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Lecture #15. The Kinetic Molecular Theory of Gases and Effusion and Diffusion. Chemistry 142 B Autumn Quarter, 2004 J. B. Callis, Instructor. The Ideal Gas Law. A purely empirical law - solely the consequence of experimental observations
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Lecture #15 The Kinetic Molecular Theory of Gasesand Effusion and Diffusion Chemistry 142 B Autumn Quarter, 2004 J. B. Callis, Instructor
The Ideal Gas Law • A purely empirical law - solely the consequence of experimental observations • Explains the behavior of gases over a limited range of conditions. • A macroscopic explanation. Says nothing about the microscopic behavior of the atoms or molecules that make up the gas.
The Kinetic Theory • Starts with a set of assumptions about the microscopic behavior of matter at the atomic level. • Supposes that the constituent particles (atoms) of the gas obey the laws of classical physics. • Accounts for the random behavior of the particles with statistics, thereby establishing a new branch of physics - statistical mechanics. • Offers an explanation of the macroscopic behavior of gases. • Predicts experimental phenomena that haven't been observed. (Maxwell-Boltzmann Speed Distribution)
The Four Postulates of the Kinetic Theory • A pure gas consists of a large number of identical molecules separated by distances that are great compared with their size. • The gas molecules are constantly moving in random directions with a distribution of speeds. • The molecules exert no forces on one another between collisions, so between collisions they move in straight lines with constant velocities. • The collisions of molecules with the walls of the container are elastic; no energy is lost during a collision. Collisions with the wall of the container are the cause of pressure.
Calculation of the Force Exerted on a Container by Collision of a Single Particle
Calculation of the Pressure in Terms of Microscopic Properties of the Gas Particles
The Kinetic Theory Relates the Kinetic Energy of the Particles to Temperature
Mean Square Speed and Temperature By use of the facts that (a) PV and nRT have units of energy, and (b) the square of the component of velocity of a gas particle striking the wall is on average one third of the mean square speed, the following expression may be derived: where R is the gas constant (8.31 J/mol K), T is the temperature in kelvin, and M is the molar mass expressed in kg/mol (to make the speed come out in units of m/s).
Finally, we obtain the relationship of kinetic energy of one gas phase particle to temperature: Note that: (a) Temperature is a measure of the molecular motion, and (b) the kinetic energy is only a function of temperature. At the same temperature, all gases have the same average kinetic energy.
Molecular Speed Distribution Thus far we have discussed the random nature of molecular motion in terms of the average (root mean square) speed. But how is this speed distributed? The kinetic theory predicts the distribution function for the molecular speeds. Below we show the distribution of molecular speeds for N2 gas at three temperatures.
The Mathematical Description of the Maxwell-Boltzmann Speed Distribution f(u)du gives the fraction of molecules that have speeds between u and u + Du.
The distribution of molecular speeds for N2 at three temperatures
Features of the Speed Distribution The most probable speed is at the peak of the curve. The most probable speed increases as the temperature increases. The distribution broadens as the temperature increases.
Relationship between molar mass and molecular speed
Features of the Speed Distribution • The most probable speed increases as the molecular mass decreases. The distribution broadens as the molecular mass decreases.
Calculation of Molecular Speeds and Kinetic Energies T = 300 K
Problem 15-1: Calculate the Kinetic Energy of (a) a Hydrogen Molecule traveling at 1.57 x 103 m/sec, at 300 K. Mass = KE = KE = KE =
Problem 15-1 Kinetic Energies for (b) CH4 and (c) CO2 at 200 K (b) For Methane, CH4 ,u =5.57 x 102 m/s KE = (c) For Carbon Dioxide, CO2 ,u =3.37 x 102 m/s KE =
Note • At a given temperature, all gases have the same molecular kinetic energy distributions, and • the same average molecular kinetic energy.
Effusion Effusion is the process whereby a gas escapes from its container through a tiny hole into an evacuated space. According to the kinetic theory a lighter gas effuses faster because the most probable speed of its molecules is higher. Therefore more molecules escape through the tiny hole in unit time. This is made quantitative in Graham's Law of effusion: The rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Effusion Calculation Problem 15-2: Calculate the ratio of the effusion rates of ammonia and hydrochloric acid. Approach: The effusion rate is inversely proportional to square root of molecular mass, so we find the molar ratio of each substance and take its square root. The inverse of the ratio of the square roots is the effusion rate ratio. Numerical Solution: HCl = NH3 = RateNH3 =
Diffusion • The movement of one gas through another by thermal random motion. • Diffusion is a very slow process in air because the mean free path is very short (for N2 at STP it is 6.6x10-8 m). Given the nitrogen molecule’s high velocity, the collision frequency is very high also (7.7x109 collisions/s). • Diffusion also follows Graham's law:
Diffusion of a gas particle through a space filled with other particles
NH3(g) + HCl(g) = NH4Cl(s) HCl = 36.46 g/mol NH3 = 17.03 g/mol Problem 15-3: Relative Diffusion Rate of NH3 compared to HCl: RateNH3 =
The inverse relation between diffusion rate and molar mass. Due to it’s light mass, ammonia travels 1.46 times as fast as hydrogen chloride NH3(g) + HCl(g) NH4Cl(s)
Problem 15-4: Gaseous Diffusion Separation of Uranium 235 / 238 235UF6 vs 238UF6 Separation Factor = after Two runs after approximately 2000 runs 235UF6 is > 99% Purity ! Y - 12 Plant at Oak Ridge National Lab
Answers to Problems in Lecture #15 • (a), (b) and (c) are all the same value: 4.13 x 10- 21 J / molecule • RateNH3 = RateHCl x 1.463 • RateNH3 = RateHCl x 1.463 • S = 1.0086