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CHAPTER 15 : KINETIC THEORY OF GASSES

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## CHAPTER 15 : KINETIC THEORY OF GASSES

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**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)**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.**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.