Transport Phenomena

The equilibrium state of any gas is most probable state but if the gas is not in an equilibrium state, we may have any of the following cases:

1. The different parts of the gas may have different velocities so there will be relative velocities of layers of gas with respect to one another in such case faster moving layer impart momentum to slower moving layer to bring equilibrium state, this gives rise to phenomenon of viscosity.
2. The different parts of the gas may have different temperatures. If so, the molecules of gas will carry kinetic energy from regions of higher temperature to the regions of lower temperature to bring the equilibrium state and gives rise to conduction phenomenon.
3. The different parts of the gas may have different molecular concentration so, molecules of gas carry the mass from region of higher concentration to lower concentration to bring equilibrium and this gives rise to phenomenon of diffusion.

Hence viscosity, conduction and diffusion represent their transportation of momentum, energy and mass respectively.

1. Viscosity of Gases

   Consider a gas flowing along a horizontal surface. The layer is in contact with the wall of the surface at rest. The velocity of the layer increases with increase in distance from the fixed layer. Viscosity is defined as the tangential force per unit area required to maintain a unit velocity gradient.

   

   
   $$F = -\eta A \frac {du}{dx} \dots (i) $$
   Here \(\eta \) is the coefficient of viscosity and A is the area of the layer and \(\frac {du}{dx}\) is the velocity gradient.
   
   The velocity of a molecule at any layer at a height x above the fixed layer \(= x\frac {du}{dx}\). If the mean free path of a molecule is \(\lambda \) then the distance for the first collision to occur \(= x -\lambda \). The average velocity of these molecules \(= (x - \lambda \frac {du}{dx} \). About \(\frac 13rd\) of the molecules are moving along each axis. As the molecules in a given spae will be free to move in any direction, \(\frac 16 th\) can be moving in one direction and the other \(\frac 16 th \) in the opposite direction.
   let be the r.m.s velocity of the molecules, n the number of molecules per c.c. and m the mass of each molecule. Then, the number of molecules crossing unit in one direction in one second \(= \frac 16nc\)
   Mass of the molecule \(=\frac 16 mnc = M\)
   Momentum in the forward direction \(=M\left [(x - \lambda \frac {du}{dx} \right ] \)
   Momentum in the opposite direction \(=M\left [(x + \lambda \frac {du}{dx} \right ] \)
   \begin{align*} \text {Total change in momentum in one second is force,} \\ F &= M\left [(x - \lambda) \right ] \frac {du}{dx}- M\left [(x + \lambda ) \frac {du}{dx} \right ]\\ \text {or,}\: F &= -2M\lambda \frac {du}{dx} \\ \text {or,}\: F &= -2\left [\frac 16 mnc \right ]\lambda \frac {du}{dx} \\ &=-\frac 13 mnc\lambda \frac {du}{dx} \dots (ii) \\ \text {From equation}\: (i)\:\text {and}\: (ii), \\ \eta A \frac {du}{dx} &= \frac 13mn\lambda \frac {du}{dx} \\ \text {But,}\: A = 1\: cm^2 \\ \eta \frac {du}{dx} &= \frac 13mn\lambda \frac {du}{dx} \dots (iii)\\ \text {But}\: \lambda &= \frac {1}{\sqrt 2 \pi d^2n} \\ \therefore \eta &= \frac 13mn.\frac {1}{\sqrt {2} \pi d^2n} = \frac {mc}{3\sqrt {2}\pi d^2 }\dots (iv) \\ \text {From kinetic theory of gas,}\: C\propto \sqrt {T} \\ \therefore n \propto \sqrt {T} \dots (v) \end{align*}
   It shows that coefficient of viscosity of gas is independent of pressure and directly proportional to square root of absolute temperature of gas.

2. Thermal Conductivity of Gases

   The thermal conductivity of gases an be explained on the basis of kinetic theory of gases. The gas molecules are free to move between the hot end and cold end. They are also free to move in any direction. Due to different temperature of gas molecules in hot and old end thus they have different energy.
   The phenomenon of thermal conduction may be regarded as transport of energy. Let \(\frac {d\theta }{dx}\) be the temperature gradient and \(\lambda \) be mean free path. The temperature decreases from the hot to the cold end and when the steady state is reached, the temperature gradient will be uniform. The mean temperature of the molecule crossing the layer P from the old side \(= \left (\theta - \frac {d\theta }{dx} \lambda \right )\)
   Let n be the number of molecules per c.c., m the mass of each molecule and C the r.m.s velocity of the molecules. On the average only \(\frac 16\) of the molecules will be moving from the hot to the cold end and \(\frac 16\) of the molecules will be moving from the cold to the hot end.

   

   
   Consider one sq. cm area of cross-section at the layer P. the mass of the molecules crossing in one second \(= \frac 16 mn = M.\)
   Therefore, heat energy carried by the molecules crossing the unit area of the layer in one second from the hot to the old end\(=MC_v\left (\theta - \frac {d\theta }{dx}\right )\lambda \)
   Therefore, the net heat conducted per unit area per second from the hot to the cold end \begin{align*}= MC_v \left (\theta + \frac {d\theta }{dx} \right ) \lambda – MC_v\left (\theta - \frac {d\theta }{dx} \right ) \lambda \\ Q &= 2MC_v\frac {d\theta }{dx} \dots (i) \\ \therefore Q 2\left [ \frac 16 mnc\right ]C_v \lambda \frac {d\theta }{dx} \\ Q &= \frac 13 mnc. C_v \lambda \frac {d\theta }{dx} \dots (ii) \\ \text {From the definition of thermal conductivity,} \\ Q = kA \frac {d\theta }{dx} t \\ \text {Here,}\: A = 1 \text {sq. m},\: t = 1\: \text {second} \\ \therefore Q &= k\frac {d\theta }{dx} \dots (iii)\\ \text {From equation}\: (ii) \: \text {and}\: (iii) \\ k &= \frac 13 mn.\: C_v\lambda \dots (iv) \\ k &= \frac 13mn.\: C_v.\frac {1}{\sqrt {2} \pi d^2 n} \\ &= \frac {mC.C_v}{3\sqrt 2\pi d^2}\dots (v) \\ \text {But,}\: C\alpha \sqrt {T} \\ \therefore k \propto \sqrt T \dots (vi) \\ \end{align*}
   Thermal conductivity of gas is directly proportional to square root of absolute temperature.

3. Diffusion {Transport of Mass}

   Let us consider a mass of the gas moving in parallel layers between the planes AB and CD. Here MN is intermediate plane. Let us suppose that the concentration increases in the vertical direction as we go from the plane AB to CD through an intermediate plane MN. Here the concentration of the gas would be more above the plane MN than below MN. So, to bring equilibrium, the molecules of gas due to thermal agitation will move from the plane CD to the plane AB crossing MN and from the plane AB to plane D crossing MN until the concentration in every region is same.

   

   As the molecules are moving in all possible direction due to thermal agitation, it may be supposed that \(\frac 13 \: \text {th} \) of the molecules are moving in each of three directions parallel to artesian coordinate axis. That means on the average \(\frac 16 \: \text {th} \) of the molecules move parallel to any one axis in one particular direction.
   let n be the concentration at the plane MN and \(\frac {dn}{dz}\) the rate of the change of the concentration with distance in an upward direction perpendicular to MN. Let eah of the planes AB and CD be at a distance \(\lambda \) from MN, where \(\lambda \) is the mean free path. Then concentration at the plane,
   \begin{align*} CD &= N + \lambda \frac {dn}{dz} \\ AB &= n - \lambda \frac {dn}{dz} \end{align*}
   The number of molecules coming from the plane CD and crossing the plane MN downward per unit area per second.
   $$ =\frac {1}{6} \vec {v} \left (n + \lambda \frac {dn}{dz} \right )$$
   The number of molecules coming from the plane CD and crossing the plane MN downward per unit area per second is
   $$ =\frac {1}{6} \vec {v} \left (n - \lambda \frac {dn}{dz} \right ) $$
   \(\therefore \) The net number of molecules crossing per unit area of the plane MN downward per second
   \begin{align*}&=\frac {1}{6} \vec {v} \left (n + \lambda \frac {dn}{dz}\right ) \\ &=\frac {1}{6} \vec {v} \left (n - \lambda \frac {dn}{dz}\right ) \\&= \frac 13 \vec v \lambda \frac {dn}{Dz} \\ \text { The coefficient of diffusion,} \\ D &= \frac {\frac 13 \rho \vec v \lambda}{dn/dz}\\ &= \frac 13 \vec v \lambda \dots (i) \\ \frac 13\frac {\rho \vec v \lambda}{3 \rho} \\ &= \frac {\eta }{\rho } \dots (ii) \\ \text {Where}\: \eta =\frac 13 \rho \vec v \lambda = \text {coefficient of viscosity} \\ \text {or,}\: \lambda \propto \frac 1n \propto \frac TP \: \text {and}\: v \propto \sqrt T \\ \therefore D &= P^{-1} T ^{1/2} \\\end{align*}
   The coefficient of diffusion is directly proportional to \(T^{1/2} \) where T is absolute temperature and is inversely proportional to pressure.

Bibliography

S.S. Singhal, J.P. Agarwal, Satya Prakash. heat and thermodynamics and statistical physics. pragati prakashan, 2010.

—. Heat and Thermodynamics and Statistical Physics. Pragati Prakashan, 2010.

Vatsyayan, Dr. Rakesh Ranjan. Refresher Course in Physics. kathmandu: Surya Book Traders, 2015.