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Topic 3.2 Modelling a gas The Ideal Gas Law and Kinetic Theory The Mole, Avogadro's Number, and Molecular Mass Atomic Mass Unit, U By international agreement, the reference element is chosen to be the most abundant type of carbon, called carbon-12, and its atomic mass is defined to be exactly twelve atomic mass units, or 12 u. Molecular Mass The molecular mass of a molecule is the sum of the atomic masses of its atoms. For instance, hydrogen and oxygen have atomic masses of 1.007 94 u and 15.9994 u, respectively. The molecular mass of a water molecule (H2O) is: 2(1.007 94 u) + 15.9994 u = 18.0153 u. Avogadro's Number NA The number of atoms per mole is known as Avogadro's number NA, after the Italian scientist Amedeo Avogadro (1776–1856): An example • The molar mass of helium is 4.0 g. Determine the mass of a single atom of helium in kilograms. An example • The molar mass of helium is 4.0 g. Determine the mass of a single atom of helium in kilograms. There are 6.02 × 1023 atoms in 4.0 g of helium. 0.004 mass of atom 6.02 10 23 = 6.645 × 10–27 kg » 6.6 × 10–27 kg An example 2 3 The molar mass of uranium is 238 g. a) Calculate the mass of one atom of uranium. b) A small rock contains 0.12 g of uranium. For this rock, calculate the number of: i moles of uranium ii atoms of uranium. 3 a There are 6.02 × 1023 atoms in 0.238 kg of uranium. mass of atom 0.238 23 6.02 10 bi = 3.95 × 10−25 kg 4.0 × 10−25 kg number of moles number of moles mass of uranium molar mass of uranium 0.12 238 5.04 × 10–4 » 5.0 × 10–4 ii number of atoms number of moles × NA number of atoms 5.04 × 10–4 × 6.02 × 1023 3.06 × 1020 » 3.1 × 1020 Number of Moles, n The number of moles n contained in any sample is the number of particles N in the sample divided by the number of particles per mole NA (Avogadro's number): The number of moles contained in a sample can also be found from its mass. The Ideal Gas Law An ideal gas is an idealized model for real gases that have sufficiently low densities. The Ideal Gas Law An ideal gas is an idealized model for real gases that have sufficiently low densities. The condition of low density means that the molecules of the gas are so far apart that they do not interact (except during collisions that are effectively elastic). The Ideal Gas Law An ideal gas is an idealized model for real gases that have sufficiently low densities. The condition of low density means that the molecules of the gas are so far apart that they do not interact (except during collisions that are effectively elastic). The ideal gas law expresses the relationship between the absolute pressure (P), the Kelvin temperature (T), the volume (V), and the number of moles (n) of the gas. PV nRT Where R is the universal gas constant. R = 8.31 J/(mol · K). Ideal gas assumptions • The particles of gas (atoms or molecules) obey Newton’s laws of motion. You should know these by now! Ideal gas assumptions • The particles in a gas move with a range of speeds Ideal gas assumptions • The volume of the individual gas particles is very small compared to the volume of the gas Ideal gas assumptions • The collisions between the particles and the walls of the container and between the particles themselves are elastic (no kinetic energy lost) Ideal gas assumptions • There are no forces between the particles (except when colliding). This means that the particles only have kinetic energy (no potential) Do you remember what internal energy is? Ideal gas assumptions • The duration of a collision is small compared to the time between collisions. An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa At seas level these values are 300K and 1. x 105 Pa respectively. If the density of air sea level is 1.2 kg.m-3, what is the density the air on Mount Everest? “Physics”, Patrick Fullick, Heinemann Example question At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest? An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest? Take 1kg of air at sea level Volume = mass/density = 1/1.2 = 0.83 m3. Therefore at sea level p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K. An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest? Therefore at sea level p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K. At the top of Mount Everest p2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K. An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest? Therefore at sea level p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K. At the top of Mount Everest p2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K. p1V1/T1 = p2V2/T2 (1.0 x 105 Pa x 0.83 m3)/300K = (3.3 x 104 Pa x V2)/250K V2 = 2.1 m3, This is the volume of 1kg of air on Everest Density = mass/volume = 1/2.1 = 0.48 kg.m-3. Sample question • A container of hydrogen of volume 0.1m3 and temperature 25°C contains 3.20 x 1023 molecules. What is the pressure in the container? K.A.Tsokos “Physics for the IB Diploma” 5th Edition Sample question • A container of hydrogen of volume 0.1m3 and temperature 25°C contains 3.20 x 1023 molecules. What is the pressure in the container? # moles = 3.20 x 1023/6.02 x 1023 = 0.53 K.A.Tsokos “Physics for the IB Diploma” 5th Edition Sample question • A container of hydrogen of volume 0.1m3 and temperature 25°C contains 3.20 x 1023 molecules. What is the pressure in the container? # moles = 3.20 x 1023/6.02 x 1023 = 0.53 P = RnT/V = (8.31 x 0.53 x 298)/0.1 = 1.3 x 104 N.m-2 K.A.Tsokos “Physics for the IB Diploma” 5th Edition • The kinetic theory now links temperature with the microscopic energies of the gas molecules • The equation resembles the kinetic energy formula. • Adjusting for N molecules gives 3/2 NkBT • This represents the total internal energy of an ideal gas (only considering translational motion of molecules of monoatomic gases) This means; No intermolecular forces between molecules between collisions i.e. energy is completely kinetic • Gas consists of large number of identical tiny particles- molecules in constant random motion • Molecules undergo perfectly elastic collisions between each other and with the walls of the container; momentum is reversed during collision • Duration of collision negligible compared with the time between collisions • Each molecule produces a force on the wall of the container • The forces of individual molecules will average out to produce a uniform pressure throughout the gas- ignoring the effect of gravity The Ideal Gas Law The constant term R/NA is referred to as Boltzmann's constant, in honor of the Austrian physicist Ludwig Boltzmann (1844–1906), and is represented by the symbol k: PV = NkT Kinetic Theory of Gases Kinetic Theory of Gases The pressure that a gas exerts is caused by the impact of its molecules on the walls of the container. Kinetic Theory of Gases The pressure that a gas exerts is caused by the impact of its molecules on the walls of the container. It can be shown that the average translational kinetic energy of a molecule of an ideal gas is given by, where k is Boltzmann's constant and T is the Kelvin temperature. Derivation of, Consider a gas molecule colliding elastically with the right wall of the container and rebounding from it. The force on the molecule is obtained using Newton’s second law as follows, F P , t The force on one of the molecule, According to Newton's law of action–reaction, the force on the wall is equal in magnitude to this value, but oppositely directed. The force exerted on the wall by one molecule, mv 2 L If N is the total number of molecules, since these particles move randomly in three dimensions, one-third of them on the average strike the right wall. Therefore, the total force is: Vrms = root-mean-square velocity. Pressure is force per unit area, so the pressure P acting on a wall of area L2 is Pressure is force per unit area, so the pressure P acting on a wall of area L2 is Since the volume of the box is V = L3, the equation above can be written as, PV = NkT An example The Speed of Molecules in Air Air is primarily a mixture of nitrogen N2 (molecular mass = 28.0 u) and oxygen O2 (molecular mass = 32.0 u). Assume that each behaves as an ideal gas and determine the rms speed of the nitrogen and oxygen molecules when the temperature of the air is 293 K. The Internal Energy of a Monatomic Ideal Gas