Presentation on theme: "The atomic mass unit u is the standard unit of measure for masses as small as those of atoms. One u is 1/12 the mass of a carbon-12 atom."— Presentation transcript:
1The atomic mass unit u is the standard unit of measure for masses as small as those of atoms. One u is 1/12 the mass of a carbon-12 atom.
2The molecular mass of a molecule is the sum of the masses of the atoms.
3One gram-mole (or mole) of a substance contains as many particles (atoms or molecules) as there are atoms in 12 grams of the isotope carbon-12.
4This number of particles is 6 This number of particles is x 1023 and is known as Avogadro’s number NA. One mole of any substance contains Avogadro’s number of particles.
5The mole is the SI base unit of “the amount of a substance The mole is the SI base unit of “the amount of a substance.” Since the masses of these particles is different, the mass of a mole is different for different substances.
6One mole of a substance has a mass in grams that is equal to the atomic or molecular mass of the substance.
7Ex 1. - The Hope diamond (44. 5 carats) is almost pure carbon Ex 1. - The Hope diamond (44.5 carats) is almost pure carbon. The Rosser Reeves ruby (138 carats) is mostly Al2O3. One carat is equal to g. Determine (a) the number of carbon atoms in the diamond and (b) the number of Al2O3 molecules in the ruby.
8An ideal gas is an ideal model for gases that relates absolute pressure, Kelvin temperature, volume and the number of moles of the gas. PV = nRT is called the ideal gas law.
9PV = nRT P is the absolute pressure V is the volume n is the number of moles T is the temperature in Kelvins R is the universal gas constant and has a value of J/(mol•K)
10This law can be expressed in terms of the total number of particles N, instead of the number of moles. The right side of PV = nRT is multiplied by NA/NA, PV = nNA•(R/NA)•T. The value nNA is moles multiplied by particles per mole, so it is equivalent to number of particles N.
11The equation then becomes: PV = N•(R/NA)•T The equation then becomes: PV = N•(R/NA)•T. The constant term R/NA called Boltzman’s constant, is represented by the symbol k, and has a value of x J/K. The ideal gas law then becomes PV = NkT.
12Ex 2. - In the lungs, the respiratory membrane separates tiny sacs of air (absolute pressure = x 105 Pa) from the capillary blood. The average radius of these alveoli is mm, and the air inside contains 14% oxygen. Assuming the air acts as an ideal gas at body temperature (310 K), find the number of oxygen molecules in one of these sacs.
13Using the ideal gas law, one mole of an ideal gas can be shown to have a volume of 22.4 liters at 0°C and one atmosphere of pressure (standard temperature and pressure, STP).
14Ex 3. - Bubbles in a carbonated drink grow in size as they move upward, some doubling in size. They also increase in speed. Why does a bubble grow and move faster as it ascends?
15Boyle’s law states that, if temperature and number of moles are held constant, the pressure and volume vary inversely; PiVi =PfVf.
17The curve that passes through the initial and final points is called an isotherm. An isotherm at a different temperature would not intersect.
18Ex 4. - A particular set of scuba gear consists of a 0 Ex 4. - A particular set of scuba gear consists of a m3 tank filled with compressed air at an absolute pressure of 2.02 x 10 7 Pa. Assuming that air is consumed at a rate of m3 per minute and that the temperature is the same at all depths, determine how long the diver can stay under seawater at a depth of (a) 10.0 m and (b) 30.0 m.
19Charles law states that if the pressure and number of molecules are held constant; the volume is directly proportional to the temperature, Vi/Ti = Vf/Tf .
20The ideal gas law provides no information as to how pressure and temperature are related to properties of the molecules themselves.
21A study of kinetic theory reveals the following formulas: A study of kinetic theory reveals the following formulas: KEave = 3/2 •kT KEave is the average kinetic energy of a particle, k is Boltzman’s constant, T is the Kelvin temperature.
22Kinetic energy also can be calculated by: Kinetic energy also can be calculated by: KEave = 1/2 • mv2rms The value mv2rms is the root-mean-square speed, the average value of the squared speed.
23U = 3/2 •nRT U is the internal energy of a U = 3/2 •nRT U is the internal energy of a monatomic ideal gas, n is the number of moles, R is the universal gas constant, T is the Kelvin temperature.
24Ex 6. - Air is primarily a mixture of N2 (molecular mass = 28 Ex 6. - Air is primarily a mixture of N2 (molecular mass = 28.0 u) and O2 (molecular mass = 32.0 u). Assume 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.
25Ex 7. - Hydrogen is the most plentiful element in the universe Ex 7. - Hydrogen is the most plentiful element in the universe. Why doesn’t our atmosphere contain hydrogen?
26Brownian motion is the irregular, zig-zag motion of larger, observable particles in suspension; pollen grains in water, fine smoke particles in air, peanut butter on Albertson’s shelves. These particles have the same KEave as the suspension particles, but have less velocity because of their greater mass.
27Diffusion is the process by which molecules of a solute in a solvent move from an area of lower concentration to an area of higher concentration.
28Ex 8. - A gas molecule has a translational rms speed of hundreds of meters per second at room temperature. So, why does it take so long for the fragrance from perfume to reach the other side of a room when it should take a fraction of a second?
29Fick’s law of diffusion: Fick’s law of diffusion: m = (DA ∆C)t / L m is the mass of solute that diffuses, t is the time of diffusion, D is the diffusion constant (m2/s), ∆C is the difference in concentration between the ends of the channel, A is the cross-sectional area, L is the length of the channel.
30Ex 9. - Water vapor exits the stomatal pores of leaves Ex 9. - Water vapor exits the stomatal pores of leaves. The diffusion constant for water vapor is D = 2.4 x 10-5 m2/s. A stomatal pore has a cross-sectional area A = 8.0 x m2 and a length of L = 2.5 x 10-5 m. The concentration of water vapor on the interior side is C2 = kg/m3; on the outside it is C1 = kg/ m3. Find the mass of water vapor that passes through a stomatal pore in one hour.