CHAPTER 15 : KINETIC THEORY OF GASSES

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Presentation transcript:

CHAPTER 15 : KINETIC THEORY OF GASSES

15.1 : Ideal Gas Equation Pressure (P) inversely proportional with Volume (V) at constant Temperature : Boyle’s law. Volume (V) directly proportional with Temperature (T) at constant Pressure : Charle’s law. Pressure (p) directly proportional with Temperature (T) at constant Volume : Gay-Lussac’s law. Iv. Relationships between three equations can be combine into single equation:- PV = nRT (Ideal gas equation) Unit R, for 1kg gas is Jkg-1K-1

Definition Definition of ideal gas Obey the gas equation in any pressure and temperature condition. Practically, ideal gas do not exist Real gas at low pressure and high temperature nearly an ideal gas.

Examples 1. An ideal gas is held in a container at constant volume. Initially, its temperature is 10.0 oC and its pressure is 2.50 atm. What is it s pressure when its temperature is 80.0 oC?

15.2 : Kinetic Theory Of Gases Introduction : A gas contains large molecules that moves in random directions :- velocity momentum continuous collisions that exerts forces per unit area is called pressure.

The assumptions of the kinetic of an ideal gas for molecular model are: A container with volume V contains a very large number N of identical molecules, each of mass m. The size of each molecule is small compared with the average distance between them and the dimensions of the container. The molecules are in constant motion. The molecules undergo perfectly elastic collisions with each other and also with the wall of the container. The container walls are perfectly rigid and do not move as a result of the collisions.

Newton’s laws of motion The molecules are in constant Newton’s laws of motion Elastic collisions. The container walls are perfectly :- =} rigid and infinitely massive and do not move.

15.3 : Gas Pressure Derivation of ideal gas equation pV= N/3 Nm<v2> and p = 1/3<v2>

Examples In a period of 1.00 s, 5.00 x 1023 nitrogen molecules strike a wall with an area of 8.00 cm2. If the molecules move with the speed of 300 m/s and strike the wall head-on in a perfectly elastic collisions, what is the pressure exerted on the wall? ( The mass of one N2 is 4.68 x 10-26 kg.)

15.4 : Root Mean Square (Rms) Speed Of Gas Molecule As known pV = 2/3 N [ ½ m<v2>]...............(1) Derivation of vrms; By multiplying (4) with M= NAm, we obtain the average translational kinetic energy per mole, NA1/2 m<v2> = ½ M <v2> = 3/2 RT.............(5) From equation (5), we can obtain the root mean square speed (vrms)

Root mean square speed (vrms)

15.5 : Molecular Kinetic Energy (Average Translational Kinetic Energy) Each molecules of gas has a true speed of its own due to moving randomly. We use V rms to determine kinetic energies of all those molecules, which is a kind of an average value Hence <K> = ½mo<c2> where mo = mass of molecule <c2> = square of the V rms

Relationship Between Thermodynamic Heat and Kinetic Energy of Molecule From the ideal gas equation, pV = nRT with substitution of n = N/ NA the ideal gas equation will be pV = (N/NA) RT ..............................(2) we know that R/ NA = k thus, pV = NkT .............................................(3)

As known pV = 2/3 N [ ½ m<v2>]...............(1) (1)= (3) 2/3 N [ ½ m<v2>] = NkT we obtain [ ½ m<v2>]= 3/2 kT...................(4) where Ktr = ½ m<v2> therefore Ktr = 3/2 kT

Examples What is the average translational kinetic energy of a molecule of an ideal gas at a temperature of 20 oC.

15.6 : Principle Of Equipartition Of Energy Definition of degrees of freedom The number of velocity components needed to describe the motion of a molecule completely.

Law of equipartition of energy Average energy that relate with every rotational and translational degree of freedom will have the same average value that is 1/2kT and this energy depends only to the absolute temperature. Therefore, K = ( f T+ f R) ½kT Where as f = degree of freedom T = absolute temperature k = Boltzmann constant Rotational kinetic energy per mole (at absolute temperature) E = f x 1/2RT whereas R = gas constant = kNA

Translational degree of freedom(Monatomic) In ideal gas molecule motion, every molecule contribute translational kinetic energy that is Ktr=3/2kT Monatomic gas such as Helium, Neon and Argon. It has kinetic energy at 3 degree of freedom in the direction of x, y and z axes.

Rotational degree of freedom(Diatomic) Involving diatomic gas such as H2, O2, and Cl2. Undergoes additional rotational motion at the x and y axes. It has 5 total degrees of freedom with 3 translational kinetic energy of degree of freedom and 2 rotational kinetic energy of degree of freedom.

Vibrational degree of freedom Involving the diatomic and polyatomic which contribute the additional 2 degree of freedom. The vibrational factor is very small and can be neglected/ignored.

Polyatomic Polyatomic gas such as H2O, CO2, NH3 and N2O4. Undergoes rotational motion at x, y and z axes. It has 6 total degrees of freedom with 3 translational kinetic energy of degree of freedom and 3 rotational kinetic energy of degree of freedom.

Examples

15.7 : Internal Energy Principle of equipartition of energy Any amount of energy absorb by the molecule of gas will be distributed equally between every translational and rotational degree of freedom of the molecule

Internal energy of ideal gas Definition : Internal energy of a gas U is equal to the total amount of average kinetic energy and potential energy which contains in any gasses.

Potential energy For the ideal gas: Potential energy can be neglected because A) the force of the exertion between the molecules do not exist so B) U = Total of average kinetic energy of the molecule = K.

Examples A tank of volume 0.300 m3, contains 1.50 mol of Neon gas at 220C. Assuming the Neon behaves like an ideal gas, find the total internal energy of the gas.

The end