Advanced Thermodynamics: Lecture 14 Shivasubramanian Gopalakrishnan - PowerPoint PPT Presentation

Advanced Thermodynamics: Lecture 14 Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Kinetic Theory Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The Properties of bulk matter are studied using two approaches 1 Kinetic Theory - which applies laws of mechanics to individual molecules and then derives expression such as the pressure of the system from such laws. 2 Statistical Mechanics - ignores detailed considerations of individual molecules and applies consideration of probabilities to very large number of molecules. Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Kinetic Theory Any macroscopic volume of a gas contains a very large number of molecules. The number molecules in one mole of a gas is 6 . 023 x 10 23 . (Avogadro’s number) The molecules themselves are separated by large distances as compared to their own sizes. The first approximation made is that, the molecules exert no force on each other except when they collide with each other. We also assume that collisions are perfectly elastic. In the absence of external forces, the molecules are uniformly distributed in the container of the gas. If N represents the total number of molecules in a volume V , the average number of molecules per unit volume, n , is given by, n = N (1) V Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The assumption of uniform distribution then implies that in any element of the volume ∆ V , the number of molecules ∆ N is ∆ N = n ∆ V (2) It is obvious that ∆ V cannot be very small since N is finite. The directions of the molecular velocities are assumed to distributed uniformly. Assume that a vector representing a velocity (magnitude and direction) is attached to every molecule. And all these vectors are transferred to a common origin and a sphere is constructed with a radius r and center as origin. The velocity vectors intersect the surface of the sphere in as many points as there are molecules. Assuming that molecules are uniformly distributed, the average number of points per unit area is N (3) 4 π r 2 Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

image: Sears and Salinger Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

and the number in an element area ∆ A is N ∆ N = (4) 4 π r 2 ∆ A The area ∆ A is nearly equal to ∆ A = ( rsin θ ∆ θ )( r ∆ φ ) = r 2 sin θ ∆ θ ∆ φ (5) The number of points in this area 4 π r 2 r 2 sin θ ∆ θ ∆ φ = N N ∆ N θφ = (6) 4 π sin θ ∆ θ ∆ φ If both sides of equation are divided by the volume V . ∆ n θφ = n (7) 4 π sin θ ∆ θ ∆ φ Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Molecular Flux At any wall of the the container of the gas or at any imagined surface within the gas molecules are continuously arriving due to random motion. Let ∆ N be the total number molecules arriving from all directions and with all speeds at one side of an element of surface area ∆ A during a time interval ∆ t . The molecular flux at the surface is defined as the total number of molecules arriving at the surface, per unit area and per unit time, is given by ∆ N Φ = (8) ∆ A ∆ t For an imagined surface within the gas, all molecules arriving at one side will cross the surface and there will be no net motion. For a wall, molecules arriving do not cross but rebound instead. Here there are two fluxes one of arriving molecules and the other of rebounding molecules. Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

image: Sears and Salinger Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

All θφ v molecules in the cylinder, and only those molecules, will reach the shaded surface during the time ∆ t , traveling in the θφ direction with a speed v . Let ∆ n v represent the number density of the molecules with speeds between v and v + ∆ v . Then the number density of the θφ v molecules is ∆ n θφ v = 1 (9) 4 π ∆ n v sin θ ∆ θ ∆ φ The volume of the slant cylinder is ∆ V = ( ∆ Acos θ )( v ∆ t ) (10) The number of θφ v molecules in the cylinder therefore is ∆ N θφ v = 1 (11) 4 π v ∆ n v sin θ cos θ ∆ θ ∆ φ ∆ A ∆ t Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

and the flux of these molecules is ∆ A ∆ t = 1 ∆Φ θφ v = ∆ N θφ v 4 π v ∆ n v sin θ cos θ ∆ θ ∆ φ (12) The flux ∆ θ v , due to molecules arriving at an angle θ with speed v , but including all angles φ , is ∆Φ θ v = 1 2 v ∆ n v sin θ cos θ ∆ θ (13) The flux ∆ θ , due to molecules arriving at an angle θ with all speeds v and including all angles φ , is ∆Φ θ = 1 X (14) 2 sin θ cos θ ∆ θ v ∆ n v Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The flux ∆ v , with speed v , but all angles θ and φ , is ∆Φ v = 1 4 v ∆ n v (15) Finally, the total flux Φ is given by Φ = 1 X (16) v ∆ n v 4 The average speed of the molecules is given by P v v = ¯ (17) N But breaking up into discrete velocities, one can write v = v 1 ∆ N 1 + v 2 ∆ N 2 + . . . = 1 X ¯ v ∆ N v (18) N N Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

In a summation convention this can be written as v = 1 X ¯ (19) v ∆ n v n and it follows, X v ∆ n v = ¯ vn (20) Hence the molecular flux can be written as, Φ = 1 4¯ vn (21) The molecules arriving at the area in the θφ direction are those coming in within the small cone, whose base is the shaded area ∆ A on the spherical surface. This area is ∆ A = r 2 sin θ ∆ θ ∆ φ (22) Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The solid angle of the cone is ∆ ω = ∆ A r 2 = sin θ ∆ θ ∆ φ (23) The flux Φ θφ v can be written as ∆ θφ v = 1 4 π v ∆ n v cos θ ∆ ω = ∆Φ ω v (24) The flux per unit solid angle, of molecules with speed v , is = 1 ∆Φ ω v 4 π v ∆ n v cos θ (25) ∆ ω And the total flux per solid angle is. ∆Φ ω ∆ ω = 1 4¯ (26) vncos θ Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Equation of state of an ideal gas image: Sears and Salinger Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

If m is the mass of a colliding molecule, the change in the normal component of momentum of a θφ v collision is mvcos θ − ( − mvcos θ ) = 2 mvcos θ (27) The rate of change of momentum per unit area due to all molecules arriving at an angle θ with speed v , or the pressure ∆ P θ v , equals the product of ∆Φ θ v and the change in momentum of a θ v molecule ∆ P θ v = (1 2 v ∆ n v sin θ cos θ ∆ θ )(2 mvcos θ ) = mv 2 ∆ n v sin θ cos 2 θ ∆ θ (28) To find the pressure ∆ P v due to molecules of speed v , we integrate, giving us ∆ P v = 1 3 mv 2 ∆ n v (29) Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Finally summing over all values of v , we have the total pressure P , P = 1 X v 2 ∆ n v 3 m (30) The average value of the square of the speed of all molecules, or the mean square speed, is found by P v 2 ¯ v 2 = (31) N Using discrete velocities we have P v 2 ∆ N v v 2 = ¯ (32) N or P v 2 ∆ n v v 2 = ¯ (33) n Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Then v 2 ∆ n v = n ¯ X v 2 (34) and P = 1 3 nm ¯ v 2 (35) Since n represents the number of molecules per unit volume N / V , we can write the equation as PV = 1 3 Nm ¯ v 2 (36) This looks similar to the ideal gas equation PV = nRT (37) Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Where n represents the number of kilo moles, equal to the total number of molecules divided by the number of molecules per kilo mole, or Avogadro number N A . We can therefore write PV = N R T (38) N A The quotient R / N A is called the universal gas constant per molecule, or Boltzmann’s constant, and is represented by k k ≡ R (39) N A In MKS system k = R = 1 . 381 × 10 − 23 Jmolecule − 1 K − 1 N A Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

In terms of the Boltzmann constant, the equation of state of an ideal gas becomes. PV = NkT (40) This will agree with equation derived from kinetic theory, if we set NkT = 1 3 Nm ¯ v 2 (41) or v 2 = 3 kT ¯ (42) m This gives us the molecular interpretation of temperature T , as a quantity proportional to the mean square speed of the molecules of the gas. 1 v 2 = 3 2 m ¯ (43) 2 kT Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The product of the one–half mass of a molecule and the mean square speed is the same the mean translational kinetic energy and is proportional to the absolute temperature. The factor 3 k / 2 being constant for all molecules dictates the face the mean KE depends only on the temperature and not one any other properties of the gas. Let T = 300K 3 2 kT = 3 2 × 1 . 38 × 10 − 23 × 300 = 6 . 21 × 10 − 21 J (44) If molecules are oxygen molecules, the mass m is 5 . 31 × 10 − 26 kg, and the mean square speed is v rms = 482 ms − 1 . Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661