AP TOPIC 6: Gases. Revised August General properties and kinetic theory

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AP OPIC 6: Gases General properties and kinetic theory Gases are made up of particles that have (relatively) large amounts of energy. A gas has no definite shape or volume and will expand to fill as much space as possible. As a result of the large amount of empty space in a volume of gas, gases are easily compressed. Pressure A pressure is exerted when the gas particles collide with the walls of any container it is held in. Pressure can be measured in a number of units, for example;.00 atm = 760. mmhg = 760. orr ask 6a Perform the following conversions 657 mmhg to atm. 830 torr to atmospheres C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page of 3

Kinetic theory he Kinetic theory is the basis for many properties of gases. he five postulates are;. Gases are composed of tiny particles (atoms or molecules) whose size is negligible compared to the average distance between them. his means that the volume of the actual individual particles in a gas can be assumed to be negligible compared to the volume of the container, and therefore the total volume that the gas fills is almost all empty space. he observation that gases are compressible agrees with the assumption that the actual gas particles have a small volume compared to the total volume (the volume of the container).. he gas particles move randomly, in straight lines in all directions and at various speeds. 3. he forces of attraction or repulsion between two gas particles are extremely weak or negligible, except when they collide. 4. When particles collide with one another, the collisions are elastic (no kinetic energy is lost). he collisions with the walls of the container create the gas pressure. Elastic collisions agree with the observation that gases, when left alone in a container, do not seem to lose energy and do not spontaneously convert to the liquid. 5. he average kinetic energy of a molecule is proportional to the Kelvin temperature and as a result, all calculations involving gases should be carried out with temperatures converted to K. (Using Kelvin eliminates the problem of negative temperatures leading to negative energies and negative volumes etc.) hese assumptions have limitations. For example, gases can be liquefied if cooled enough. his means real gas particles DO attract one another to some extent; otherwise the particles would never stick to one another and therefore never condense to form a liquid. C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page of 3

Pressure and olume relationships: Boyle s Law Boyle s Law states that, at constant temperature, pressure is inversely proportional to volume. his means that as the pressure increases the volume decreases and vice versa. P a constant his makes sense. If the volume is increased the gas particles collide with the walls of the container less often and the pressure is reduced. If we know the volume and pressure of a gas at a given temperature, and then volume or pressure is changed, Boyle s law allows us to calculate the new volume or pressure by applying the simple relationship below. P P P and are the original conditions and P and are the new conditions. ask 6b. If a.3 L sample of a gas at 53.0 torr is put under pressure up to a value of 40. torr at a constant temperature, what is the new volume?. he pressure on a 4 ml sample of gas is decreased from 8 mmhg to 790 mmhg. What will the new volume of the gas be? C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page 3 of 3

olume and emperature relationships: Charles Law Charles Law states that, at constant pressure, volume is directly proportional to temperature. his means the volume of a gas increases with increasing temperature and vice versa. a constant his makes sense. If the temperature is increased the gas particles gain kinetic energy, move around more and occupy more space. If we know the volume and temperature of a gas at a given pressure, and then volume or temperature is changed, Charles law allows us to calculate the new volume or temperature by applying the simple relationship below. and are the original conditions and and are the new conditions. ask 6c. An.0 L sample of a gas is collected at 76 K and then cooled by 4 K. he pressure is held constant at.0 atm. Calculate the new volume of the gas.. A gas has a volume of 0.57 L at 35.0 o C and.00 atm pressure. What is the temperature inside a container where this gas has a volume of 0.535 L at.00 atm? C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page 4 of 3

olume and Moles relationships: Avogadro s Law Avogadro s Law states that, at constant temperature and pressure, volume is directly proportional to the number of moles of gas present. his means the volume of a gas increases with increasing number of moles and vice versa. n a constant his makes sense. As more moles of a gas are placed into a container if conditions of temperature and pressure are to remain the same, the gas must occupy a larger volume. If we know the volume and number of moles of a gas at a given temperature and pressure and then volume or the number of moles is changed, Avogadro s law allows us to calculate the new volume or number of moles by applying the simple relationship below. n n and n are the original conditions and and n are the new conditions. ask 6d. A 3. L sample of 0.50 moles of O is held under conditions of.00 atm and 5.0 o C. If all of the O is then converted to Ozone (O 3 ) what will be the volume of ozone?. If. g of Neon gas occupies a volume of.0 L at 8.0 o C. What volume will 6.58 g of Neon occupy under the same conditions? C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page 5 of 3

Pressure and emperature relationships: Gay-Lussac s Law Gay-Lussac s Law states that, at constant volume, pressure is directly proportional to temperature. his means that temperature increases with increasing pressure and vice versa. P a constant his makes sense. If the temperature of a gas is raised then the particles will have more energy and collisions with the walls of the container will be more forceful and the pressure will increase. If we know the pressure and temperature of a gas at a given volume and then pressure or the temperature is changed, Gay-Lussac s law allows us to calculate the new pressure or temperature by applying the simple relationship below. P P = P and are the original conditions and P and are the new conditions. ask 6e. A gas at 5 o C in a closed container has its pressured raised from 50. atm to 60. atm. What is the final temperature of the gas?. A gas exerts a pressure of 900 mmhg at 0 o C. What temperature would be required to lower the pressure to.00 atm? C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page 6 of 3

he Ideal Gas Law he combination of laws above leads to the formulation of the Ideal Gas Law. Most gases obey this law at temperatures above 73 K and at pressures of.00 atm or lower. P nr R = the universal gas constant = 0.08 L atm K - mol -. his equation is useful because it can be manipulated to include other variables. For example, n can be replaced by mass molarmass and then manipulated further to include density. In addition to the Ideal gas equation other equations can be derived such as the General Gas equation P n P n If the number of moles of gas in an experiment is constant (frequently the case) the expression becomes the Combined Gas Law equation; P P ask 6f. Assuming ideal behavior, how many moles of Helium gas are in a sample that has a volume of 8. L at a temperature of 0.00 o C and a pressure of.0 atm?. A sample of aluminum chloride weighing 0.00 g was vaporized at 350. o C and.00 atm pressure to produce 9. ml of vapor. Calculate a value for the Molar Mass of aluminum chloride. C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page 7 of 3

Dalton s Law of Partial Pressures Dalton s Law states that in a mixture of gases the total pressure exerted by the mixture is equal to the sum of the individual partial pressures of each gas. Ptotal P P P3 etc. Assuming ideal behavior the equation can be simplified to; P total n total R ask 6g. Calculate the number of moles of Hydrogen present, in a 0.64 L mixture of Hydrogen and water vapor, at.0 o C, that has a total pressure of 750. torr, given that the vapor pressure of water at this temperature is 0.0 torr.. 3.00 L of Carbon monoxide gas at a pressure of 99. kpa, and.00 L of Carbon dioxide gas at a pressure of 300. kpa are injected into a.5 L container. Assuming no reaction between the two gases, what is the total pressure in the container? C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page 8 of 3

Deviations from ideal behavior At high pressures and low temperatures gas particles come close enough to one another to make the two postulates of the Kinetic heory below, invalid. (i) Gases are composed of tiny particles whose size is negligible compared to the average distance between them. When the gas is pressurized into a small space the gas particles size becomes more significant compared to the total volume. (ii) he forces of attraction or repulsion between two particles in a gas are very weak or negligible. Low temperature means less energy, so the particles are attracted to one another more. Under these conditions (high P and low ) gases are said to behave non-ideally or like real gases. his has two consequences. (i) When gases are compressed to high pressures, the size of the gas particles is no longer negligible compared to the total space occupied by the gas (its total volume). herefore, the observed total volume occupied by the gas under these real conditions is artificially large since the gas particles are now occupying a significant amount of that total volume. Previously, under ideal conditions, the gas particles size were negligible compared to the total volume occupied by the gas and therefore the observed total volume was approximately correct. he real gas situation necessitates the need to correct the volume by subtracting a factor. (ii) he actual pressure of a gas is lower than one would expect when assuming there were no attractive forces between the particles. Because, in a real gas, the particles are attracted to one another, they collide with the walls with less force, and the observed pressure is less than in an ideal gas. his necessitates the need to correct the pressure by adding a factor. hese corrections lead to the van der Waals equation below. Where the terms a and b are constants for a particular gas. an P nb n R C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page 9 of 3

Molar olume and gas stoichiometry We have seen how Avogadro's law states that equal volumes of all gases at constant temperature and pressure will contain equal numbers of moles. he volume of one mole of any gas is called its molar volume and can be calculated using the ideal gas equation. P nr By applying the data, pressure (P) =.00 atm, temperature () = 73 K, the gas constant (R) = 0.08 L atm K - mol -, number of moles (n) =.00 mol, the volume () can be found. A simple calculation finds its value to be =.4 L. hat is to say; One mole of any ideal gas at standard temperature and pressure (s.t.p) will occupy a volume of.4 L ask 6h Assume conditions of standard temperature and pressure in the questions below.. Calculate the mass of ammonium chloride required to produce.6 L of ammonia gas in the reaction below. NH 4 CI (s) + Ca(OH) (s) NH 3(g) + CaCI (s) + H O (g). Calculate the volume of carbon dioxide gas produced when 9.85 g of barium carbonate is completely decomposed in the reaction below. BaCO 3(s) BaO (s) + CO (g) C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page 0 of 3

Distribution of molecular speeds he kinetic theory introduces the idea that all of the gas particles in a sample move at different velocities, and that the velocity, and therefore the kinetic energy, is dependent upon the temperature. A typical plot showing the variation in particle speeds is shown below for hydrogen gas at 73 K. he vertical line on the graph represents the root-mean-square-speed (u rms ) and can be calculated using u rms 3R M Where R = universal gas constant = 8.345 kg m s - mol - K -, = temperature in Kelvin, M = molar mass of the gas in kg mol -. he root-mean-square-speed is the square root of the averages of the squares of the speeds of all the particles in a gas sample at a particular temperature. C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page of 3

ask 6i. Determine the u rms of the following gases at 5.0 o C. (Include units). (a) Helium (b) Oxygen (c) Radon. Determine the u rms of the following gases at 50.0 o C. (Include units). (a) Helium (b) Oxygen (c) Radon 3. What can be said quantitatively, about the u rms of a gas in relation to its molar mass and in relation to its temperature? C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page of 3

Grahams Law of Effusion and Diffusion Effusion is the process in which a gas escapes from one vessel to another by passing through a very small opening. Grahams Law of effusion states that the rate of effusion is inversely proportional to the square root of the density of the gas at constant temperature. Rate of effusionof A Rate of effusionof B density of B density of A Molarmass of B Molarmass of A Diffusion is the process by which a homogeneous mixture is formed by the random mixing of two different gases. Grahams Law of diffusion states that the rate of diffusion is inversely proportional to the square roots of their respective densities and molecular masses. Rate Rate density of density of Molarmass of Molarmass of Henry s Law Henry s law describes the relationship between the solubility of a gas in a solvent and pressure. c = k P Where; c = molar concentration of gas dissolved in the solvent in mol L - P = pressure in atm k = a constant. his makes sense. At higher pressures more gas particles strike the surface of the solution and enter the solvent meaning the concentration of the gas is higher. Note: Gas solubility usually decreases with increase in temperature since the gas particles have more energy and can escape from the solvent meaning less gas is dissolved in solution. C:\www.adriandingleschemistrypages.com\ap\apnotes\apnotes06.doc Page 3 of 3