Transport Phenomena

a 1.1 Mean Free PathFree path between two successive collisions will be a straight path with unchanging velocity since the molecules exert no force on one another except when they collide. As a result, a single molecule's journey is made up of a succession of small zig-zag trajectories of varying lengths. These pathways of varying lengths are known as molecular free paths, and their mean is known as the mean free path.

Let l1,l2,l3................ln are successive free path travelled in total time t, then we havel1+l2+l3+...........................ln= ⏨V twhere ⏨V is the average speed of the molecule.If nis the total number of collision suffered byt eh molecule in one second, λ be the mean free path thenλ= l1+l2+l3+.......................ln

n= ⏨V t

nor, λ= ⏨Vt

n Expression for Mean Free path Let us consider a gas containing n number of molecules per unit volume. Let only one molecule under consideration is in motion ad all other at rest.

If 𝜎 is the diameter of each molecule then the moving molecule will collide with all these molecules whose centres lie on the circumference of the cylinder as shown in figure.If V is the veloeity of the moving molecule. then in one sec it will collide with all the molecules whese centre lie in a cylinder of radius 𝜎 and lengh v with in the volume 𝜋𝜎2v.Thus the number of molecules per unit volume. Now the mean free path is given by λ = Distance travelled by molecule in one second

Number of collision suffered by molecules in one second = v

𝜋𝜎2vn=1

𝜋𝜎2n The expression has been derived assuming that only one molecule under consideration in moving while all the others are at rest. Maxwell assumed the realistic assumption than all the molecules are moving with all possible velocities in all possible directions and found the result applying distribution law of molecular speeds. i.c. 𝜆=1

2𝜋𝜎2nIf m is the mass of a molecule then mn=𝜌. Where 𝜌 is the density of the gas∴𝜆=m

2𝜋𝜎2𝜌Thus the mean free path is inversely proportional to the density of gas. By the kinetic molecular theory of gases.P=1

3𝜌⏨v2where ⏨v2 is the mean square speed of the molecules.we have ⏨v2=3KT

m,m is the mass of a molecule.where K is Boltzmann's constanta

P	=1 3𝜌3KT m
or. m 𝜌	=KT P

∴ equation (4) reduces toa

𝜆	=1 2𝜋𝜎2m 𝜌=1 2𝜋𝜎2KT P
∴𝜆	=KT 2𝜋𝜎2P

So the mean free path is directly proportional to the absolute temperature and inversely proportional to the pressure of the gas. TRANSPORT PHENOMENONIf the gas is not in equilibrium state, we may have any of the following three cases:(1) The different parts of gas may have different velocities so there will be a relative motion of the layer of gas with respect to one another. This gives rise to the phenomenon of viscosity.(2) The different parts of gas may have different temperatures so the molecules of the gas will carry kinetic enengy from region of higher temperature to the region of lower temperature to hring the equilibrium state. This gives to the phenomenon of conduction.(3) The different parts of the gas may lave different molecular concentration i.e. the number of molecules per unit volume so the molecules of gas will carry the mass from regions of higher concentrations to those of lower concentrations to bring the equilibrium state. It gives the phenomenon of diffusion. These all three phenomena are called transport phenomenon. ViscosityLet us consider a mass of gas moving in parallel layers between two horizontal planes AB and CD. Suppose the velocity of the layer of gas in contact with the plane AB is zero and increases as we pass towards the plane CD. Also consider an intermediate plane MN.Let n be the number of molecules per unit volume and ⏨v their average speed. Since molecules are moving due to thermal agitation in all possible directions, it may be supposed that one third of the molecules are moving in each of three directions parallel to three co-ordinate axis, so that an average one sixth of the molecule move parallel to any one axis in particular direction.

The number of molecules crossing the plane MN upwards of downwards per unit area per second =n⏨v

6. Let G be the momentum of each molecule in the plane MN and if the momentum changes according to distance then the momentum gradient =dG

dz where dG is the change in momentum in distance z.∴ Change in momentum for a distance=𝜆dG

dzLet each of the planes AB and CD at a distance 𝜆 from MN. Where 𝜆 is mean free path. So the momentum of each molecule at the planeCD=G+𝜆dG

dzAnd the momentum of each molecule at the lower planeAB=G-𝜆dG

dz∴ The momentum carried by molecules crossing per unit area per second from upper layer CD in downward direction per unit area per second from upper layer CD in downward direction.=1

6n⏨v(G+𝜆dG

dz)Similarly, the momentum carried by the molecules crossing per unit area per second from lower layer AB in upward direction.=1

6n⏨v(G-𝜆dG

dz)∴ The net momentum transferred downward per unit area per second through the intermediate plane MN.=1

6n⏨v(G+𝜆dG

dz)-1

6nv(G-𝜆dG

dz)=1

6n⏨v(G+𝜆dG

dz-G+𝜆dG

dz)=1

3n⏨vdG

dz=1

3⏨vn𝜆d

dz(mu)Since G=mu, u being the velocity of molecules at plane MN.=1

3nm⏨v𝜆du

dzSince the rate of change of momentum is known as force. So the above expression is a measure of force exerted on the plane MN by the upper layers. We are considering unit area. So force cames and tangential stress.∴ Coefficient of viscosity 𝜂= Tangential stress

Velocity gradient =1

3mn⏨v𝜆du

dz

du

dz=1

3mn⏨v𝜆But 𝜆=1

2𝜋𝜎2n ∴ 𝜂=m⏨v

32𝜋𝜎2Since ⏨v𝛼T∴ The coefficient of viscosity is directly proportional to the square root of the absolute temperature of the gas,But also, 𝜆=KT

2𝜋𝜎2P∴ 𝜂=1

3mn⏨v𝜆=1

3𝜌⏨v𝜆[ Where, nm=𝜌]or, 𝜆=3𝜂

𝜌⏨v Thermal ConductionLet's consider a system of gas which have different temperatures at different regions. Due to this equality in temperature heat energy is transformed from region at higher temperature to the region at lower temperature. This exchange of heat energy gives the phenomenon of thermal conductivity.

Suppose AB and CD be two parallel layers in which a system of gas is enclosed which is shown in figure. Let n be the number molecule and ⏨v beverage velocity of gas molecules. Then number of molecules moving in particular direction of any one axis per unit area per second is 1

6n⏨v.Let E be the energy of each particle in MN layer and 𝜆 be the separation between AB,MN the ener direction free path Now,Energy of each molecule at CD=E+𝜆dE

d𝜆Energy of each molecule at AB=E-𝜆dE

d𝜆Then amount of energy transferred per unit area of MN in downward direction per second from CD=1

6n⏨v(E+𝜆dE

d𝜆). Amount of energy transferred per unit area of MN is upward direction per second from AB=1

6n⏨v(E-𝜆dE

d𝜆). So the net energy crossing unit area of MN per second in downward directiona

=1 6n⏨v(E+𝜆dE d𝜆)-1 6n⏨v(E-𝜆dE d𝜆)

=1 3n⏨v𝜆dE d𝜆

where E is thermal energy =mCvT[a

where m= mass of molecule

Cr= Specific heat capacity at

constant volume

T= Absolute temperature

]∴ Net transfer energy per umit area per second =1

3mn⏨𝜆,Cyd~T

d𝜆[ where dE

d𝜆=d(mCvT)

d𝜆=mCvdT

d𝜆]We know,Coefficient of thermal conductivity (K) is given byK= Net energy transferred per unit area per second

Temperature gradient or, K=1

3mn⏨v𝜆CvdT

d𝜆

dT

d𝜆∴ K=1

3mn⏨v𝜆CvBut we know, coefficient of viscosity (𝜂)=1

3mn⏨v𝜆This equation (i) becomes K=nCvAgain we know, mean free path (𝜆)=1

2𝜋𝜎2n and ⏨v=8KT

𝜋m So putting these values in equation (i), we geta

K=1 3mn8KT 𝜋m×1 2𝜋𝜎2n.CV

=2 3(KTm 𝜋)×Cy 𝜋𝜎2

This is the required expression of thermal conductivity of gas.DiffusionConsider a mass of the gas is moving in parallel layers between the plane AB and CD. MN is an intermediate plane. Let us suppose that the concentration i.e. number of molecules per unit volume increases in vettical direction as we go from plane AB to CD, So that the molecules will move from CD to AB to establish equillibrium. This phenomenon is called transport of mass.

Let n be the concentration at the plane MN and dn

dz is the rate of change of concentration with distance in an upward direction. Let AB and CD be at a distance 𝜆 from the plane MN where 𝜆 is the mean free path. The concentration at the plane CD=n+𝜆dn

dzThe concentration at the plane AB=n-𝜆dn

dzThe number of molecules coming from the plane CD crowsing MN downwards per unit area per second =1

6n⏨v=1

6n(n+2dn

dz)And the number of molecules coming from AB and crossinig unit area of molecules upwards per second is =1

6𝜋⋅(n-𝜆⋅dn

dz)The net number of molecules crossing unit area of plane MN downwards per second =1

6n(n+2dn

dz)-1

6𝜋⋅(n-𝜆⋅dn

dz) =1

3𝜋⋅dn

dz𝜆The coefficient of diffusion is the ratio of the number of molecules across unit area in one sec to the concentration gradient.i.e. coefficient of diffusion D=1/3⏨v𝜆(dn/dz)

(dn/dz)=1

3a

∴D=1 3⏨v.

or. D=1 3𝜌⏨v𝜆 𝜌=𝜂 𝜌 [∵1 3𝜌⏨v=n]

or. D=𝜂𝜌

where |𝜂=1

3𝜌v>= coefticient of viscositySince 𝜆𝛼1

n𝛼T

P and ⏨v𝛼TFrom equation (5).a

D𝛼P-1~T12

Also K/n=C, f.e. n=KC, and n=DP

∴K/C=D𝜌

} aNumericals1. Calculate the mean diameter of a molecule of benzene if there are 2.79×1019 molecules /cc and the mean free path of benrene =2.2×10-4cm.[T.U. 2055, 2064]Solution:Here.Number of molecules per unit volume (n)=2.79×1019 molecules/ cc Mean free path of benzene (𝜆)=2.2×10-6cmMean diameter of a molecule of benzene (𝜎)= ?We have, mean free path (𝜆)=1

2𝜋𝜎2nar, 𝜆=1

2𝜋𝜎2nor, 𝜎2=1

2𝜆n𝜋or, 𝜎2=1

2×(2.2×10-6)×(2.79×1019)×𝜋or, 𝜎2=3.69×10-15or, 𝜎=6.07×10-8cmThus mean diameter of a molecule is 6.07×10-8cm.2. The density of nitrogen at NTP is 1.25kgm-3 and its coeflicient of viscosity is 1.66×10-5Nsm-2. Calculate the mean free path of a molecule of nitrogen gas.Solution: [T.U.2058]Here, density of nitrogen at NTP (𝜌)=1.25kgm-3Coefficient of viscosity (𝜂)=1.66×10-5Nsm-2.Mean free path of molecule of nitrogen gas (𝜆)= ?We have,a

Velocity (v) of molecule =8KT 𝜋m

∴v=8KT 𝜋m

And mean free path (𝜆)=3𝜂

𝜌vor, 𝜆=3𝜂

𝜌8KT

𝜋m=3𝜂

𝜌m𝜋

8KT∴ 𝜆=3𝜂

𝜌m𝜋

8KTWe have,Mass of oxygen molecule (m)= Molecular mass

Avogadro's Number =M

NAor, m=28

6.023×1023gram=28

6.023×1023×10-3 kg =4.65×10-26kgThen putting all values in equation (i) we geta

𝜆	=3×1.66×10-5 1.25×4.65×10-26×3.14 8×1.38×10-2×273
	=3.98×10-5×2.2×10-3
	=8.76×10-8m=8.76×10-8m

Thus the mean free path of molecule of nitrogen gas is 8.76×10-7m3. The mean free path of oxygen molecule of STP is 9.37×10-6cm. Calculate the molecular diameter of oxygen.Given, Avogadro's number =6.02×1023mole-1Volume of 1 mole of oxygen gas at STP =22.4 liters.[T.U. 2059]Sulution:Here, mean free path of oxygen molecules (𝜆)=9.37×10-6cmAvogadro's number (NA)=6.02×1023mole-1Molecular diameter (𝜎) of oxygen = ?We know that,Volume of 1 mole of oxygen gas at STP =22.4 liters=22.4×103ccThus the number of moleules per ec of oxygen gas at STP is given by n=6.02×1023

22.4×107=2.69×1019 molecules /ccBy formula, we have 𝜆=1

2𝜋𝜎2na

or, 𝜎2=1 2𝜋𝜆n

or, 𝜎=1 2𝜋𝜆n=1 2×22 7×9.37×10-6×2.69×1019

=2.99×10-4cm

Thus the molecular diameter of oxygen molecule is 2.99×10-5cm.4. The molecular diameter of nitrogen is 3.5×10-4cm. Calculate the mean free path of temperature 27∘C and pressure 1 atmosphere.[T.U. 2062]Solution:Here, molecular diameter of nitrogen (𝜎)=3.5×10-8~cmTemperature (T)=27∘C=27+273=300~KPressure (P)=1~atm=1.01×103~N/m2=1.01×106 dyne /cm2Boltzmann constant (K) in CGS unit =1.38×10-16erg/ degree Mean free path (𝜆)= ?a

We have, mean free path (𝜆)=KT 2𝜋𝜎2P

=1.38×10-16×300 2×22 7×(3.5×10-4)2×1.01×106

=7.53×10-6cm

Thus the mean free path of nitrogen molecule is 7.53×10∘cm5. Calculate the mean free path of gas molecules in a chamber of 10-6~mm of mercury pressure, assuming molecular diameter to be 2A. Take temperature of chamber to be 273 K and one gram molecules of the gas oceupies 22.4 litres at NTP.Solution: [T.U. 2063, 2065, 2067]Here preasure of chamber (P)=10∘mm of Hg=1.33×10-Nm2Molecular diameter (𝜎)=2A-2×10-18mMean free paih (𝛬)= ?We know, Volume at STP (| or |0∘C)=22.4×10-3~m3 and pressure is 1.01×105N/m2. Therefore, number of molecules in 22.4×103m3 at 1.01×103N/m2 and 273K is 6.023×1023.Hence, number of molecules in 1~m3 at 1~N/m2 and 273Ka

=6.023×1027 22.4×10-2×1.01×103

=6.023×1.01 22.4×1.01×1023-4.3-5

=0.269×1017=2.69×1016 molecules

Now, the mean free path (𝜆)=1

2𝜋n𝜎2a

=1 2×22 7×2.69×1016×(2×10-19)2

=2.09×102

Therefore mean free path is 2.09×102 metres.6. Calculate the mean free path of oxygen molecule at STP. Given density =1.40kg/m3 and coefficient of viscosity =1.92×10-5Nsm-2.[T.U. 2068]Solution:Here, density of oxygen gas (𝜌)=1.40kg/m3Coefficient of viscosity (𝜂)=1.93×10-5Nsm-2Mean free path (d)= ?We know,a

Mass of the oxygen molecule (m)	= Molecular mass Avogadro's number
	=32×10-3 6.023×102=5.31×10-3/2kg

We have, 𝜂=1

3𝜌⏨v [Where ⏨v is velocity of molecule] or, 𝜆=3n

𝜌VWe also have,⏨v= SKT

m𝜋Putting this value of ⏨v in equation (i), we geta

𝜆	=3𝜂 𝜌8KT m𝜋=3n 𝜌m𝜋 8KT
	=3×1.92×10-5 1.4×5.31×10-26×22 7 8×1.38×10-23×273
	=4.11×10-5×2.35×10-3=9.66×10-8m

Therefore mean free path of oxygen molecule is 9.66×10-5m.

Brownian Motion

Brownian motion is the seemingly random motion of particles, atoms, or molecules that emerges out of the random collisions of those particles. Brownian motion can be observed as light shines through a window. Particles of dust or pollen can be seen in the light floating in the air, following what seems to be random jittery patterns. The dust particles aren't moving on their own, but are colliding with molecules of the air keeping the dust in motion.

Causes of Brownian motion

Brownian motion is caused by the structure and physics of fluids; i.e., liquids and gases. According to kinetic theory, all matter is in motion; atoms and molecules especially within liquids and gases are in constant vibrating motion. These particles will travel in straight lines until redirected by a collision. Particles within gases and liquids are constantly moving, colliding, and moving toward equilibrium. A common example of Brownian motion is dust particles floating in the air. When light is shining through a window, dust particles can be observed floating in the air and following seemingly random paths of jittery motion. This random jittery movement is called Brownian motion.