E6 PROPERTIES OF GASES Flow-times, density, phase changes, solubility Introduction Kinetic-Molecular Theory The kinetic energy of an object is dependent on its mass and its speed. The relationship, given by the equation E k ½mv 2 shows that if two objects have the same kinetic energy, the heavier object must be travelling more slowly. Kineticmolecular theory uses statistics (due to the large number of molecules involved) to explain the properties of gases. It is based on three assumptions about the particles that make up the gas: Assumption 1 Assumption 2 Assumption 3 The particles in a gas have mass, but no volume. The particles of a gas are in constant, random, straight-line motion, changing direction when they collide with each other or with the walls of the container. Particle collisions are elastic: the gas particles exchange kinetic energy between one another, but the total kinetic energy of the sample does not change. The corresponding statistical equation for the average kinetic energy of a gas molecule is E k ½mv 2, where E k is the average kinetic energy, m is the molecular mass and v 2 is the average of the squares of the molecular speeds. Gases at the same temperature have the same average kinetic energy, so, on average, molecules with a lower speed have a higher mass. If the gas is pure, all molecules will have the same mass, but at any instant the direction of movement will be different for each. A container of gas should be considered as mainly empty space with particles of gas, all travelling at different speeds in different directions, spread randomly throughout. Each collision of a particle with the container exerts a force, and many such collisions add up to give the observed pressure. Increasing the number of collisions per unit time increases the pressure and can be effected in a number of different ways: Decrease the volume - less time elapses before a particle collides with the container. Increase the amount of gas present - more molecules are present to collide with the container. Increase the temperature - the molecular speeds increase so less time elapses between each collision. Diffusion and Effusion Diffusion describes the movement of one gas into another, whereas effusion (see figure E6-1) is the movement of a gas into an empty space. If a sample of gas in a confined space is connected to another space in which gas is present at a lower pressure, there will be a net movement of gas molecules into this other space until the pressure is uniform throughout. Figure E6-1 If the two spaces are connected by a narrow path, and if the pressure difference is maintained, movement of molecules in the reverse direction can be ignored. For a particular pair of spaces, at a particular temperature and pressure, the time required for the movement of gas into the second space is not the same for all gases. For any one gas this depends on a number of factors, including the mass of a molecule of the gas. The time required is thus related to the molar mass of the gas. This is because, at a particular
temperature, the speeds of the gas molecules depend on their mass - the heavier the gas, the slower the average speed of the molecules. The factors influencing the flow-time, other than molar mass, become negligible when the pressure-difference between the two spaces is very small and the narrow path very short. In this experiment the pressure-difference and the length of the narrow path are relatively large, but despite these limitations, a simple relationship between flow-time and relative molecular mass can be shown. See Example 1 at the end of this document. Buoyancy As well as the downward force of gravitational attraction, objects immersed in a fluid (the Earth s atmosphere, for example) experience an upward force equal to the weight * of fluid displaced - ships would not float, nor hydrogen balloons rise through the atmosphere if this were not so. Thus, whenever an object is weighed in air, the recorded weight will be less than the true weight (measured in a vacuum) by an amount equal to the weight of air displaced. When weighing solids and liquids in air, the weight of air displaced is small compared to the weight of the object, and is usually neglected. For gases, however, the weight of air displaced is of a similar order of magnitude as the weight of the gas, and therefore cannot be ignored. System in Air Apparent Mass of System System A w System B w System C x Figure E6-3 First, consider Systems A and B. Assume for the moment that the syringes are sealed. The syringe in System B contains 50 ml of air, so the weight of syringe plus air has increased over System A by an amount equal to the weight of 50 ml of air. However, the total volume of air displaced by the syringe in System B has also increased by 50 ml. Thus the upward buoyancy force for System B has increased over System A by an amount equal to the weight of 50 ml of air. Because the same gas and volumes of gas are involved, the two opposing weights cancel, and the weight of the syringe will be the same regardless of the position of the plunger in the syringe barrel (the same conclusion is reached if the syringe is open to the atmosphere). If time permits, weigh the syringe with and without air in the barrel, to confirm this for yourself. Now consider System C where the syringe contains 50 ml of an unknown gas. The weight of the syringe in System C has increased over the weight of the syringe in System A by an amount equal to the weight of 50 ml of unknown gas, but decreased (due to buoyancy) by an amount equal to the weight of 50 ml of air. Because the weights of air and gas will be different, the two opposing weights no longer cancel. Thus, Weight of syringe with gas, x weight of empty syringe, w + weight of 50 ml of unknown gas - weight of 50 ml of air * Distinction Between Mass And Weight The mass of a volume of unknown gas is determined by a weighing procedure in Experiment 2. The terms mass and weight are often used interchangeably, but are fundamentally different. Mass is an intrinsic property of matter, depending only on the types and numbers of atoms constituting the matter. Weight on the other hand is a force (usually the force of gravitational attraction) acting on an object and is proportional to the mass of the object. A balance measures the force needed to oppose equally the gravitational attraction force, and so keep the object at rest, i.e. a balance measures the weight of an object.
Rearranging, and replacing the weight terms by the masses which produce the same force, gives mass of 50 ml of unknown gas x - w + (mass of 50 ml of air). An important note about units: The SI units of energy, pressure and volume are the joule (J), the pascal (Pa) and the cubic metre (m 3 ) respectively. These are related as follows: 1 Pa x 1 m 3 1 J. Measurements in the laboratory are usually made in kpa and L where so 1 kpa 10 3 Pa and 1 L 10 3 m 3 1 kpa x 1 L 10 3 Pa x 10 3 m 3 1 J So although they are not SI units, the use of kpa and L in the Ideal Gas Equation is allowed as they are self-consistent. An ideal gas is one which obeys the general gas law, PV nrt, where n amount (in moles) contained in the volume V, at absolute temperature T and pressure P. R is the universal gas constant. For a sample of gas with mass m and with molar mass M, n m/m, and hence PV m M RT This can be rearranged to M PV m R T Equation 1 In Experiment 2, the mass (and therefore weight) of air displaced will be estimated by using the ideal gas equation. As the composition of air is approximately 78% N 2, 21% O 2 and 1% Ar, the "molar mass" of air can be calculated as: M 28.02 x 0.78 + 32.00 x 0.21 + 39.95 x 0.01 28.98 g mol 1. The mass of 50 ml of air at 20 C and 101 kpa can thus be calculated from Equation (1). MPV m R T 1 28.98 g mol 101 kpa 0.050 L 1 1 8.314 J K mol 293 K 0.060 g Calculation of Molar Mass from Density By rearranging Equation 1 we get, M where d m/v, the absolute density of the gas. mrt RT ( ) V P P Equation 2 d Phase changes There are three commonly recognised physical states of matter: solid, liquid and gas. Most substances can exist in all three states depending on the temperature of the surroundings, e.g. ice, water and steam. Solids are confined to a set volume and shape. Liquids have a definite volume and take on the shape of their container to the limit of the liquid s volume. Gases completely fill the volume of their containers and have no boundaries of their own. Most substances change from a solid to a liquid and then to a gas as the temperature rises. However, a few substances such as iodine and carbon dioxide pass straight from solid to gas at atmospheric pressure. This process is known as sublimation. The melting point of a solid is the lowest temperature at which the solid changes to a liquid. The same temperature is also known as the freezing point of the liquid. Liquids may be converted to gases by boiling or
evaporation. The boiling point is a characteristic temperature at a given pressure for a pure substance. Both the melting point and boiling point may be used as tests of a substance s purity. Solubility of gases The solubility of a gas in a liquid depends on the temperature and pressure of the system. It is increased by raising the pressure and/or lowering the temperature. The solubility of any substance, solid, liquid or gas can be expressed in molarity. A useful alternative to molarity, however, is the mole fraction, a dimensionless number that expresses the number of moles of solute as a fraction of the total number of moles present. For example, if a solution contains n 1 mole of solvent and n 2 mole of solute, the mole fraction of solute (designated x 2 ) is expressed as n2 x2 n + n When only solvent is present (i.e. no solute), x 2 0 and when only solute is present, x 2 1. See Example 2 at the end of this document. Reaction of carbon dioxide in water When CO 2 (g) dissolves in water, the following equilibria are established: 1 2 CO 2 (g) CO 2 (aq) CO 2 (aq) + H 2 O(l) H 2 CO 3 (aq) H + (aq) + HCO 3 (aq) Addition of OH (aq) will cause the equilibria to move to the right by removing H + (aq), while adding H + (aq) will push the equilibria to the left. As a consequence of these equilibria, the solubility of CO 2 (g) in water is dependent on the ph of the solution. Experiment 4 will investigate this effect.
EXAMPLE 1 A series of flow-times were measured for a number of gases of known molar mass, producing the results shown below Gas Molar Mass (M) / g mol 1 Table E6-1. Flow Times Flow-time (t) / s Flow-time 2 (t 2 ) / s 2 CO 2 44 6.9 47.6 CHClF 2 86 9.6 92.2 CCl 2 F 2 120 11.4 130 CClF 2 CF 3 154 12.9 166 C 4 F 8 200 14.8 219 The plot of molar mass versus the square of the flow-time is shown in Figure E6-2. From the graph, it can be seen that the molar mass is directly proportional to the square of the flow-time, or, expressed mathematically: M kt 2. Figure E6-2 (flow-time) 2, t 2 (s 2 ) molar mass, M (g mol 1 ) Figure E6-2 Flow Times The constant, k, in the above expression is given by the slope of the line in Figure E6-2. It relates to the specific piece of apparatus as well as the temperature and pressure difference. In a rigorous experiment to measure the molar mass of an unknown gas by this method it is necessary to obtain the flow-times for a number of known gases, and then find the best value for the constant, k, in the expression above. However, here time does not permit more than a duplicate measurement of flow-time for one known gas, in this case N 2. Consider two gases of molar masses M 1 and M 2 that have flow times of t 1 and t 2 respectively under identical experimental conditions. From the above relationship, Dividing the first of these equations by the second we get M 1 kt 1 2 and M 2 kt 2 2 which is a general relationship between molar mass and flow-time for a pair of gases at constant temperature and pressure difference.
Another expression shows the general relationship between molar mass and the rate of flow of gas through a narrow path. The rate of flow is inversely proportional to the flow-time. (Graham s Law of Effusion, which this experiment demonstrates, is normally expressed as a relationship between molar mass and rate of effusion.) EXAMPLE 2 In an experiment at 20 C and 101 kpa, 40 ml of CO 2 is introduced into a 50 ml syringe at room temperature. Water (5.0 ml) is then added, and the system brought to equilibrium by shaking. The final volume V final was 39 ml. The volume of CO 2 that dissolved (V sol ) is obtained from the final volume of the contents in the syringe (V final ), as shown below. V sol 40 ml + 5 ml - V final The amount (n 2 mol) of dissolved solute can be calculated from the volume V sol by assuming that the gas obeys the ideal gas law; that is, PV sol n 2 RT. In other words, V sol (volume of CO 2 dissolved) 45 ml 39 ml 6 ml n 2 (mol of CO 2 dissolved) n 1 (mol of water present) x 2 (mol fraction of CO 2 in solution at 20 C and 101 kpa) Note: For this practical exercise, we are concerned with dilute water solutions, the density of which may be taken to be 1.0 g ml -1.