Properties of gases Chapter 6 Gases Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1050.php Gases expand to fill and take the shape of the container they are in. Gases can diffuse into each other and mix in all proportions. We can fully describe the physical state of a gas by knowing any three of four variables related to each other by a gas equation of state. All media copyright of their respective owners 2 State variables of a gas Amount - we express this in moles (mol) which is related to the number of gas molecules through Avogadro s number (N A = 6.022142 x 10 23 mol -1 ) Recall that the molar mass (M) relates the mass of the gas to the number of moles. Equal masses of different gases represent a different number of moles! State variables of a gas Volume - this is the size of the container holding the gas. While we often express volume in liters (L), this unit for volume is derived from the base SI unit for distance, the meter (m) where 1 L = 1 dm 3. All media copyright of their respective owners 3 All media copyright of their respective owners 4 1
State variables of a gas Temperature - this is a reflection of the average energy available to a molecule in the system. Recall that in chemistry that we ALWAYS express temperature in kelvin (K). T(K) = t(ºc) + 273.15 All media copyright of their respective owners 5 State variables of a gas Pressure the molecules of the gas are in constant motion and often collide with the walls of the container and each other. The collisions exert a net force on the wall of the container. The SI unit of force, the newton (N) is another derived unit based on force = mass x acceleration 1 N = 1 kg m s -2 All media copyright of their respective owners 6 State variables of a gas Pressure is then the net force the gas molecules exert on a given area of the container wall. The SI derived unit for pressure is the pascal (Pa) Pressure = force / area 1 Pa = 1 N m -2 = 1 (kg m s -2 ) m -2 1 Pa = 1 kg m -1 s -2 Most of the time we would talk about pressure in kilopascal (kpa) Liquid pressure We can compare gas pressure to the pressure exerted by a liquid, since assessing the force exerted by a column of liquid is easy. All media copyright of their respective owners 7 All media copyright of their respective owners 8 2
Liquid pressure Imagine a cylinder of liquid, like a drinking straw. The liquid in the cylinder will have a certain total mass (m) determined by the density (d) of the liquid and the total volume (V) of the liquid. m = d x V All media copyright of their respective owners 9 Liquid pressure Now, the force (F) exerted at the bottom of the cylinder by the liquid is the mass multiplied by the acceleration due to gravity (g). F = m x g so F = d x V x g This force is what we think of as the weight of the liquid. All media copyright of their respective owners 10 Liquid pressure However, the volume of the liquid is related to the cross-sectional area (A) of the cylinder and the height (h) it rises up from the bottom of the cylinder V = h x A so F = d x h x A x g Liquid pressure and since P = F / A then P = (d x h x A x g) / A so P = d x h x g The pressure depends on the density of the liquid and the height it rises up the cylinder, but NOT on the area (shape) of the cylinder. All media copyright of their respective owners 11 All media copyright of their respective owners 12 3
Measuring gas pressure by liquid pressure Measuring gas pressure by liquid pressure A barometer can be used to measure the barometric pressure (P bar ) of the atmosphere when it forces a liquid up a cylinder with a closed end. Most barometers use mercury as the liquid because it is quite dense. Even so, the atmosphere of the Earth can push about 760 millimeters of mercury (mmhg) up the tube. All media copyright of their respective owners 13 All media copyright of their respective owners 14 Measuring gas pressure by liquid pressure The atmosphere pushes mercury over two and a half feet up the tube. If we used water as the liquid, it would rise about 34 feet up the tube! All media copyright of their respective owners 15 Drawback of using liquid pressure The density of a liquid changes with temperature and the acceleration of gravity can change with your specific location on Earth. We define 1 mmhg of pressure as the liquid pressure of exactly 1 mm of Hg at 273.15 K and g = 9.80655 m s -2 All media copyright of their respective owners 16 4
Problem A barometer is filled with diethylene glycol (d = 1.118 g cm -3 ) instead of mercury (d = 13.5951 g cm -3 ). The liquid height is found to be 9.25 m. What is the barometric pressure in mmhg? Problem answer P bar = 760. 7 mmhg The answer should have 3 significant figures (s.f.) since the liquid height had 3 s.f. (multiplication/division rule - text page 20). The subscript 7 is called a guard digit. It IS NOT significant, but would be used to help avoid rounding errors in later calculations involving this calculated value of P bar. We would report the answer to the correct number of s.f. as 761 mmhg. All media copyright of their respective owners 17 All media copyright of their respective owners 18 Units of pressure Units of pressure Pressure can be measured in many different units. You should practice unit conversion calculations. One atmosphere of pressure is 1 atm 1 atm = 760 Torr 760 mmhg (to 4 d.p.) 1 atm = 101,325 Pa = 101.325 kpa (these equalities are exact by def n) Some textbooks treats 1 atm as the commonly used standard unit of pressure. IT ISN T! The standard unit of pressure in chemistry is 1 bar = 10 5 Pa (exactly) 0.98692 atm All media copyright of their respective owners 19 All media copyright of their respective owners 20 5
Measuring gas pressure by liquid pressure Measuring gas pressure by liquid pressure A manometer can be used to measure a gas pressure (P gas ) compared to the atmospheric pressure (P bar ) when it forces a liquid up a cylinder with an end open to the atmosphere. P gas = P bar + P P = g x d x h h = h open - h closed All media copyright of their respective owners 21 All media copyright of their respective owners 22 Problem Suppose we have a mercury manometer (d = 13.6 g cm -3 ) like in Figure 6.5(b) of the previous slide where the height of mercury in the open arm is 7.83 mm higher than in the closed arm. If P bar is 748.2 mmhg, then what is P gas in the manometer? Problem answer P gas = 756.0 3 mmhg The answer should have 4 significant figures (s.f. to the 1 st decimal place) since P bar had significant figures to the least decimal places (addition/subtraction rule - text page 20). The subscript 3 is a guard digit. It IS NOT significant. It s good for avoiding rounding errors in later calculations using P gas. We would report the answer to the correct number of s.f. as 756.0 mmhg. All media copyright of their respective owners 23 All media copyright of their respective owners 24 6
Boyle s Law For a fixed amount of gas at a constant temperature, the gas volume is inversely proportional to the gas pressure. P Ñ 1/V or PV = a a is a constant Boyle s Law Therefore for constant n and constant T P 1 V 1 = P 2 V 2 All media copyright of their respective owners 25 All media copyright of their respective owners 26 Charles s Law For a fixed amount of gas at a constant gas pressure, the gas volume is directly proportional to the absolute (Kelvin) temperature. V Ñ T or V = bt b is a constant Charles s Law Therefore for constant n and constant P V 1 /T 1 = V 2 /T 2 All media copyright of their respective owners 27 All media copyright of their respective owners 28 7
Standard conditions Equal volumes equal numbers hypothesis To be sure people are talking about the same conditions when discussing gases, it s often useful to define a standard. When discussing gases, the set of standard conditions often used are standard temperature and pressure (STP) where T = 0 ºC = 273.15 K and P = 1 bar Equal numbers of molecules of different gases at the same T and P occupy equal volumes. All media copyright of their respective owners 29 All media copyright of their respective owners 30 Avogadro s Law Molar volume at STP At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas. V Ñ n or V = c n c is a constant Therefore for constant T and constant P V 1 /n 1 = V 2 /n 2 If we choose our constant temperature and pressure to be STP (273.15 K and 1 bar) then Avogadro s Law tells us the volume of some given amount of gas will be a constant. If we choose 1 mole of gas as this amount, then the molar volume (V m ) of a gas at STP is V m = 22.711 L mol -1 at STP All media copyright of their respective owners 31 All media copyright of their respective owners 32 8
Gas law assumptions Properties of an ideal gas Charles s Law ignores the fact that all gases will eventually condense to a liquid if the T is low enough, and then eventually form a solid if the T is lowered further. Therefore, the gas law behaviors are actually reflecting an idealized picture of a gas the ideal gas! 1) The molecules of the gas occupy no volume. (They can occupy the same space!) 2) There are no intermolecular forces between gas molecules. (No attractive or repulsive interactions!) 3) The gas molecules are in constant random motion and collide elastically. (No net change in total energy of the gas!) 4) The gas has achieved a state of equilibrium (n, P, T and V are all constant!). All media copyright of their respective owners 33 All media copyright of their respective owners 34 Combining the gas laws If V Ñ 1/P (since P Ñ 1/V) and V Ñ T and V Ñ n then V Ñ nt/p or V = RnT/P where R is gas constant Ideal gas law! PV = nrt All media copyright of their respective owners 35 Value of the gas constant R We always measure T in kelvin and n in moles, but P and V can often be expressed in many different units, therefore the gas constant R can APPEAR to have many different values depending on the units of P and V. All media copyright of their respective owners 36 9
Value of the gas constant R Consider the two statements: I have 1 dozen eggs AND I have 12 individual eggs Both statements say the SAME THING, but have different numbers appearing in them due to different units. UNITS MATTER! ALWAYS SHOW THEM! All media copyright of their respective owners 37 Value of the gas constant R Therefore, R always represents the SAME constant, even if the numbers don t always look the same. All media copyright of their respective owners 38 R in J K -1 mol -1 The first four values of R look straight forward. We have just measured P and V in various units. P in atm or bar or kpa or Pa V in L or m 3 R in J K -1 mol -1 We can convert any value of R to any other by unit conversion: 0.0820574 101325 1 1 1000 =8.31447 All media copyright of their respective owners 39 All media copyright of their respective owners 40 10
R in J K -1 mol -1 R in J K -1 mol -1 What about R in J K -1 mol -1? How do we get from P and V to energy units? Recall that 1 Pa = 1 kg m -1 s -2 and volume can be expressed as a cubic distance like m 3 All media copyright of their respective owners 41 Therefore PV can have units of kg m -1 s -2 m 3 = kg m 2 s -2 The formula for kinetic energy (E k ) is E k = ½ m v 2 If we measure mass in kg and velocity in m s -1 then E k has units of kg (m s -1 ) 2 = kg m 2 s -2 = J (joule) All media copyright of their respective owners 42 R in J K -1 mol -1 A pressure multiplied by a volume has units equivalent to energy units! 1 Pa m 3 = 1 J We ll see more in the next chapter! For now consider blowing up a balloon. To increase the volume of the balloon against the external atmospheric pressure requires us to blow air into the balloon. This takes energy to accomplish. All media copyright of their respective owners 43 General gas equation Since PV = nrt then R = PV/nT = constant For any sample of gas undergoing any sort of change in any of these variables = f is after the change final state i is before the change - initial state All media copyright of their respective owners 44 11
Problem What is the volume occupied by 20.2 g of ammonia [NH 3 (g)] at -25 ºC and 752 mmhg? All media copyright of their respective owners 45 Problem answer M = 17.0291 7 g mol -1 P = 0.989 5 atm T = 248.15 K R = 0.082057 L atm K -1 mol-1 V = 24.4 1 L = 24.4 L n = 1.18 6 mol In this problem we see the importance of keeping track of units so we can convert when necessary, using guard digits to avoid rounding errors and choosing an appropriate value of R to use based on the units of our variables. All media copyright of their respective owners 46 Problem How many molecules of N 2 (g) remain in an ultrahigh vacuum chamber of 3.45 m 3 in volume when the pressure is reduced to 6.67 x 10-7 Pa at 25 ºC? Problem answer N A = 6.022 x 10 23 (N 2 molecules) mol -1 T = 298.15 K R = 8.3145 Pa m 3 K -1 mol -1 n = 9.28 3 x 10-10 mol = 5.59 0 x 10 14 N 2 molecules = 5.59 x 10 14 N 2 molecules All media copyright of their respective owners 47 All media copyright of their respective owners 48 12
Problem A 1.00 ml sample of N 2 (g) at 36.2 ºC and 2.14 atm is heated to 37.8 ºC and the pressure is changed to 1.02 atm. What volume does the gas occupy at this final temperature and pressure? Problem answer T i = 309.35 K and T f = 310.95 K V f = 2.10 9 ml = 2.11 ml All media copyright of their respective owners 49 All media copyright of their respective owners 50 Using the ideal gas equation Finding molar mass (M) of a gas since PV = nrt and M = m/n (or n = m/m) then PV = mrt/m leading to M = mrt/(pv) Using the ideal gas equation Finding gas density (d) since M = mrt/(pv) = (RT/P)(m/V) but since m/v = d then M = (RT/P)(m/V) = drt/p leading to d = PM/RT All media copyright of their respective owners 51 All media copyright of their respective owners 52 13
Problem A 1.27 g sample of an oxide of nitrogen, believed to be either NO or N 2 O occupies a volume of 1.07 L at 25 ºC and 737 mmhg. Which oxide is it? Problem answer M NO = 30.0061 g mol -1 M N2O = 44.0128 g mol -1 M calculated = 29.9 4 g mol -1 = 29.9 g mol -1 The oxide of nitrogen must be NO. All media copyright of their respective owners 53 All media copyright of their respective owners 54 Problem The density of a gas sample is 1.00 g L -1 at 109 ºC and 745 mmhg. What is the molar mass of the gas? Problem answer M = 31.9 8 g mol -1 = 32.0 g mol -1 All media copyright of their respective owners 55 All media copyright of their respective owners 56 14
Gases in chemical reactions Problem If we treat any gases in a chemical reaction of interest as ideal gases, then we can use the ideal gas equation and the balanced equation of the reaction to solve reaction stoichiometry problems, if we know something about the state variables of the gas. How many grams of Na (l) are produced per liter of N 2 (g) at 25 ºC and 1.0 bar. 2 2 +3 ( ) All media copyright of their respective owners 57 All media copyright of their respective owners 58 Problem answer n = 0.040 3 mol of N 2 (g) in 1 L at the given T and P m = 0.61 8 g of Na (l) per liter of N 2 (g) produced Law of combining volumes In reactions involving ALL gases (treated ideally) at a given T and P, since V Ñ n then the ratios of volumes of the gases match the stoichiometric ratios for the same gases All media copyright of their respective owners 59 All media copyright of their respective owners 60 15
Law of combining volumes Law of combining volumes If 2 NO (g) + O 2 (g) 2 NO 2 (g) then 2 moles of NO react with 1 mole of O 2 to form 2 moles of NO 2 (2:1:2 stoichiometry) but the law of combining volumes tells us that 2 L of NO will react with 1 L of O 2 to give 2 L of NO 2 (still 2:1:2, but in terms of volume) More generally, if 2 NO (g) + O 2 (g) 2 NO 2 (g) then 2V m NO (g) + 1V m O 2 (g) 2V m NO 2 (g) where V m is the molar volume of an ideal gas at the given T and P of the reaction All media copyright of their respective owners 61 All media copyright of their respective owners 62 Problem If all gases are measured at the same temperature and pressure, what volume of NH 3 (g) is produced when 225 L of H 2 (g) are consumed in the reaction + ( ) ( ) not a balanced equation! Problem answer Since balancing the equation shows us that 3 moles of H 2 are required to produce 2 moles of NH 3, then the law of combining volumes tells us that 3 L of H 2 will produce 2 L of NH 3. Since we used 225 L of H 2 we produced 2/3 (225 L) = 150 L NH 3 All media copyright of their respective owners 63 All media copyright of their respective owners 64 16
Mixtures of gases If we treat all gases in a mixture as ideal, then by the properties of an ideal gas, none of the gases behave any differently from any of the others, and each gas can be treated separately and the properties of mixture can be treated as a sum of the same properties of the individual gases. Law of partial pressures Dalton s law of partial pressures - The total pressure of a mixture of (ideal) gases P tot is the sum of the pressures of each gas treated individually (partial pressure) at the same temperature and volume as the mixture. In a mixture of gases A, B, C and D P tot = P A + P B + P C + P D here P A is the partial pressure of A All media copyright of their respective owners 65 All media copyright of their respective owners 66 Law of partial pressures Law of partial pressures In a mixture of gases A, B, C and D P tot = P A + P B + P C + P D here P A is the partial pressure of A Be careful here! Since the mixture occupies a fixed volume, then V tot = V A = V B = V C = V D All media copyright of their respective owners 67 All media copyright of their respective owners 68 17
Mixtures of gases In a mixture of gases A, B, C and D V tot = V A + V B + V C + V D so percent volume of A is V A /V tot x 100% Be careful here! The mixture is at a fixed total pressure so Mixtures of gases In a mixture of gases A, B, C and D n tot = n A + n B + n C + n D leads to mole fraction of A (x A ) n A /n tot = x A P tot = P A = P B = P C = P D All media copyright of their respective owners 69 All media copyright of their respective owners 70 Mole fraction in ideal gas mixture Problem = = = A gas mixture is made when 2.0 L of O 2 and 8.0 L of N 2, each at 0.00 ºC and 1.00 atm, are mixed together. This non-reactive mixture is then compressed to occupy a total volume of 2.0 L at 298 K. What is the pressure of the mixture at these new conditions? All media copyright of their respective owners 71 All media copyright of their respective owners 72 18
Problem answer Collecting a gas over water n O2 = 0.089 2 mol n N2 = 0.35 7 mol n tot = 0.44 6 mol P tot = 5.4 5 atm = 5.5 atm We can collect a gas over water (or another liquid) via a setup like we see here. As more gas is collected, the liquid is displaced downwards as the pressure of the gases in the container must match P bar. All media copyright of their respective owners 73 All media copyright of their respective owners 74 Collecting a gas over water Collecting a gas over water We must be careful though! The gas we re interested in IS NOT THE ONLY GAS IN THE CONTAINER. The liquid water can evaporate to give gaseous water in the container as well! Therefore P tot = P gas + P water = P bar Therefore, for the gas we re collecting P gas = P bar - P water All media copyright of their respective owners 75 All media copyright of their respective owners 76 19
So what is P water? The partial pressure of the water that must be subtracted from P bar to get P gas is called the vapour pressure of water. Vapour pressure changes as a function of temperature and identity of the liquid. Problem The reaction of aluminum with hydrochloric acid produces hydrogen gas: 2 Al (s) + 6 HCl (aq) 2 AlCl 3 (aq) + 3 H 2 (g) If 35.5 ml of H 2 is collected over water at 26 ºC and a barometric pressure of 755 mmhg, how many moles of HCl must have been consumed? P water = 25.2 mmhg at 26 ºC All media copyright of their respective owners 77 All media copyright of their respective owners 78 Problem answer n HCl = 2.77 7 x 10-3 mol = 0.00278 mol Kinetic molecular theory of gases This is a theory we use to explain the behaviour of gases, based on a model with certain properties (compare these to the description of an ideal gas in slide 34). All media copyright of their respective owners 79 All media copyright of their respective owners 80 20
Properties of the model Properties of the model i. A gas is made up of an extremely large number of very small particles (molecules or atoms) moving in constant, random, straight-line motion. ii. The distance between any two molecules is vast in comparison to the size of the molecules that most of the space is empty and the molecules can be treated as point masses (no volume) All media copyright of their respective owners 81 iii. Molecules collide fleetingly with each other and the container walls. iv. No intermolecular forces EXCEPT during collisions (no action at a distance of one molecule on another). v. A given molecule may gain or lose energy during collisions, but if the temperature is constant then the TOTAL energy is ALSO constant. All media copyright of their respective owners 82 Deriving Boyle s Law We ve seen PV = a or P = a / V Here the constant a will depend on the number of molecules N = n x N A (extremely large number of molecules! see note i. on slide 81) and the temperature (see note v. on slide 82) Deriving Boyle s Law If a given molecule of mass m is travelling with a speed u x in the x direction (speed is just the size [magnitude] of the vector quantity velocity) towards a wall perpendicular to it s path (that is, the yz plane), then the force this molecule exerts on the wall depends on 2 factors: All media copyright of their respective owners 83 All media copyright of their respective owners 84 21
Deriving Boyle s Law Frequency of collisions - measures the number of collisions of molecules with the walls of the container in a given time period. More collisions per unit time implies a greater net force from collisions: collision frequency Ñ molecular speed x molecules per unit volume collision frequency Ñ u x x (N/V) Deriving Boyle s Law Momentum transfer (impulse) in a collision the molecule transfers momentum impulse Ñ molecular speed x mass impulse Ñ u x x m Now P Ñ collision frequency x impulse P Ñ u x x (N/V) x u x x m Ñ (N/V) m u x 2 All media copyright of their respective owners 85 All media copyright of their respective owners 86 Deriving Boyle s Law Deriving Boyle s Law Remember, we have a very large number of molecules, each with its own u x2. So it makes sense to talk about the average mean-square speed in the x- direction is and Molecules travel in 3 dimensions! So average mean-square speed is = + + and if = since = = = 1 3 All media copyright of their respective owners 87 All media copyright of their respective owners 88 22
Distribution of molecular speeds We can statistically predict what fraction F(u) of our N molecules have a given speed u =4 2 Distribution of molecular speeds Here we have a plot of the percentage (fraction) of molecules that have a given speed. Notice that no molecules have zero speed. We ll talk about other features soon. All media copyright of their respective owners 89 All media copyright of their respective owners 90 Distribution of molecular speeds =4 2 As temperature increases the average speed of any given molecule increases. (Constant M) As the molar mass increases, the average speed of any given molecule decreases. (Constant T) All media copyright of their respective owners 91 Distribution of molecular speeds This figure shows three different characteristic speeds of the distribution. The most probable speed u m is the speed at the peak of the distribution, and more molecules have this speed compared to any other speed. All media copyright of their respective owners 92 23
Distribution of molecular speeds The average speed u av is literally the sum of the speeds of all molecules divided by the total number of molecules. u m u av because the distribution curve is not symmetric Distribution of molecular speeds The root-meansquare speed u rms is found by taking the square root of the average mean-square speed. = All media copyright of their respective owners 93 All media copyright of their respective owners 94 Deriving Boyle s Law Consider = 1 3 For N = N A molecules (so we have 1 mol) = 1 3 Deriving Boyle s Law But So = for 1 mol = 1 3 But N A m is the mass of one mole of molecules the molar mass M! 3 = All media copyright of their respective owners 95 All media copyright of their respective owners 96 24
Deriving Boyle s Law Deriving Boyle s Law 3 = Taking the square root of both sides 3 = = 3 = = Heavier molecules (larger M) have a smaller u rms than lighter molecules at the same T. The u rms of a sample of molecules increases with the square root of temperature. All media copyright of their respective owners 97 All media copyright of their respective owners 98 Problem Problem answer Which has a greater u rms at 25 ºC, NH 3 or HCl? Calculate u rms for the gas with the higher u rms. NH 3 has the smaller molar mass, so it will have the larger u rms. At 25 ºC u rms (NH 3 ) = 660. 8 m s -1 = 661 m s -1 All media copyright of their respective owners 99 All media copyright of their respective owners 100 25
The meaning of temperature The meaning of temperature The kinetic energy of a moving object is E k = ½ mv 2 Here v is the velocity of the object, but remember, the speed (u) is just the magnitude (size) of the velocity, so for a given molecule, the kinetic energy is E k = ½ mu 2 But if we look at the mean-square speed u 2 of a group of molecules then the average kinetic energy of a molecule is But = = 1 3 = 2 3 1 2 = 2 3 All media copyright of their respective owners 101 All media copyright of their respective owners 102 The meaning of temperature 3 = 2 This is why on slide 5 I said the temperature was a reflection of the average (translational kinetic) energy available to a molecule in the system! All media copyright of their respective owners 103 Diffusion and effusion Because molecules are constantly colliding, it takes time for them to travel large distances, even if their average speed is pretty fast. Diffusion is the name we give to this molecular migration due to the random motion of the colliding molecules. Effusion is the escape of the randomly moving gas molecules through a small hole in the container. All media copyright of their respective owners 104 26
Diffusion and effusion Rate of effusion Since the rate of effusion is directly proportional to molecular speed, which is inversely proportional to the square root of molar mass, lighter molecules diffuse faster than heavier molecules rate effusion A rate effusion B =,, = 3 3 = All media copyright of their respective owners 105 All media copyright of their respective owners 106 In general Since many gas properties are directly proportional to molecular speed, which is inversely proportional to the square root of molar mass, the ratio of those properties for two gases is property of A property of B = The properties include molecular speed, effusion rate, distance travelled and amount of gas effused. All media copyright of their respective owners 107 Time of effusion Since the time it takes for a specific amount of gas to effuse is inversely proportional to the rate of effusion AND rate is inversely proportional to the molar mass, then the time of effusion must be directly proportional to the molar mass. That is time of effusion A time of effusion B = All media copyright of their respective owners 108 27
Problem If 2.2 x 10-4 mol of N 2 effuse through a hole in 105 s, then how long would it take for the same amount of H 2 to effuse through the same hole? Problem answer time = 28.2 s All media copyright of their respective owners 109 All media copyright of their respective owners 110 Problem A sample of Kr(g) effuses through a hole in 87.3 s. The same amount of an unknown gas escapes in 131.3 s through the same hole. What is the molar mass of the unknown gas? Problem answer M = 189. 5 g mol -1 = 1.90 x 10 2 g mol -1 All media copyright of their respective owners 111 All media copyright of their respective owners 112 28
Nonideal (real) gases The properties of an ideal gas on slide 34 include some properties that cannot be really true for gases in reality namely that real molecules must have size and must have intermolecular interactions (attractions and repulsions) with each other. First consider the molar volume at STP Compressibility factor Recall the ideal gas law is PV = nrt So for an ideal gas (PV)/(nRT) = 1 = z for ALL conditions We call z the compressibility factor. However, for real gases at almost any set of conditions z 1 due to real behaviour! All media copyright of their respective owners 113 All media copyright of their respective owners 114 Compressibility factor Compressibility factor Real gases show very close to ideal gas behaviour at high T and low P (or large V) where the chance for intermolecular interactions (attractions and repulsions) is fleeting due to fast moving molecules that are very far apart on average. When T is low and/or P is high (or V is low), the chance for interactions is much greater, leading to nonideal behaviour. All media copyright of their respective owners 115 All media copyright of their respective owners 116 29
Molecular volume The volume of the container available to a given molecule is actually the volume of the container MINUS the total volume occupied by all the other molecules. V available = V container - nb Here b represents the excluded volume occupied by 1 mol of the real gas molecules Attractive intermolecular forces Real molecules can attract each other over reasonable distances. These attractions serve to decrease the rate AND force of collisions with the walls, lowering pressure P lowering = an 2 /V 2 Here a represents the attractive forces There is a dependence on n 2 and V 2 because both the rate and force of collisions are affected All media copyright of their respective owners 117 All media copyright of their respective owners 118 The van der Waals equation If the ideal gas law is an equation of state for an ideal gas, then real gases cannot be described adequately by this equation of state. One equation of state used to describe real gas behaviour is the van der Waals (vdw) equation + = All media copyright of their respective owners 119 The van der Waals equation Rearranging the vdw equation makes it a bit easier to see the result of intermolecular forces = The attractive forces reduce the pressure, while the excluded volume (repulsive forces) increase the pressure compared to an ideal gas at the same temperature and volume. On average, ONLY attractive OR repulsive forces dominate at a given set of conditions. All media copyright of their respective owners 120 30
vdw factors a and b Problem We see that the values of a and b increase with molar mass, for the most part. For molecules with similar molar masses, like N 2 and CO or N 2 O and CO 2, the less symmetric molecule usually has a larger a. Given the vdw factors for CO 2 from the table on the previous slide, what is the pressure and compressibility factor of 1.00 mol of CO 2 in a 2.00 L container at 273 K? All media copyright of their respective owners 121 All media copyright of their respective owners 122 Problem answer P = 10.6 8 bar z = 0.941 All media copyright of their respective owners 123 31