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

6.023x1023. (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 V (1)

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 4πr2 (3)

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 4πr2 ∆A (4) The area ∆A is nearly equal to ∆A = (rsinθ∆θ)(r∆φ) = r2sinθ∆θ∆φ (5) The number of points in this area ∆Nθφ = N 4πr2 r2sinθ∆θ∆φ = N 4πsinθ∆θ∆φ (6) If both sides of equation are divided by the volume V . ∆nθφ = n 4πsinθ∆θ∆φ (7)

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 Φ = ∆N ∆A∆t (8) For an imagined surface within the gas, all molecules arriving at

• ne 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 θφv 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 ∆nv represent the number density of the molecules with speeds between v and v + ∆v. Then the number density of the θφv molecules is ∆nθφv = 1 4π∆nvsinθ∆θ∆φ (9) The volume of the slant cylinder is ∆V = (∆Acosθ)(v∆t) (10) The number of θφv molecules in the cylinder therefore is ∆Nθφv = 1 4πv∆nvsinθcosθ∆θ∆φ∆A∆t (11)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

and the flux of these molecules is ∆Φθφv = ∆Nθφv ∆A∆t = 1 4πv∆nvsinθcosθ∆θ∆φ (12) The flux ∆θv, due to molecules arriving at an angle θ with speed v, but including all angles φ, is ∆Φθv = 1 2v∆nvsinθcosθ∆θ (13) The flux ∆θ, due to molecules arriving at an angle θ with all speeds v and including all angles φ, is ∆Φθ = 1 2sinθcosθ∆θ X v∆nv (14)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The flux ∆v, with speed v, but all angles θ and φ, is ∆Φv = 1 4v∆nv (15) Finally, the total flux Φ is given by Φ = 1 4 X v∆nv (16) The average speed of the molecules is given by ¯ v = P v N (17) But breaking up into discrete velocities, one can write ¯ v = v1∆N1 + v2∆N2 + . . . N = 1 N X v∆Nv (18)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

In a summation convention this can be written as ¯ v = 1 n X v∆nv (19) and it follows, X v∆nv = ¯ 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 = r2sinθ∆θ∆φ (22)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The solid angle of the cone is ∆ω = ∆A r2 = sinθ∆θ∆φ (23) The flux Φθφv can be written as ∆θφv = 1 4πv∆nvcosθ∆ω = ∆Φωv (24) The flux per unit solid angle, of molecules with speed v, is ∆Φωv ∆ω = 1 4πv∆nvcosθ (25) And the total flux per solid angle is. ∆Φω ∆ω = 1 4¯ vncosθ (26)

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 θφv collision is mvcosθ − (−mvcosθ) = 2mvcosθ (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 ∆Pθv, equals the product of ∆Φθv and the change in momentum

• f a θv molecule

∆Pθv = (1 2v∆nvsinθcosθ∆θ)(2mvcosθ) = mv2∆nvsinθcos2θ∆θ (28) To find the pressure ∆Pv due to molecules of speed v, we integrate, giving us ∆Pv = 1 3mv2∆nv (29)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Finally summing over all values of v, we have the total pressure P, P = 1 3m X v2∆nv (30) The average value of the square of the speed of all molecules, or the mean square speed, is found by ¯ v2 = P v2 N (31) Using discrete velocities we have ¯ v2 = P v2∆Nv N (32)

• r

¯ v2 = P v2∆nv n (33)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Then X v2∆nv = n ¯ v2 (34) and P = 1 3nm ¯ v2 (35) Since n represents the number of molecules per unit volume N/V , we can write the equation as PV = 1 3Nm ¯ v2 (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 NA. We can therefore write PV = N R NA T (38) The quotient R/NA is called the universal gas constant per molecule, or Boltzmann’s constant, and is represented by k k ≡ R NA (39) In MKS system k = R NA = 1.381 × 10−23Jmolecule−1K −1

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 3Nm ¯ v2 (41)

• r

¯ v2 = 3kT m (42) 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 2m ¯ v2 = 3 2kT (43)

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 3k/2 being constant for all molecules dictates the face the mean KE depends

• nly on the temperature and not one any other properties of the

gas. Let T = 300K 3 2kT = 3 2 × 1.38 × 10−23 × 300 = 6.21 × 10−21J (44) If molecules are oxygen molecules, the mass m is 5.31 × 10−26 kg, and the mean square speed is vrms = 482ms−1.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661