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Gases: Their Properties and Behavior Chapter 9. 2 Properties of Gases There are 5 important properties of gases: –Confined gases exerts pressure on the.

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Presentation on theme: "Gases: Their Properties and Behavior Chapter 9. 2 Properties of Gases There are 5 important properties of gases: –Confined gases exerts pressure on the."— Presentation transcript:

1 Gases: Their Properties and Behavior Chapter 9

2 2 Properties of Gases There are 5 important properties of gases: –Confined gases exerts pressure on the wall of a container uniformly –Gases have low densities –Gases can be compressed –Gases can expand to fill their contained uniformly –Gases mix completely with other gases in the same container

3 Chapter 93 Kinetic Molecular Theory of Gases The Kinetic Molecular Theory of Gases is the model used to explain the behavior of gases in nature. This theory presents physical properties of gases in terms of the motion of individual molecules: 1.Average Kinetic Energy  Kelvin Temperature 2.Gas molecules are points separated by a great distance 3.Particle volume is negligible compared to gas volume 4.Gas molecules are in rapid random motion 5.Gas collisions are perfectly elastic 6.Gas molecules experience no attraction or repulsion

4 Chapter 94 Properties that Describe a Gas These properties are all related to one another. When one variable changes, it causes the other three to react in a predictable manner.

5 Chapter 95 Gas Pressure (P) Gas pressure (P) is the result of constantly moving gas molecules striking the inside surface of their container.

6 Chapter 96 Atmospheric Pressure Atmospheric pressure is the pressure exerted by the air on the earth. Evangelista Torricelli invented the barometer in 1643 to measure atmospheric pressure. Atmospheric pressure is 760 mm of mercury or 1 atmosphere (atm) at sea level. What happens to the atmospheric pressure as you go up in elevation?

7 Chapter 97 Measurement of Gas Pressure Traditionally, the gas pressure inside of a container is measured with a manometer

8 Chapter 98 Units of Pressure Standard pressure is the atmospheric pressure at sea level, 760 mm of mercury. –Here is standard pressure expressed in other units:

9 Chapter 99 Gas Pressure Conversions The barometric pressure is torr. What is the barometric pressure in atmospheres? In mm Hg? In Pascals (Pa)?

10 Chapter 910 Gas Law Problems

11 Chapter 911 Boyle’s Law (V and P) Mathematically, we write: For a before and after situation: Boyle’s Law states that the volume of a gas is inversely proportional to the pressure at constant temperature. P  1. V P 1 V 1 = P 2 V 2

12 Chapter 912 Boyle’s Law Problem A 1.50 L sample of methane gas exerts a pressure of 1650 mm Hg. What is the final pressure if the volume changes to 7.00 L? (1650 mm Hg )(1.50 L) 7.00 L = 354 mm Hg P 1 V 1 = P 2 V 2 rearranges to P 1 V 1 V2V2 = P 2

13 Chapter 913 Charles’ Law (V and T) In 1783, Jacques Charles discovered (while hot air ballooning) that the volume of a gas is directly proportional to the temperature in Kelvin. Mathematically, we write: For a before and after situation: T  V V1V1 T1T1 V2V2 T2T2 =

14 Chapter 914 Charles’ Law Problem A 275 L helium balloon is heated from 20  C to 40  C. What is the final volume at constant P? (275 L)(313 K) 293 K = 294 L V 1 T 2 T1T1 = V 2 V1V1 T1T1 V2V2 T2T2 = rearranges to

15 Chapter 915 Gay-Lussac’s Law (P and T) In 1802, Joseph Gay-Lussac discovered that the pressure of a gas is directly proportional to the temperature in Kelvin. Mathematically, we write: For a before and after situation: T  P P1P1 T1T1 P2P2 T2T2 =

16 Chapter 916 Gay-Lussac’s Law Problem A steel container of nitrous oxide at 15.0 atm is cooled from 25  C to –40  C. What is the final volume at constant V? (15.0 atm)(298 K) 233 K = 11.7 atm P 1 T 2 T1T1 = P 2 P1P1 T1T1 P2P2 T2T2 = rearranges to

17 Chapter 917 Avogadro’s Law (n and V) In the previous laws, the amount of gas was always constant. However, the amount of a gas (n) is directly proportional to the volume of the gas, meaning that as the amount of gas increases, so does the volume. Mathematically, we write: For a before and after situation: n  V V1V1 n1n1 V2V2 n2n2 =

18 Chapter 918 Avogadro’s Law Problem A steel container contains 2.6 mol of nitrous oxide with a volume 15.0 L. If the amount of nitrous oxide is increased to 8.4 mol, what is the final volume at constant T and P? (15.0 L)(8.4 mol) 2.6 mol = 48.5 L V 1 n 2 n1n1 = V 2 V1V1 n1n1 V2V2 n2n2 = rearranges to

19 Chapter 919 Combined Gas Law When we introduced Boyle’s, Charles’, and Gay- Lussac’s Laws, we assumed that one of the variables remained constant. Experimentally, all three (temperature, pressure, and volume) usually change. By combining all three laws, we obtain the combined gas law: P1V1P1V1 T1T1 P2V2P2V2 T2T2 =

20 Chapter 920 Combined Gas Law Problem 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?

21 Chapter 921 Molar Volume and STP Standard temperature and pressure (STP) are defined as 0  C and 1 atm. At standard temperature and pressure, one mole of any gas occupies 22.4 L. The volume occupied by one mole of gas (22.4 L) is called the molar volume. 1 mole Gas = 22.4 L

22 Chapter 922 Molar Volume Calculation – Volume to Moles A sample of methane, CH 4, occupies 4.50 L at STP. How many moles of methane are present? 4.50 L CH 4 ×= mol CH 4 1 mol CH L CH 4 Volume Molar Volume Moles

23 Chapter 923 Mole Unit Factors We now have three interpretations for the mole: 1 mol = 6.02 × particles 1 mol = molar mass 1 mol = 22.4 L (at STP for a gas) This gives us 3 unit factors to use to convert between moles, particles, mass, and volume.

24 Chapter 924 Mole Calculation - Grams to Volume What is the mass of 3.36 L of ozone gas, O 3, at STP? = 7.20 g O L O 3 ×× 22.4 L O 3 1 mol O g O 3 1 mol O 3 Grams Molar Mass Moles Molar Volume

25 Chapter 925 Mole Calculation – Molecules to Volume How many molecules of hydrogen gas, H 2, occupy L at STP? L H 2 × 1 mol H L H ×10 23 molecules H 2 1 mole H 2 × 1.34 × molecules H 2 Volume Molar Volume Moles Avogadro’s Number Atoms

26 Chapter 926 Gas Density and Molar Mass The density of a gas is much less than that of a liquid. We can calculate the density of any gas at STP easily. You can rearrange this equation to find the Molar mass of an unknown gas too! = density, g/L molar mass in grams (MM) molar volume in liters (MV)

27 Chapter 927 Calculating Gas Density What is the density of ammonia gas, NH 3, at STP? 1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass? = g/L g/mol 22.4 L/mol

28 Chapter 928 The Ideal Gas Law When working in the lab, you will not always be at STP. The four properties used in the measurement of a gas (Pressure, Volume, Temperature and moles) can be combined into a single gas law: Here, R is the ideal gas constant and has a value of: Note the units of R. When working problems with the Ideal Gas Law, your units of P, V, T and n must match those in the constant! PV = nRT atm  L/mol  K

29 Chapter 929 Ideal Gas Law Problem 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.

30 Chapter 930 Density and Molar Mass Calculations: You can calculate the density or molar mass (M) of a gas. The density of a gas is usually very low under atmospheric conditions. Ideal Gas Law and Molar Mass

31 Chapter 931 What is the molar mass of a gas with a density of g/L –1 at STP? What is the density of uranium hexafluoride, UF 6, (MM = 352 g/mol) under conditions of STP? The density of a gaseous compound is 3.38 g/L –1 at 40°C and 1.97 atm. What is its molar mass? Ideal Gas Law and Molar Mass

32 Chapter 932 Gases in Chemical Reactions Gases are involved as reactants and/or products in numerous chemical reactions. Typically, the information given for a gas in a reaction is its Pressure (P), volume (V) (or amount of the gas (n)) and temperature (T). We use this information and the Ideal Gas Law to determine the moles of the gas (n) or the volume of the gas (V). Once we have this information, we can proceed with the problem as we would any other stoichiometry problem. A (g) + X (s) → B (s) + Y (l)

33 Chapter 933 Reaction with a Gas Hydrogen gas is formed when zinc metal reacts with hydrochloric acid. How many liters of hydrogen gas at STP are produced when 15.8 g of zinc reacts? Zn (s) + 2HCl (aq) → H 2 (g) + ZnCl 2 (aq) Grams of ZnMoles of ZnMoles of H h Liters of H 2 Molar Mass of Zn Molar Volume Mole Ratio Can use because at STP!

34 Chapter 934 Reaction with a Gas Hydrogen gas is formed when zinc metal reacts with hydrochloric acid. How many liters of hydrogen gas at a pressure of 755 atm and 35°C are produced when 15.8 g of zinc reacts? Zn (s) + 2HCl (aq) → H 2 (g) + ZnCl 2 (aq) Grams of ZnMoles of ZnMoles of H h Liters of H 2 Molar Mass of Zn Ideal Gas Law Mole Ratio Use because not at STP!

35 Chapter 935 Dalton’s Law of Partial Pressures In a mixture of gases the total pressure, P tot, is the sum of the partial pressures of the gases: Dalton’s law allows us to work with mixtures of gases. P tot = P 1 + P 2 + P 3 + etc.

36 Chapter 936 For a two-component system, the moles of components A and B can be represented by the mole fractions (X A and X B ). 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 ? Dalton’s Law of Partial Pressures 1 BA BA B B BA A A      XX nn n X nn n X

37 Chapter 937 Dalton’s Law of Partial Pressures Mole fraction is related to the total pressure by: 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?

38 Chapter 938 Dalton’s Law of Partial Pressures Exactly 2.0 moles of Ne and 3.0 moles of Ar were placed in a 40.0 L container at 25°C. What are the partial pressures of each gas and the total pressure? 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?

39 Chapter 939 Kinetic Molecular Theory of Gases The Kinetic Molecular Theory of Gases is the model used to explain the behavior of gases in nature. This theory presents physical properties of gases in terms of the motion of individual molecules. 1.Average Kinetic Energy  Kelvin Temperature 2.Gas molecules are points separated by a great distance 3.Particle volume is negligible compared to gas volume 4.Gas molecules are in rapid random motion 5.Gas collisions are perfectly elastic 6.Gas molecules experience no attraction or repulsion

40 Chapter 940 Kinetic Molecular Theory of Gases

41 Chapter 941 Average Kinetic Energy (KE) is given by:  Kinetic Molecular Theory of Gases

42 Chapter 942 Kinetic Molecular Theory of Gases The Root–Mean–Square Speed (u RMS ): is a measure of the average molecular speed of a particle of gas. Taking square root of both sides gives the equation R = J/mol K

43 Chapter 943 Calculate the root–mean–square speeds (u RMS ) of helium atoms and nitrogen molecules in m/s at 25°C. Kinetic Molecular Theory of Gases

44 Chapter 944 Diffusion is the mixing of different gases by random molecular motion and collision. Graham’s Law: Diffusion and Effusion Effusion is when gas molecules escape without collision, through a tiny hole into a vacuum.

45 Chapter 945 Graham’s Law: The rate of effusion is proportional to its RMS speed (u RMS). For two gases at same temperature and pressure: Graham’s Law: Diffusion and Effusion Rate 

46 Chapter 946 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? What are the relative rates of diffusion of the three naturally occurring isotopes of neon: 20 Ne, 21 Ne, and 22 Ne? Graham’s Law: Diffusion and Effusion

47 Chapter 947 Deviations from Ideal behavior result from two key 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. These assumptions breakdown at high pressures, low volumes and low temperatures. Behavior of Real Gases

48 Chapter 948 At STP, the volume occupied by a single molecule is very small relative to its share of the total volume –For example, a He atom (radius = 31 pm) has roughly the same space to move about as a pea in a basketball Let’s say we increase the pressure of the system to 1000 atm, this will cause a decrease in the volume the gas has to move about in –Now our He atom is like a pea in a ping pong ball Therefore, at high pressures, the volume occupied by the gaseous molecules is NOT negligible and must be considered. So the space the gas has to move around in is less than under Ideal conditions! Behavior of Real Gases V Real > V Ideal

49 Chapter 949 At low volumes, particles are much closer together and attractive forces become more important than at high volumes. This increase in intermolecular attractions pulls the molecules away from the walls of the containers, meaning that they do not hit the wall with as great a force, so the pressure is lower than under ideal conditions. Behavior of Real Gases P Real < P Ideal

50 Chapter 950 A similar phenomenon is seen at low temperatures (aka. The Flirting Effect) –As molecules slow down, they have more time to interact therefore increasing the effect of intermolecular forces. Again, this increase in intermolecular attractions pulls the molecules away from the walls of the containers, meaning that they do not hit the wall with as great a force, so the pressure is lower than under ideal conditions. Behavior of Real Gases P Real < P Ideal

51 Chapter 951 Behavior of Real Gases

52 Chapter 952 Corrections for non-ideality require the van der Waals equation. Correction for Intermolecular Attractions Correction for Molecular Volume Behavior of Real Gases n = moles of gas a and b are constants given in the problem P Real < P Ideal V Real > V Ideal

53 Chapter 953 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 = ) Calculate the pressure exerted by 4.37 moles of molecular chlorine confined in a volume of 2.45 L at 38°C. Compare the pressure with that calculated using the ideal gas equation. (a = 6.49 and b = ) Behavior of Real Gases


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