Chapter 10. Gases THREE STATES OF MATTER. Chapter 10 Problems 6/29/2012. Problems 16, 19, 26, 33, 39,49, 57, 61
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1 Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 10 John Bookstaver St. Charles Community College Cottleville, MO Chapter 10 Problems Problems 16, 19, 26, 33, 39,49, 57, 61 THREE STATES OF MATTER 1
2 General Properties of There is a lot of free space in a gas. can be expanded infinitely. occupy containers uniformly and completely. diffuse and mix rapidly. : What Are They Like? Composed of widely separated particles in constant, random motion Flow readily and occupy the entire volume of their container Vapor is the term used to denote the gaseous state of a substance existing more commonly as a liquid e.g., water is a vapor, oxygen is a gas Many low molar mass molecular compounds are gases methane (CH 4 ), carbon monoxide (CO) EOS Common EOS 2
3 Properties of Gas properties can be modeled using math. Model depends on V = volume of the gas (L) T = temperature (K) n = amount (moles) P = pressure (atmospheres) Properties of : Gas Pressure The gaseous states of three halogens. Most common gases are colorless H 2, O 2, N 2, CO and CO 2 The Concept of Pressure The pressure exerted by a solid. Both cylinders have the same mass They have different areas of contact Force (N) P (Pa) = Area (m2 ) 3
4 Gas Pressure Pressure is the force per unit area consider the unit pounds per square inch SI units express pressure in Newtons (N) per square meters (m 2 ) -- or N m 2 a.k.a. Pascals (Pa) A barometer is an instrument used to measure atmospheric pressure EOS Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643) Units of Pressure mm Hg or torr These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury. Atmosphere 1.00 atm = 760 torr 4
5 Pressure Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). P of Hg pushing down related to Hg density column height Pressure Column height measures P of atmosphere 1 standard atm = 760 mm Hg = 29.9 inches Hg = about 34 feet of water SI unit is PASCAL, Pa, where 1 atm = kpa Examples of Pressure Units Given these values, one can generate conversion factors to switch between units: e.g., 760 mmhg = bar 760 mmhg bar or bar 760 mmhg EOS 5
6 Liquid Pressure The pressure exerted by a liquid depends on: The height of the column of liquid. The density of the column of liquid. P = g h d Barometers Used to measure atmospheric pressure The pressure exerted by a column of mercury exactly 760 mm high is defined as 1 atmosphere (atm) tend to settle under the effects of gravity pressure as altitude EOS Barometric Pressure Standard Atmospheric Pressure 1.00 atm, 760 mm Hg, 760 torr, kpa, bar 6
7 Manometer This device is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel. Open-Ended Manometers Open-ended manometers compare gas pressure to barometric pressure Column height differences are proportional to gas pressure P gas = P bar + Dh EOS Manometers 7
8 Standard Pressure Normal atmospheric pressure at sea level is referred to as standard pressure. It is equal to 1.00 atm 760 torr (760 mm Hg) kpa Boyle 1662 Simple Gas Laws P 1 V PV = constant Pressure-Volume Relationship: Boyle s Law For a given amount of a gas at constant temperature, the volume of the gas varies inversely with its pressure i.e., if V, then P EOS 8
9 Boyle s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure. As P and V are inversely proportional A plot of V versus P results in a curve. Since PV = k V = k (1/P) This means a plot of V versus 1/P will be a straight line. EXAMPLE Relating Gas Volume and Pressure Boyle s Law. The volume of a large irregularly shaped, closed tank can be determined. The tank is first evacuated and then connected to a 50.0 L cylinder of compressed nitrogen gas. The gas pressure in the cylinder, originally at 21.5 atm, falls to 1.55 atm after it is connected to the evacuated tank. What is the volume of the tank? 9
10 EXAMPLE P 1 V 1 = P 2 V 2 V 2 = P 1V 1 P 2 = 694 L V tank = 644 L Charles s Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. i.e., V T = k A plot of V versus T will be a straight line. Temperature-Volume Relationship: Charles s Law The volume of a fixed amount of a gas at constant pressure is directly proportional to its Kelvin temperature i.e., if V, then T or V/T = k EOS 10
11 Temperature-Volume Relationship: Charles s Law Absolute zero is the temperature obtained by extrapolation to zero volume Absolute zero on the Kelvin scale = C and K = 0 C EOS Avogadro s Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Mathematically, this means V = kn Avogadro s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related. twice as many molecules 11
12 Molar Volumes and Standard Pressure and Temperature Standard Temperature and Pressure (STP) is defined as T = 0 o C and P = 1 atm The molar volume of a gas is the volume occupied by one mole of the gas at STP EOS Avogadro s Law At an a fixed temperature and pressure: V n or V = c n At STP 1 mol gas = 22.4 L gas Ideal-Gas Equation So far we ve seen that V 1/P (Boyle s law) V T (Charles s law) V n (Avogadro s law) Combining these, we get V nt P 12
13 The Ideal Gas Law The constant of proportionality (k) is given the symbol R PV k R nt For 1 mol of an ideal gas at STP R = L atm mol 1 K 1 EOS Ideal-Gas Equation The constant of proportionality is known as R, the gas constant. Ideal-Gas Equation The relationship V nt P then becomes V = R or nt P PV = nrt 13
14 The Combined Gas Law Given the various gas laws, all can be combined into a single form V = k/p, V = kt, and V = kn V a (nt)/p PV k nt For initial and final conditions: PV PV k nt nt 1 2 EOS The General Gas Equation P R = = P 1 V 1 2V 2 n 1 T 1 n 2 T 2 If we hold the amount and volume constant: P 1 T 1 = P 2 T 2 Using the Gas Laws 14
15 6-4 Applications of the Ideal Gas Equation Molar Mass Determination PV = nrt and n = m M PV = m M RT M = m RT PV EXAMPLE Determining a Molar Mass with the Ideal Gas Equation. Polypropylene is an important commercial chemical. It is used in the synthesis of other organic chemicals and in plastics production. A glass vessel weighs g when clean, dry and evacuated; it weighs when filled with water at 25 C (δ water = g cm -3 ) and g when filled with propylene gas at mm Hg and 24.0 C. What is the molar mass of polypropylene? Strategy: Determine V flask. Determine m gas. Use the Gas Equation. EXAMPLE Determine V flask : V flask = m H2O d H2O = ( g g) ( g cm -3 ) = cm 3 = L Determine m gas : m gas = m filled - m empty = ( g g) = g 15
16 EXAMPLE Use the Gas Equation: PV = nrt PV = m M RT M = m RT PV M = ( g)( L atm mol -1 K -1 )(297.2 K) ( atm)( L) M = g/mol Densities of If we divide both sides of the ideal-gas equation by V and by RT, we get n V = P RT Densities of We know that moles molecular mass = mass n = m So multiplying both sides by the molecular mass ( ) gives m V = P RT 16
17 Densities of Mass volume = density So, d = m V = P RT Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas. GAS DENSITY PV = nrt n V = P RT m M V = P RT where M = molar mass and density (d) = m/v d = m V = PM RT d and M proportional Gas Densities are much less dense than liquids and solids Because of the magnitude of the value, densities of gases are reported in g/l M At STP, d 22.4 L At other conditions, use the combined gas law MP d RT the density of a gas is directly proportional to molar mass and pressure, and inversely proportional to its Kelvin temperature EOS 17
18 in Chemical Reactions Stoichiometric factors relate gas quantities to quantities of other reactants or products. Ideal gas equation relates the amount of a gas to volume, temperature and pressure. Law of combining volumes can be developed using the gas law. and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself. and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution Strategy: Calculate moles of H 2 O 2 and then moles of O 2 and H 2 O. Finally, calc. P from n, R, T, and V. 18
19 and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution 1.1 g H 2 O 2 1 mol 34.0 g mol mol H 2 O 2 1 mol O 2 = mol O 2 mol H 2 O 2 2 and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution of O 2 = nrt/v (0.016 mol)( L atm/k mol)(298 K 2.50 L P of O 2 = 0.16 atm and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) What is P of H 2 O? Could calculate as above. But recall Avogadro s hypothesis. V n at same T and P P n at same T and V There are 2 times as many moles of H 2 O as moles of O 2. P is proportional to n. Therefore, P of H 2 O is twice that of O 2. P of H 2 O = 0.32 atm 19
20 Mixtures of Gas laws apply to mixtures of gases. Simplest approach is to use n total, but... Partial pressure Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone. Dalton s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, P total = P 1 + P 2 + P 3 + Dalton s Law John Dalton
21 Partial Pressure P tot = P a + P b + V a = n a RT/P tot and V tot = V a + V b + V a V tot n a RT/P = tot = n tot RT/P tot n a n tot Recall n a n tot = a P a P tot n a RT/V = tot = n tot RT/V tot n a n tot Mole Fraction Consider the ratio of a component s partial pressure to total pressure V, R, and T are all constant and drop out P1 n1 n1 x P tot n1 n 2... n tot 1 The mole fraction (x 1 ) is the fraction of all the molecules in a mixture that are of a given type EOS Collection of over Water As essentially insoluble gas is passed into a container of water, the gas rises because its density is much less than that of water and the water must be displaced EOS 21
22 Collection of over Water Assuming the gas is saturated with water vapor, the partial pressure of the water vapor is the vapor pressure of the water. P total = P gas + P H2O(g) P gas = P total P H2O(g) = P bar P H2O(g) EOS Pneumatic Trough P tot = P bar = P gas + P H2O Partial Pressures When one collects a gas over water, there is water vapor mixed in with the gas. To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure. 22
23 Vapor Pressure as a Function of Temperature The combined gas law shows the relationship between P and T at constant n and V: As with Charles s law for V and T, P and T are directly proportional P nr T V EOS Kinetic-Molecular Theory This is a model that aids in our understanding of what happens to gas particles as environmental conditions change. Main Tenets of Kinetic- Molecular Theory consist of large numbers of molecules that are in continuous, random motion. 23
24 Main Tenets of Kinetic- Molecular Theory The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained. Main Tenets of Kinetic- Molecular Theory Attractive and repulsive forces between gas molecules are negligible. KINETIC MOLECULAR THEORY (KMT) Theory used to explain gas laws. KMT assumptions are consist of molecules in constant, random motion. P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible. 24
25 Kinetic Molecular Theory Because we assume molecules are in motion, they have a kinetic energy. KE = (1/2)(mass)(speed) 2 At the same T, all gases have the same average KE. As T goes up for a gas, KE also increases and so does speed. Kinetic Molecular Theory At the same T, all gases have the same average KE. As T goes up, KE also increases and so does speed. Main Tenets of Kinetic- Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant. 25
26 Kinetic Molecular Theory Maxwell s equation u 2 3RT M root mean square speed where u is the speed and M is the molar mass. speed INCREASES with T speed DECREASES with M Distribution of Gas Molecule Speeds Boltzmann plots Named for Ludwig Boltzmann doubted the existence of atoms. This played a role in his suicide in Velocity of Gas Molecules Molecules of a given gas have a range of speeds. 26
27 Velocity of Gas Molecules Average velocity decreases with increasing mass. Main Tenets of Kinetic- Molecular Theory The average kinetic energy of the molecules is proportional to the absolute temperature. Effusion Effusion is the escape of gas molecules through a tiny hole into an evacuated space. 27
28 Effusion The difference in the rates of effusion for helium and nitrogen, for example, explains a helium balloon would deflate faster. Diffusion Diffusion is the spread of one substance throughout a space or throughout a second substance. GAS DIFFUSION AND EFFUSION Graham s law governs effusion and diffusion of gas molecules. Rate for A Rate for B M of B M of A Rate of effusion is inversely proportional to its molar mass. Thomas Graham, Professor in Glasgow and London. 28
29 Effusion At a given temperature, the rates of effusion of gas molecules are inversely proportional to the square roots of their molar masses rate rate 3RT M 3RT M M M EOS rateof effusion of rateof effusion of B Graham s Law A (u (u ) ) rms A rms B 3RT/M A 3RT/MB M M B A Only for gases at low pressure (natural escape, not a jet). Tiny orifice (no collisions) Does not apply to diffusion. Ratio used can be: Rate of effusion (as above) Distances traveled by Molecular speeds molecules Effusion times Amounts of gas effused. Real Under many conditions, real gases do not follow the ideal gas law Intermolecular forces of attraction cause the measured pressure of a real gas to be less than expected -- When molecules are close together, the volume of the molecules themselves becomes a significant fraction of the total volume of a gas 2 na P 2 V nb nrt V van der Waals equation EOS 29
30 Deviations from Ideal Gas Law Real molecules have volume. There are intermolecular forces. Otherwise a gas could not become a liquid. Deviations from Ideal Gas Law Account for volume of molecules and intermolecular forces with VAN DER WAALS s EQUATION. Measured V = V(ideal) Measured P ( n 2 a P ) V 2 intermol. forces V - nb vol. correction nrt J. van der Waals, , Professor of Physics, Amsterdam. Nobel Prize Deviations from Ideal Gas Law Cl 2 gas has a = 6.49, b = For 8.0 mol Cl 2 in a 4.0 L tank at 27 o C. P (ideal) = nrt/v = 49.3 atm P (van der Waals) = 29.5 atm 30
31 van der Waals Equation Corrections for real gas behavior are made using the parameters a and b a accounts for intermolecular attractions in real gases b accounts for the real volumes of gases EOS Real In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure. Real Even the same gas will show wildly different behavior under high pressure at different temperatures. 31
32 Deviations from Ideal Behavior The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure and/or low temperature. Corrections for Nonideal Behavior The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account. The corrected ideal-gas equation is known as the van der Waals equation. The van der Waals Equation n 2 a (P + ) (V nb) = nrt V 2 32
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