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**Gases: Their Properties and Behavior**

Chapter 9 Gases: Their Properties and Behavior

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**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 Chapter 9

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**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: 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 rapid random motion Gas collisions are perfectly elastic Gas molecules experience no attraction or repulsion Chapter 9

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**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. Chapter 9

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Gas Pressure (P) Gas pressure (P) is the result of constantly moving gas molecules striking the inside surface of their container. Chapter 9

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**What happens to the atmospheric pressure as you go up in elevation?**

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? Chapter 9

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**Measurement of Gas Pressure**

Traditionally, the gas pressure inside of a container is measured with a manometer Chapter 9

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Units of Pressure Standard pressure is the atmospheric pressure at sea level, 760 mm of mercury. Here is standard pressure expressed in other units: Chapter 9

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**Gas Pressure Conversions**

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

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Gas Law Problems Chapter 9

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**Boyle’s Law (V and P) 1 . P V P1V1 = P2V2**

Boyle’s Law states that the volume of a gas is inversely proportional to the pressure at constant temperature. Mathematically, we write: For a before and after situation: P 1 . V P1V1 = P2V2 Chapter 9

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**Boyle’s Law Problem P1 V1 rearranges to = P2 V2 (1650 mm Hg )(1.50 L)**

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? P1V1 = P2V2 rearranges to P1 V1 V2 = P2 (1650 mm Hg )(1.50 L) = 354 mm Hg 7.00 L Chapter 9

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**Charles’ Law (V and T) T V V1 T1 V2 T2 =**

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 V1 T1 V2 T2 = Chapter 9

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**Charles’ Law Problem V1 T1 V2 T2 V1 T2 = rearranges to = V2 T1**

A 275 L helium balloon is heated from 20C to 40C. What is the final volume at constant P? V1 T1 V2 T2 V1 T2 = rearranges to = V2 T1 (275 L)(313 K) 293 K = 294 L Chapter 9

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**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 P1 T1 P2 T2 = Chapter 9

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**Gay-Lussac’s Law Problem**

A steel container of nitrous oxide at 15.0 atm is cooled from 25C to –40C. What is the final volume at constant V? P1 T1 P2 T2 = P1 T2 rearranges to = P2 T1 (15.0 atm)(298 K) = 11.7 atm 233 K Chapter 9

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**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 V1 n1 V2 n2 = Chapter 9

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**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? V1 n1 V2 n2 = V1 n2 rearranges to = V2 n1 (15.0 L)(8.4 mol) = 48.5 L 2.6 mol Chapter 9

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**Combined Gas Law P1V1 T1 P2V2 T2 =**

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: P1V1 T1 P2V2 T2 = Chapter 9

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**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 20oC to 35oC? Chapter 9

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**Molar Volume and STP 1 mole Gas = 22.4 L**

Standard temperature and pressure (STP) are defined as 0C 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 Chapter 9

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**Molar Volume Calculation – Volume to Moles**

A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present? Volume Molar Moles 4.50 L CH4 × = mol CH4 1 mol CH4 22.4 L CH4 Chapter 9

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**Mole Unit Factors 1 mol = 6.02 × 1023 particles 1 mol = molar mass**

We now have three interpretations for the mole: 1 mol = 6.02 × 1023 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. Chapter 9

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**Mole Calculation - Grams to Volume**

What is the mass of 3.36 L of ozone gas, O3, at STP? Grams Molar Mass Moles Volume = 7.20 g O3 3.36 L O3 × × 22.4 L O3 1 mol O3 48.00 g O3 Chapter 9

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**Mole Calculation – Molecules to Volume**

How many molecules of hydrogen gas, H2, occupy L at STP? Volume Molar Moles Avogadro’s Number Atoms 0.500 L H2 × 1 mol H2 22.4 L H2 6.02×1023 molecules H2 1 mole H2 × 1.34 × 1022 molecules H2 Chapter 9

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**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! molar mass in grams (MM) = density, g/L molar volume in liters (MV) Chapter 9

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**Calculating Gas Density**

What is the density of ammonia gas, NH3, at STP? 1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass? = g/L 17.04 g/mol 22.4 L/mol Chapter 9

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**The Ideal Gas Law PV = nRT 0.0821 atmL/molK**

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 atmL/molK Chapter 9

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Ideal Gas Law Problem Sulfur hexafluoride (SF6) 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. Chapter 9

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**Ideal Gas Law and Molar Mass**

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. Chapter 9

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**Ideal Gas Law and Molar Mass**

What is the molar mass of a gas with a density of g/L–1 at STP? What is the density of uranium hexafluoride, UF6, (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? Chapter 9

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**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) Chapter 9

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**Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2 (aq)**

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) → H2 (g) + ZnCl2 (aq) Grams of Zn Moles of Zn Moles of Hh Liters of H2 Molar Mass of Zn Volume Mole Ratio Can use because at STP! Chapter 9

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**Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2 (aq)**

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) → H2 (g) + ZnCl2 (aq) Grams of Zn Moles of Zn Moles of Hh Liters of H2 Molar Mass of Zn Ideal Gas Law Mole Ratio Use because not at STP! Chapter 9

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**Dalton’s Law of Partial Pressures**

In a mixture of gases the total pressure, Ptot, is the sum of the partial pressures of the gases: Dalton’s law allows us to work with mixtures of gases. Ptot = P1 + P2 + P3 + etc. Chapter 9

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**Dalton’s Law of Partial Pressures**

For a two-component system, the moles of components A and B can be represented by the mole fractions (XA and XB). What is the mole fraction of each component in a mixture of g of H2, g of N2, and 2.38 g of NH3? n n X = A X = B X + X = 1 A n + n B n + n A B A B A B Chapter 9

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**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 H2O (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 H2O in the air? Chapter 9

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**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 (CH4), moles of ethane (C2H6), and moles of propane (C3H8). If the total pressure of the gas is 1.50 atm, what are the partial pressures of the gases? Chapter 9

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**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. 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 rapid random motion Gas collisions are perfectly elastic Gas molecules experience no attraction or repulsion Chapter 9

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**Kinetic Molecular Theory of Gases**

Chapter 9

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**Kinetic Molecular Theory of Gases**

Average Kinetic Energy (KE) is given by: Chapter 9

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**Kinetic Molecular Theory of Gases**

The Root–Mean–Square Speed (uRMS): 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 Chapter 9

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**Kinetic Molecular Theory of Gases**

Calculate the root–mean–square speeds (uRMS) of helium atoms and nitrogen molecules in m/s at 25°C. Chapter 9

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**Graham’s Law: Diffusion and Effusion**

Diffusion is the mixing of different gases by random molecular motion and collision. Effusion is when gas molecules escape without collision, through a tiny hole into a vacuum. Chapter 9

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**Graham’s Law: Diffusion and Effusion**

Graham’s Law: The rate of effusion is proportional to its RMS speed (uRMS). For two gases at same temperature and pressure: Rate Chapter 9

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**Graham’s Law: Diffusion and Effusion**

Under the same conditions, an unknown gas diffuses times as fast as sulfur hexafluoride, SF6 (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: 20Ne, 21Ne, and 22Ne? Chapter 9

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Behavior of Real Gases Deviations from Ideal behavior result from two key assumptions about ideal gases. Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another. Volume of the molecules is negligibly small compared with that of the container. These assumptions breakdown at high pressures, low volumes and low temperatures. Chapter 9

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**Behavior of Real Gases VReal > VIdeal**

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! VReal > VIdeal Chapter 9

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**Behavior of Real Gases PReal < PIdeal**

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. PReal < PIdeal Chapter 9

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**Behavior of Real Gases PReal < PIdeal**

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. PReal < PIdeal Chapter 9

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Behavior of Real Gases Chapter 9

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Behavior of Real Gases Corrections for non-ideality require the van der Waals equation. Correction for Intermolecular Attractions Correction for Molecular Volume VReal > VIdeal PReal < PIdeal n = moles of gas a and b are constants given in the problem Chapter 9

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Behavior of Real Gases Given that 3.50 moles of NH3 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 = ) Chapter 9

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