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1 Vanessa Prasad-Permaul Valencia Community College CHM 1045

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2 Elements that exist as gases at 25 0 C and 1 atmosphere

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3 Physical Characteristics of Gases Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids.

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4 a) Gas is a large collection of particles moving at random throughout a volume b) Collisions of randomly moving particles with the walls of the container exert a force per unit area that we perceive as gas pressure

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5 HOW IS PRESSURE DEFINED? The force the gas exerts on a given area of the container in which it is contained. The SI unit for pressure is the Pascal, Pa. Pressure = Force Area If you’ve ever inflated a tire, you’ve probably made a pressure measurement in pounds (force) per square inch (area).

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6 Units of Pressure 1 pascal (Pa) = 1 kg/m · s 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 1 bar = 10 5 Pa 1 atm = lb/in 2 Barometer 760mm mercury Height is proportional to the barometric pressure Hg is used instead of H 2 O : more dense better visibility accuracy

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EXERCISE 5.1 A GAS CONTAINER HAD A MEASURED PRESSURE OF 57kPa. CALCULATE THE PRESSURE IN UNITS OF ATM AND mmHg. First, convert to atm (57 kPa = 57 x 10 3 Pa). 57 x 10 3 Pa x 1 atm = = 0.56 atm x 10 5 Pa Next, convert to mmHg. 57 x 10 3 Pa x 760 mmHg = = 4.3 x 10 2 mmHg x 10 5 Pa 7

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8 EMPIRICAL GAS LAWS You can predict the behavior of gases based on the following properties: Pressure Volume Amount (moles) Temperature * If two of these physical properties are held constant, it is possible to show a simple relationship between the other two…

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9 As P (h) increases V decreases Hg Boyle’s experiment: A manometer to study the relationship between pressure (P) & Volume (V) of a gas

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10 BOYLE’S LAW the volume of a sample of gas at a given temperature varies inversely with the applied pressure So…. For a given amount of gas (n) constant temperature (T) : If pressure (P) increases, the volume (V) of the gas decreases P 1 * V 1 = P 2 * V 2

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Pressure–Volume Law (Boyle’s Law): 11

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12 P a 1/ V P * V = constant P 1 * V 1 = P 2 * V 2 Boyle’s Law Constant temperature Any given amount of gas

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13 EXERCISE 5.2 A volume of carbon dioxide gas equivalent to 20.0 L was 23 o C and 1.00atm pressure. What would be the volume of gas at constant temperature and 0.830atm? P 1 * V 1 = P 2 * V 2 Application of Boyle’s law gives V 2 = V 1 x P 1 = 20.0 L x 1.00atm = = 24.1 L P atm

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14 As T increases V increases Charles’ Law: relating volume and temperature

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Temperature–Volume Law (Charles’ Law): 15 Charles’ Law

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16 For a given amount of gas at constant pressure V a T T (K) = t ( 0 C) Charles’ Law Temperature must be in Kelvin V / T = constant Variation of gas volume with temperature

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17 CHARLES’ LAW the volume occupied by any sample of gas at a constant pressure is directly proportional to the absolute temperature So…. For a given amount of gas (n) constant pressure (P) : If temperature (T) increases, the volume (V) of the gas increases V 1 V 2 T 1 = T 2

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18 EXERCISE 5.3 A chemical reaction is expected to produce 4.3dm 3 of oxygen at 19 o C and 101kPa. What will the volume be at constant pressure and 25 o C? First, convert the temperatures to the Kelvin scale. T i = ( ) = 292 K T f = ( ) = 298 K Following is the data table. V i = 4.38 dm 3 P i = 101 kPa T i = 292 K V f = ? P f = 101 kPa T f = 298 K Apply Charles’s law to obtain V f = V i x T f = 4.38 dm 3 x 298K = = 4.47 dm 3 T i 292K

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19 COMBINED GAS LAW Relating Volume, Temperature and Pressure Taking Boyle’s Law and Charles’ Law: The volume occupied by a given amount of gas is proportional to the absolute temperature divided by the pressure: V = constant x T or PV = constant (for a given amount of gas) P T P 1 V 1 = P 2 V 2 T 1 T 2

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We can combine Boyle’s and charles’ law to come up with the combined gas law Use Kelvins for temp, any pressure, any volume 20 Combined Gas Law

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21 EXERCISE 5.4 A balloon contains 5.41dm 3 of helium at 24 o C and 10.5kPa. Suppose the gas in the balloon is heated to 35 o C and the pressure is now 102.8kPa, what is the volume of the gas? First, convert the temperatures to kelvins. T i = ( ) = 297 K T f = ( ) = 308 K Following is the data table. V i = 5.41 dm 3 P i = kPa T i = 297 K V f = ? P f = kPa T f = 308 K Apply both Boyle’s law and Charles’s law combined to get V f = V i x P i x T f =5.41 dm 3 x 101.5kPa x 308K = = 5.54 dm 3 P f T i 102.8kPa 297K

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22 V number of moles ( n ) V = constant x n V 1 / n 1 = V 2 / n 2 Constant temperature Constant pressure equal volumes of any two gases at the same temperature & pressure contain the same number of molecules

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The Volume–Amount Law (Avogadro’s Law): 23 Avogadro’s Law

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The Volume–Amount Law (Avogadro’s Law): At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present. Use any volume and moles V 1 = V 2 n 1 n 2 24 Avogadro’s Law

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25 Avogadro’s Number = one mole of any gas contains the same number of molecules x Must occupy the same volume at a given temperature and pressure The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.4 L.

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26 Charles’ law: V T (at constant n and P ) Avogadro’s law: V n (at constant P and T ) Boyle’s law: V (at constant n and T ) 1 P V nT P V = constant x = R nT P P R is the molar gas constant PV = nRT

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27 PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)( K) R = L atm / (mol K) R = J / (mol K) R = cal / (mol K) *The units of pressure times volume are the units of energy joules (J) or calories (cal) IDEAL GAS EQUATION

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Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro. 1 mole of an ideal gas occupies L at STP STP conditions are K and 1 atm pressure The gas constant R = L·atm·K –1 ·mol –1 P has to be in atm V has to be in L T has to be in K 28

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29 EXERCISE 5.5 Show that the moles of gas are proportional to the pressure for constant volume and temperature Use the ideal gas law, PV = nRT, and solve for n : n = PV RT n = V x P RT Note that everything in parentheses is constant. Therefore, you can write n = constant x P Or, expressing this as a proportion, n P

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30 EXERCISE 5.6 What is the pressure in a 50.0L gas cylinder that contains 3.03kg of oxygen at 23 o C? T = 23 o C + 273K = 296K V = 50.0L R = L. atm/K. mol n = 3.03kg x 1000g x 1mol = mol O 2 1 kg g P = ? PV= nRT P = nRT = mol x L. atm x 296K V K. mol = 46.0 atm 50.0L

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Density and Molar Mass Calculations: You can calculate the density or molar mass (MM) of a gas. The density of a gas is usually very low under atmospheric conditions. 31 The Ideal Gas Law

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32 EXERCISE 5.7 Calculate the density of helium (g/L) at 21oC and 752mmHg. The density of air under these conditions is 1.188g/L. What is the difference in mass between 1 liter of air and 1 liter of helium? VariableValue P 752 mmHg x 1 atm = atm 760mmHg V1 L (exact number) T( ) = 294 K n?

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33 Using the ideal gas law, solve for n, the moles of helium. n = PV = atm x 1L = RT L. atm/ K. mol x 294K Now convert mol He to grams mol He x 4.00g He = g He = 0.164g/L 1.00 mol He 1 L Therefore, the density of He at 21°C and 752 mmHg is g/L. The difference in mass between one liter of air and one liter of He: mass air − mass He = g − g = = g difference mol

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34 EXERCISE 5.8 A sample of a gaseous substance at 25 o C and atm has a density of 2.26 g/L. What is the molecular mass of the substance? Variable Value P atm V 1 L (exact number) T ( ) = 298 K n ? From the ideal gas law, PV = nRT, you obtain n = PV = atm x 1 L = mol RT L. atm/K. Mol x 298K

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35 Dividing the mass of the gas by moles gives you the mass per mole (the molar mass). Molar mass = grams of gas = 2.26g = g/mol moles of gas mol Therefore, the molecular mass is 64.1 amu.

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36 Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2

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For a two-component system, the moles of components A and B can be represented by the mole fractions ( X A and X B ). Mole fraction is related to the total pressure by: 37 Dalton’s Law of Partial Pressures

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38 EXERCISE 5.10 A 10.0L flask contains 1.031g O 2 and 0.572g CO 2 at 18 o C. What are the partial pressures of each gas? What is the total pressure? What is the mole fraction of oxygen in this mixture? Each gas obeys the ideal gas law g O 2 x 1 mol = mol O g P = nRT = mol x L.atm/K.mol x 291K = atm V 10.0L g CO 2 x 1 mol = mol CO g P = nRT = mol x L.atm/K.mol x 291K = atm V 10.0L

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39 The total pressure is equal to the sum of the partial pressures: P T = P O 2 + P CO 2 = atm atm = = atm The mole fraction of oxygen in the mixture is Mole fraction O 2 = P O 2 = atm = = P T atm x 100% = 71.2 mole % of O 2 in this gas mixture

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This theory presents physical properties of gases in terms of the motion of individual molecules. Average Kinetic Energy Kelvin Temperature Gas molecules are points separated by a great distance Particle volume is negligible compared to gas volume Gas molecules are in constant random motion Gas collisions are perfectly elastic Gas molecules experience no attraction or repulsion 40

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Average Kinetic Energy ( KE ) is given by: 42 U = average speed of a gas particle R = J/K mol m = mass in kg MM = molar mass, in kg/mol N A = x 10 23

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The Root–Mean–Square Speed: is a measure of the average molecular speed. 43 Taking square root of both sides gives the equation

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Calculate the root–mean–square speeds of helium atoms and nitrogen molecules in m/s at 25°C. 44

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Maxwell speed distribution curves. 45 Kinetic Molecular Theory

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Diffusion is the mixing of different gases by random molecular motion and collision. 46 Graham’s Law

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Effusion is when gas molecules escape without collision, through a tiny hole into a vacuum. 47 Graham’s Law

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Graham’s Law: Rate of effusion is proportional to its rms speed, u rms. For two gases at same temperature and pressure: 48 Graham’s Law

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At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures. 49 Behavior of Real Gases

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The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value. 50 Behavior of Real Gases

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Corrections for non-ideality require van der Waals equation. Intermolecular Attractions Excluded Volume 51 Behavior of Real Gases

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A sample of argon gas has a volume of 14.5 L at 1.56 atm of pressure. What would the pressure be if the gas was compressed to 10.5 L? (at constant temperature and moles of gas) 52 Example 1: Boyle’s Law

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A sample of CO 2(g) at 35 C has a volume of 8.56 x10 -4 L. What would the resulting volume be if we increased the temperature to 85 C? (at constant moles and pressure) 53

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6.53 moles of O 2(g) has a volume of 146 L. If we decreased the number of moles of oxygen to 3.94 moles what would be the resulting volume? (constant pressure and temperature) 54

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Oxygen gas is normally sold in 49.0 L steel containers at a pressure of atm. What volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature raised from 20 o C to 35 o C? 55

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An inflated balloon with a volume of 0.55 L at sea level, where the pressure is 1.0 atm, is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? 56

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Sulfur hexafluoride (SF 6 ) is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C. 57

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What is the volume (in liters) occupied by 7.40 g of CO 2 at STP? 58

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What is the molar mass of a gas with a density of g/L at STP? 59

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What is the density of uranium hexafluoride, UF 6, (MM = 352 g/mol) under conditions of STP? 60

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The density of a gaseous compound is 3.38 g/L at 40°C and 1.97 atm. What is its molar mass? 61

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Exactly 2.0 moles of Ne and 3.0 moles of Ar were placed in a 40.0 L container at 25 o C. What are the partial pressures of each gas and the total pressure? 62

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A sample of natural gas contains 6.25 moles of methane (CH 4 ), moles of ethane (C 2 H 6 ), and moles of propane (C 3 H 8 ). If the total pressure of the gas is 1.50 atm, what are the partial pressures of the gases? 63

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What is the mole fraction of each component in a mixture of g of H 2, g of N 2, and 2.38 g of NH 3 ? 64

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On a humid day in summer, the mole fraction of gaseous H 2 O (water vapor) in the air at 25°C can be as high as Assuming a total pressure of atm, what is the partial pressure (in atm) of H 2 O in the air? 65

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In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles. Example: Assuming no change in temperature and pressure, calculate the volume of O 2 (in liters) required for the complete combustion of 14.9 L of butane (C 4 H 10 ): 2 C 4 H 10 ( g ) + 13 O 2 ( g ) 8 CO 2 ( g ) + 10 H 2 O( l ) 66

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All of the mole fractions of elements in a given compound must add up to?

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Hydrogen gas, H 2, can be prepared by letting zinc metal react with aqueous HCl. How many liters of H 2 can be prepared at 742 mm Hg and 15 o C if 25.5 g of zinc (MM = 65.4 g/mol) was allowed to react? Zn( s ) + 2 HCl( aq ) H 2 ( g ) + ZnCl 2 ( aq ) 68

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Under the same conditions, an unknown gas diffuses times as fast as sulfur hexafluoride, SF 6 (MM = 146 g/mol). What is the identity of the unknown gas if it is also a hexafluoride? 69

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What are the relative rates of diffusion of the three naturally occurring isotopes of neon: 20 Ne, 21 Ne, and 22 Ne? 70

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Deviations result from assumptions about ideal gases. 1. Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another. 2. Volume of the molecules is negligibly small compared with that of the container. 71

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Given that 3.50 moles of NH 3 occupy 5.20 L at 47°C, calculate the pressure of the gas (in atm) using (a) the ideal gas equation (b) the van der Waals equation. ( a = 4.17, b = ) 72

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Assume that you have mol of N 2 in a volume of L at 300 K. Calculate the pressure in atmospheres using both the ideal gas law and the van der Waals equation. For N 2, a = 1.35 L 2 ·atm mol –2, and b = L/mol. 73

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