory of gases, the pressure of the gas is

given by

P=1

3ρ¯

where ¯

v2is the mean squared speed of the gas molecules.

We have

Prakash Gupta Heat and Thermodynamics

v2=3kT

∴P=1

3ρ3kT

or, m

ρ=kT

substituting this value in equ (3), we get

λ=kT

√2(πσ2P)

Thus λ∝T

i.e., mean free path is directly proportional to the absolute tem-

perature of the gas.

Prakash Gupta Heat and Thermodynamics

3 Q. What is transport Phenomenon?

The equilibrium of any gas is the most probable stte but if the

gas is not in an equlibrium state, we may have any of the follow-

ing three cases:

1) The diﬀerent parts of the gas may have diﬀerent velocities.

So, there will be a relative motion of the layers of the gas with

respect to one another. This gives rise to the phenomenon of

viscosity.

2) The diﬀerent parts of gas may have diﬀerent temperature. So,

the molecules of the gas will carry KE from regions of higher

temperature to the regions of lower temperature to bring the

equlibrium state. This gives rise to the phenomenon of conduc-

tion.

3) The diﬀerent parts of gas may have diﬀerent molecular con-

Prakash Gupta Heat and Thermodynamics

centration i.e., the no. of molecules per unit volume. SO,the

molecules of the gas will carry the mass from regions of higher

concentration to those of lower concentration to bring the equi-

librium state. This gives rise to the phenomenon of diﬀusion.

These phenomena, i.e., viscosity, conduction and diﬀusion are

called transport phenomena.

4 Transport of Momentum Q. Derive an expression for coeﬃcient of vis-

cosity on the basis of kinetic theory of gases. How does the coeﬃcient of

viscosity of gas depend upon the pressure and temperature of the gas?

Let us consider a mass of gas moving in parallel layers between

two horizontal planes AB and CD. Let us suppose that the ve-

locity of the layer of gas in contact with the plane AB is zero

and increases as we pass towards the plane CD. Also, consider

Prakash Gupta Heat and Thermodynamics

an intermediate plane MN.

Let n be the no. of of molecules per unit volume and ¯

vtheir

average speed. Since the molecules are moving due to thermal

agitation in all direction, it may be supposed that one third of the

molecules are moving in each of the three directions parallel to

three co-ordinates axis. So that, on an average, one sixth of the

molecules move parallel to any one axis in particular direction.

Therefore, the

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no. of molecules crossing the plane MN upwards or downwards

per unit area per second = n¯

Let G be the momentum of each molecule in the plane MN and

dz be the momentum gradient in upward direction. Let each

plane AB and CD be at a distance λfrom MN where λis the

mean free path. Then momentum of each molecule at the plane

CD = G+λdG

and the momentum of each molecule at the lower plane AB =

G−λdG

Therefore, the net momentum transferred per unit area of the

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plane MN per second

=n¯

6G+λdG

dz −n¯

6G−λdG

dz 

3n¯

vλdG

3n¯

vλ d

dz(mu)

where G=mu, u is the velocity of molecules at plane MN.

3mn¯

vλdu

dz (1)

Prakash Gupta Heat and Thermodynamics

This gives tangential stress i.e., force per unit area.

∴Coeﬃcient of Viscosity,η=tangential stress

velocity gradient

3mn¯

vλdu

or, η=1

3mn¯

v=1

3ρ¯

vλ

But the mean free path, λ=1

√2πσ2n, σ is the diameter of

molecules.

∴η=m¯

3√2(πσ2)(2)

Prakash Gupta Heat and Thermodynamics

As the above expression does not contain n, coeﬃcient of viscosity

(η) is independent of pressure.

And ¯

v∝√T

∴η∝√T

ie., coeﬃcient of viscosity of a gas varies directly as the square

root of the absolute temperature of the gas.

5 Transport of Energy: Q. What is Thermal Conduction? Show that the

coeﬃcient of conductivity of a gas is independent of its pressure.

Let us consider that the mass of the gas is at rest. The layers

AB, MN and CD are such that the temperature increases as we

go from the plane AB to CD through MN.

Prakash Gupta Heat and Thermodynamics

Let E be en-

ergy of each molecule in the plane MN and fracdEdz be the

energy gradient in upward direction. Let each plane AB and CD

be at a distance λfrom MN where λis the mean free path.

Then, energy of each molecule at the plane CD = E+λdE

and the energy of each molecule at the plane AB = E−λdE

The no. of molecule crossing the plane MN upward or downwards

per unit area per second = n¯

Prakash Gupta Heat and Thermodynamics

Therefore, the net energy transferred across the unit area of the

plane MN per second

=n¯

6E+λdE

dz −n¯

6E−λdE

dz 

3n¯

vλdE

But dE =mcvdT

where cv= sp. heat capacity at constant volume.

3n¯

vλmcvdT

3mn¯

vλcvdT

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∴Coeﬃcient of Conductivity, K =Rate of heat transfered

temperature gradient

3mn¯

vλcvdT

∴K=1

3mn¯

vλcv=1

3ρ¯

vλcv

But

λ=1

√2πσ2n

∴K=m¯

vcv

3√2(πσ2)

Prakash Gupta Heat and Thermodynamics

This is the expression of coeﬃcient of thermal conductivity.

As the above expression doesnot contain n, coeﬃcient of thermal

conductivity is independent of pressure.

And ¯

v∝√T

∴K∝√T

i.e., coeﬃcient of thermal conductivity of a gas varies directly as

the square root of the absolute temperature of the gas.

6 Transport of Mass: Q. Find an expression for coeﬃcient of diﬀusion of

a gas and show that it is proportional to T3/2.

Let us consider a mass of gas moving in parallel layers between

the plane AB and CD. Let us suppose the concentration, i.e., the

no. of molecules per unit volume, increases in the vertical direc-

Prakash Gupta Heat and Thermodynamics

tion as we go from the plate AB to CD through an intermediate

plane MN.

Let n be the concentration at the plane MN and dη

dz is the con-

centration gradient in upward direction. Let each plane AB and

CD be at a distance λfrom MN where λis the mean free path.

Then, the concentration at the plane CD is n+λdn

Prakash Gupta Heat and Thermodynamics

and the concentration at the plane AB is n−λdn

The net no. of molecules crossing unit area of the plane MN

6¯

vn+λdn

dz −1

6¯

vn−λdn

dz 

3¯

vλdn

Prakash Gupta Heat and Thermodynamics

∴Coeﬃcient of Diﬀusion, D =no of molecules crossing per unit area per sec

concentration gradient

3¯

vλdn

∴K=1

3¯

vλ =1

ρ1

3ρ¯

vλ

=η

Prakash Gupta Heat and Thermodynamics

where η=1

3ρ¯

vλ

Also, D=1

3¯

vλ =1

3¯

√2πσ2n

∴D=¯

3√2(nπσ2)

Here, ¯

v∝√Tand 1

n∝T

∴D∝T3/2

Prakash Gupta Heat and Thermodynamics

7 What is Brownian Motion? Describe Einstein’s theory of Brownian Mo-

tion? OR Deduce an expression for the coeﬃcient of diﬀusion from

ramdom molecular motion

In a colloidal solution, if examined under an ultramicroscope,

the suspendended particles appear moving to and fro, rapidly

and continously in an entirly random way. This irregular motion

is called Brownian motion and the suspended particles are called

Brownian Particles.

The gas molecules are of the same nature as that of the Brownian

particles, the only diﬀerence being that the Brownian particles are

much larger. All the laws of kinetic theory of gases are applicable

tot he Brownian particles.

Prakash Gupta Heat and Thermodynamics

7.1 Einstein’s Theory of Brownian Motion

This theory explains the physical nature of the phenomenon.

Brownian particle due to their random motion, tend to diﬀuse

into the medium in the course of time. Einstein, therefore, related

the diﬀusion coeﬃcient to Brownian motion. Einstein calculated

the diﬀusion coeﬃcient in two ways. 1) From random molecular

motion:

Let us consider a cylinder with its axis parallel to x-axis and its

faces S1and S2separated by a distance 4. Let A be the cross-

sectional area of the cylinder and n1and n2denote the molecular

concentration at S1and S2respectively.

Prakash Gupta Heat and Thermodynamics

The no. of molecules crossing end face S2to the right in time t

is 1

2n14A.

Similarly, the no. of molecules crossing end face S1to the left in

time t is 1

2n24A.

Hence, the excess of the particle crossing a middle layer

2(n1−n2)4A

Prakash Gupta Heat and Thermodynamics

If −dn

dx is the concentration gradient, then according to the deﬁ-

nition of diﬀusion coeﬃcent, we have

2(n1−n2)4A=−Ddn

dxtA

But n1−n2=−4dn

∴1

2− 4dn

dx4A=−Ddn

dxtA

Hence, D=42

2) From diﬀerence in osmotic pressure:

Let the osmotic pressure acting on the end faces s1and s2of the

cylinder be p1and p2.

Prakash Gupta Heat and Thermodynamics

From the gas laws,

p1=n1kT

p2=n2kT

∴(p1−p2)A= (n1kT −n2kT )A

= (n1−n2)kT A

This force pushes the cylinder in +ve direction of x-axis.

no. of molecules contained in the cylinder = n4A

where n= mean concentration

Prakash Gupta Heat and Thermodynamics

Thus, force on a molecule is

f=n1−n2)kT A

n4A

=kT

nn1−n2

4

=kT

ndn

dx

∴f=−kT

ndn

dx

But this force is equal to voscous force = 6πηrv

6πηrv =−kT

ndn

dx

Prakash Gupta Heat and Thermodynamics

where r= radius of molecule and η= coeﬃcient of viscosity

According to deﬁnition of diﬀuse coeﬃcient

D=no. of molecules passing per unit area per sec

concentration gradient

=nv

−dn

=kT

6πηr

∴D=RT

6πηrNA

where NAis Avegadro’s No.

Prakash Gupta Heat and Thermodynamics

8 Calculate the mean free path of oxygen molecules at STP. (ρ= 1.40kg/m3, η =

1.92 ×10−5Nsm−2)

We have

v=r8RT

πM =r8×8.314 ×273

3.14 ×32 ×10−3

= 424.93m/s

and η=1

3ρ¯

vλ

∴λ=3η

ρ¯

=3×1.92 ×10−5

1.4×424.93 = 9.68 ×10−8m