Chemistry 11 Unit 11 Ideal Gas Law (Special Topic)
2 1. States of substances It has been studied in Unit 3 that there exist 3 states of matter in nature: gas, liquid and solid. (Technically there is the fourth one, plasma; but its discussion is beyond our scope in Chemistry 11.) Gas is the simplest state that can be well described by four parameters: 1. Volume 2. Pressure 3. Temperature 4. Amount of gas
3 Gas thermometer Measure the temperature of a gas sample Pressure gauge Measure the pressure of gas inside a container or cylinder Gas cylinders Contain the gas sample; provided with difference volumes
4 2. Gas laws Solids and liquids are incompressible; applying force and pressure cannot change their volumes. In other words, their volumes are fixed, and depend on the amounts of the substances present. Gases, on the other hand, behave very differently. Gases can be compressed or expanded readily, and they can change their shapes easily by filling them into different containers.
5 Besides external forces, volume of a gas can be changed also by means of changing temperature: It seems all the four parameters of a gas are interrelated. But how?
6 Experimentally, people have investigated the relationships between these factors and came up with three laws: (1) Boyle s law At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure. P 1 V or P 1 V 1 = P 2 V 2
7 (2) Charles law The volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature. V T or V 1 T 1 = V 2 T 2
8 (3) Avogadro s law At constant temperature and pressure, the volume of a gas is directly proportional to its number of moles. V n or V 1 n 1 = V 2 n 2
9 These three laws can be combined together to form a single relationship: PV nt = constant The constant term is called a universal gas constant R; its value is 8.314 J/K/mol in SI unit (or in another convention 0.08206 atm L/K mol). The relationship above can therefore be written in the form called the ideal gas law: PV = nrt
10 Example: Calculate the pressure inside a television picture tube which has a volume of 5.0 L at a temperature of 25 C, and it contains 0.010 mg of nitrogen. V = 5.0 L, T = 25 + 273.15 = 298.15 K R = 0.08206 atm L/K mol n = 0.010 0.001/28 = 3.57 10-7 mol By the ideal gas law, the pressure is P = nrt V = (3.57 10 7 )(0.08206)(298.15) 5.0 = 1.75 10 6 atm
11 Example: What will be the volume of a 0.70 g of nitrogen gas at standard temperature and 150 kpa pressure? Standard temperature = 273.15 K Pressure: 1 atm = 101.325 kpa n = 0.70 / 28.02 = 0.02498 mol The volume of the N 2 gas is: V = nrt P = (0.02498)(8.314)(273.15) 150000 = 3.7819 10 4 m 3 Note that 1 m 3 = 1000 L. Hence, the volume can be written as 0.378 L instead.
12 Example: 2.00 g of acyclic, saturated hydrocarbon gave 400 cm 3 of vapor at 523 K and one atmospheric pressure. Determine the formula of this compound. Since 1 L = 1000 cm 3, 400 cm 3 is equivalent to 0.4 L. By the ideal gas law: n = PV RT = (1.0)(0.4) (0.08206)(523) = 0.00932 mol The molar mass of the compound is therefore: MW = 2.00 0.00932 = 214.59 g/mol
13 Since the compound is a saturated hydrocarbon, it has a general formula C n H 2n+2. It means, n 12.011 + 2n + 2 1.00797 = 214.59 n = 15 The molecular formula of the compound is thus C 15 H 32.
14 Example: A sample of solid lead (II) nitrate was heated, yielding oxygen, nitrogen dioxide and a yellow solid. The total volume of gases collected at 200 C and 1.0 atm was 360 cm 3. What is the mass of the lead (II) nitrate sample used? The reaction is 2Pb(NO 3 ) 2 2PbO + 4NO 2 + O 2 Two moles of Pb(NO 3 ) 2 yield altogether 5 moles of gas.
15 The number of moles of gas n = PV RT = (1.00)(0.360) (0.08206)(273.15 + 200) = 9.272 10 3 mol Using the mole ratio, the mass of Pb(NO 3 ) 2 sample can be calculated: m = 2 5 9.272 10 3 331.21 = 1.228 g
16 Example: Two flasks each of equal volume are connected by a tube of negligible volume. Initially both flasks are at 27 C and contains 0.7 mol of hydrogen gas, the pressure being half atmosphere. One of the flasks is then immersed into a hot bath 127 C while the other is kept at 27 C. Calculate the final pressure of, and the number of moles of hydrogen in each flask.
17 Note that the total number of moles of hydrogen gas is preserved. Since the two flasks are connected, the pressure in flask 1 should be equal to the pressure in flask 2 at the end. By conservation of moles: 0.7 + 0.7 = n 1 + n 2 0.5 V R(273.15 + 27) + 0.5 V R(273.15 + 27) = PV R(273.15 + 127) + PV R(273.15 + 27) Cancelling V and R, we obtain 1 300.15 = P 1 400.15 + 1 300.15
18 It yields P = 0.5714 atm. Using this, we can work out the numbers of moles of hydrogen in the two flasks: and n 1 n 2 = n 1 + n 2 = 1.4 PVΤR(400.15) PVΤR(300.15) = 300.15 400.15 ቊ n 1 = 0.6 mol n 2 = 0.8 mol
19 3. Dalton s law of partial pressures For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were present alone in the container. This statement is called the Dalton s law of partial pressures. Mathematically, it states that P total = P 1 + P 2 + P 3 + where P 1, P 2, P 3, and so on, are the partial pressures of different gases in the mixture.
20 The Dalton s law of partial pressures can be interpreted using the ideal gas law. Note that the total number of moles of gases in the mixture is the sum of the number of moles of each gas. That means, n total = n 1 + n 2 + n 3 + Since the mixture is in a container kept at a constant temperature, V and T are the same for all gases. Hence, P total V RT = P 1V RT + P 2V RT + P 3V RT + So, P total = P 1 + P 2 + P 3 +
21 From this, it can be seen that for gas i: Dividing it by P total yields P i = n irt V P i = P total Rearranging: n i RTΤV n total RTΤV = n i n total P i = n i n total P total = X i P total
22 The quantity X i is called the mole fraction of i. Therefore, the partial pressure of a gas in a mixture is equal to the mole fraction of that gas multiplied by the total pressure of gas mixture.
23 Example: A mixture of 2.0 g of methane and 15.5 g of disilane (Si 2 H 6 ) is placed in a flask. The total pressure exerted by this mixture is 0.980 atm. Calculate the partial pressure of each gas. The numbers of moles of the two gases are: n CH 4 = 2.0 16.0 = 0.125 n Si 2 H 6 = 15.5 62.2 = 0.249 The total number of moles of gases: n total = 0.125 + 0.249 = 0.374
24 The partial pressure of methane: P CH 4 = 0.125 0.349 The partial pressure of disilane: 0.98 = 0.3275 atm P Si 2 H 6 = 0.249 0.374 0.98 = 0.6525 atm
25 Example: 10.0 g of hydrogen and 64.0 g of oxygen are contained in a 10.0 L flask at 200 C. Calculate the total pressure of the mixture. If a spark ignites the mixture, what will be the final pressure, if the temperature remains at 200 C? The numbers of moles of gases are: n H 2 = 10.0 2.0 = 5 n O 2 = 64.0 32.0 = 2 n total = 5 + 2 = 7
26 By the ideal gas law: P = nrt V = (7)(0.08206)(473.15) (10.0) = 27.2 atm The reaction occurring on sparking is 2H 2 + O 2 2H 2 O The amount of gas left at the end is: n total = n H 2 + n H 2 O = 1 + 4 = 5 Hence, the final pressure is: P = nrt V = (5)(0.08206)(473.15) (10.0) = 19.4 atm
27 Practice: A 2.69 g sample of PCl 5 was placed in an evacuated 1.00 L flask and completely vaporized at a temperature of 250 C. The pressure observed at this temperature was 1.00 atm. Some of the PCl 5 sample may have dissociated to PCl 3 and Cl 2. What are the partial pressure of PCl 5, PCl 3 and Cl 2 under these experimental conditions? [P(PCl 5 )=0.108 atm, P(PCl 3 )=0.446 atm, P(Cl 2 )=0.446 atm]
28 4. Van der Waals equation In the discussion of the ideal gas law, we always assume that gas behaves ideally; that is, it fulfils the following postulates: (a) Volume occupied by gas molecules is negligible. (b) No attraction between particles. (c) Elastic collision between particles. (d) The average speed of particles is directly proportional to the square root of absolute temperature. (e) The average KE is directly proportional to absolute temperature.
29 In reality: (a) Every particle occupies a certain volume. (b) Interaction exists between particles. It may be either attractive (e.g. London force) or repulsive (e.g. electrostatic). Particles are not independent. Therefore, gases behave ideally only when: (a) pressure is low (b) temperature is high
30 At low temperature and high pressure, gases deviate from ideal gas behavior. The degree of deviation is measured by a quantity called compressibility factor. Z = PV nrt
31 To modify the ideal gas law to suit the behavior of real gases, we recall: (a) Since each particle takes up some space, the actual volume available for gas particles to move is The actual free volume around the gas particles V actual = V container nb The volume of the container in which the gas is kept Number of moles of gas Volume occupied by gas particles; to be determined empirically
32 (b) Due to the attraction between particles, the actual pressure exerted by the particles on the surface of the container is reduced. The amount of reduction is proportional to the square of the density of the gas; therefore, P observed = P real a n2 V 2 Density of the gas Pressure measured in experiment The real pressure exerted by the gas The correction constant (depending on the gas)
33 Putting these modifications together yields the van der Waals equation, the equivalent of the ideal gas law to real gas. P an2 V 2 V nb = nrt The parameters a and b are determined experimentally.
34 The following is a table that summarizes the van der Waals parameters for some common gases.
35 Example: Calculate the pressure exerted by 0.2500 mol of Krypton in a 1.00 L container at 25 C. Given a = 2.32 atm L 2 /mol 2, b = 0.0398 L/mol. [6.03 atm]