# Download Lecture 4 (Dec.9)

Survey

Document related concepts

Energy applications of nanotechnology wikipedia , lookup

Alternative fuel wikipedia , lookup

Regenerative brake wikipedia , lookup

Bolivian gas conflict wikipedia , lookup

Natural gas prices wikipedia , lookup

Pickens Plan wikipedia , lookup

Aliso Canyon gas leak wikipedia , lookup

Internal energy wikipedia , lookup

Conservation of energy wikipedia , lookup

Energy Independence and Security Act of 2007 wikipedia , lookup

Transcript

Chapter 14: The Ideal Gas Law and Kinetic Theory Fluids – Chapter 11 - cover 14.1: Molecular Mass, the Mole and Avogadro’s Number. 14.2: The Ideal Gas Law. 14.3: Kinetic Theory of Gases. 14.2 Kinetic Theory of Gases The particles are in constant, random motion, colliding with each other and with the walls of the container. Each collision changes the particle’s speed. As a result, the atoms and molecules have different speeds. 14.2 Kinetic Theory of Gases THE DISTRIBUTION OF MOLECULAR SPEEDS An apparatus used to measure atomic velocities. J.A. Eldridge 1927 . (Recall problem 8-15.) 14.2 Kinetic Theory of Gases Measuring temperature The molecules of the fluid carry momentum and exert an impulse on the surface when they bounce from it. We can calculate the average force on the wall exerted by the colliding molecules. 14.2 Kinetic Theory of Gases With every collision an impulse of Δ(mv) is imparted to the wall. where v is the component of velocity in the “normal” direction to the wall. 14.2 Kinetic Theory of Gases If a single molecule repeatedly strikes the wall with a mean time of Δt between collisions, then the average force exerted on the wall by this molecule is: 14.2 Kinetic Theory of Gases If a single molecule repeatedly strikes the wall with a mean time of Δt between collisions, then the average force exerted on the wall by this molecule is: 14.2 Kinetic Theory of Gases So, for a single molecule, the average force exerted on the wall is: For N molecules in the container, one third of them (on average) strike a wall during time Δt . The total force on a wall is thus: The quantity is the average value of the squared speed. 14.2 Kinetic Theory of Gases The quantity is the average value of the squared speed since we are dealing with a (Maxwellian) distribution of speeds. We can define the “root-mean-square” or “rms” speed: 14.2 Kinetic Theory of Gases Having calculated the force on a wall, we can now calculate the pressure, P = F / A (where A is the wall’s area), inside the container: where V is the container volume. 14.2 Kinetic Theory of Gases We have the pressure: Notice: We can identify: as the average (translational) kinetic energy of a molecule in the gas. So: 14.2 Kinetic Theory of Gases We have (from kinetic theory): Previously, we had: (the ideal gas law). This gives a connection between the temperature of a gas and the average kinetic energy of the constituent molecules: 14.2 Kinetic Theory of Gases Problem 14-44: A cubical box with side of length 0.300 m contains 1.000 moles of neon gas at room temperature (293 K). What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35 x 10-26 kg. 14.2 Kinetic Theory of Gases Problem 14-44: A cubical box with side of length 0.300 m contains 1.000 moles of neon gas at room temperature (293 K). What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35 x 10-26 kg. Solution: 1.000 mole of neon → number of neon atoms inside is equal to Avogadro’s number: N = NA. The collision rate with a single wall is: where t is the (mean) time between collisions for any single atom striking the wall. 14.2 Kinetic Theory of Gases Problem 14-44: A cubical box with side of length 0.300 m contains 1.000 moles of neon gas at room temperature (293 K). What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35 x 10-26 kg. Solution: For a cubical container of side length L, the mean time between collisions is: 14.2 Kinetic Theory of Gases Problem 14-44: A cubical box with side of length 0.300 m contains 1.000 moles of neon gas at room temperature (293 K). What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35 x 10-26 kg. Solution: Thus, the collision rate in the cubical container (1.000 moles of gas) is: 14.2 Kinetic Theory of Gases Problem 14-44: A cubical box with side of length 0.300 m contains 1.000 moles of neon gas at room temperature (293 K). What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35 x 10-26 kg. Solution: Calculate the mean squared velocity using the connection between the temperature of the gas and the average kinetic energy of the constituent atoms: 14.2 Kinetic Theory of Gases Problem 14-44: A cubical box with side of length 0.300 m contains 1.000 moles of neon gas at room temperature (293 K). What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35 x 10-26 kg. Solution: We now have the collision rate: 14.2 Kinetic Theory of Gases Problem 14-44: A cubical box with side of length 0.300 m contains 1.000 moles of neon gas at room temperature (293 K). What is the average rate (in atoms/s) at which neon atoms collide with one side of the container? The mass of a single neon atom is 3.35 x 10-26 kg. Solution: Inserting numerical data: 14.2 Kinetic Theory of Gases Problem 14-38: Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomic gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom. 14.2 Kinetic Theory of Gases Problem 14-38: Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomic gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom. Solution: The ratio of the average kinetic energies is equal to the ratio of the Kelvin temperatures of the gases. 14.2 Kinetic Theory of Gases Problem 14-38: Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomic gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom. Solution: We are not given the temperatures but these are related to the given information by the ideal gas law: We have gas containers of same V with equal pressures P containing the same mass, m, of gas. We will need to determine n (ratio) for the two cylinders. 14.2 Kinetic Theory of Gases Problem 14-38: Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomic gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom. Solution: We will need to determine n (ratio) for the two cylinders. 14.2 Kinetic Theory of Gases Problem 14-38: Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomic gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom. Solution: The number of moles can be calculated from the mass m (in grams) and the mass per mole M (in grams per mole): 14.2 Kinetic Theory of Gases Problem 14-38: Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomic gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom. Solution: The kinetic energy ratio is thus given by: 14.2 Kinetic Theory of Gases Problem 14-38: Two gas cylinders are identical. One contains the monatomic gas argon (Ar), and the other contains an equal mass of the monatomic gas krypton (Kr). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio of the average kinetic energy of a krypton atom to the average kinetic energy of an argon atom. Solution: The result: 14.2 Kinetic Theory of Gases We can define the “internal energy” of the gas as the total translational energy of the N atoms in the gas: Thus, we have: (for a monoatomic gas). The internal energy of the gas is determined by the Kelvin temperature. 14.2 Kinetic Theory of Gases The internal energy of a monatomic gas is given by the total translational energy of the N atoms in the gas: In thermodynamics, we identify a monatomic gas as having 3 “degrees of freedom” corresponding to the 3 available directions (x,y,z) in space for translation. There is an internal energy of ½kT per atom per degree of freedom. 14.2 Kinetic Theory of Gases For a gas consisting of diatomic molecules ( air ie. N2 ) there are two extra degrees of freedom which can “store” internal energy: rotation about two axes. This gives 5 degrees of freedom altogether, each providing an internal energy of ½kT per molecule. Thus, for a diatomic molecule: Three translational degrees of freedom plus two rotational degrees of freedom. 14.2 Kinetic Theory of Gases Problem 14-42: Compressed air can be pumped underground into huge caverns as a form of energy storage. The volume of a cavern is 5.6 x 105 m3 and the pressure of the air in it is 7.7 x 106 Pa. Assume that air is a diatomic ideal gas whose internal energy U is given by If one home uses 30.0 kWh of energy per day, how many homes could this internal energy serve for one day? 14.2 Kinetic Theory of Gases Problem 14-42: Since we are treating the air as a diatomic ideal gas (PV = nRT), it follows that U 52 nRT 52 PV 5 2 7.7×106 Pa 5.6 105 m3 1.11013 J The number of joules of energy consumed per day by one house is 3 J h 3600 s 8 30.0 kW h 30.0 10 1.08 10 J s 1 h The number of homes that could be served for one day by 1.1 1013 J of energy is 1 home 1.1×10 J 8 1.08 ×10 13 5 = 1.0 ×10 homes J PHYS1020 General Physics I