Gases and Kinetic Molecular Theory

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CHAPTER 12 GASES AND KINETIC-MOLECULAR THEORY
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Presentation transcript:

Gases and Kinetic Molecular Theory 12 Gases and Kinetic Molecular Theory

CHAPTER GOALS Comparison of Solids, Liquids, and Gases Composition of the Atmosphere and Some Common Properties of Gases Pressure Boyle’s Law: The Volume-Pressure Relationship Charles’ Law: The Volume-Temperature Relationship; The Absolute Temperature Scale Standard Temperature and Pressure The Combined Gas Law Equation Avogadro’s Law and the Standard Molar Volume

CHAPTER GOALS Summary of Gas Laws: The Ideal Gas Equation Determination of Molecular Weights and Molecular Formulas of Gaseous Substances Dalton’s Law of Partial Pressures Mass-Volume Relationships in Reactions Involving Gases The Kinetic-Molecular Theory Diffusion and Effusion of Gases Real Gases: Deviations from Ideality

Comparison of Solids, Liquids, and Gases The density of gases is much less than that of solids or liquids. Densities (g/mL) Solid Liquid Gas H2O 0.917 0.998 0.000588 CCl4 1.70 1.59 0.00503 Gas molecules must be very far apart compared to liquids and solids.

Composition of the Atmosphere and Some Common Properties of Gases Composition of Dry Air Gas % by Volume N2 78.09 O2 20.94 Ar 0.93 CO2 0.03 He, Ne, Kr, Xe 0.002 CH4 0.00015 H2 0.00005

Pressure Pressure is force per unit area. lb/in2 N/m2 Gas pressure as most people think of it.

Pressure Atmospheric pressure is measured using a barometer. Definitions of standard pressure 76 cm Hg 760 mm Hg 760 torr 1 atmosphere 101.3 kPa Hg density = 13.6 g/mL

Boyle’s Law: The Volume-Pressure Relationship V  1/P or V= k (1/P) or PV = k P1V1 = k1 for one sample of a gas. P2V2 = k2 for a second sample of a gas. k1 = k2 for the same sample of a gas at the same T. Thus we can write Boyle’s Law mathematically as P1V1 = P2V2

Boyle’s Law: The Volume-Pressure Relationship Example 12-1: At 25oC a sample of He has a volume of 4.00 x 102 mL under a pressure of 7.60 x 102 torr. What volume would it occupy under a pressure of 2.00 atm at the same T?

Boyle’s Law: The Volume-Pressure Relationship Notice that in Boyle’s law we can use any pressure or volume units as long as we consistently use the same units for both P1 and P2 or V1 and V2. Use your intuition to help you decide if the volume will go up or down as the pressure is changed and vice versa.

Charles’ Law: The Volume-Temperature Relationship absolute zero = -273.15 0C

Charles’ Law: The Volume-Temperature Relationship Charles’s law states that the volume of a gas is directly proportional to the absolute temperature at constant pressure. Gas laws must use the Kelvin scale to be correct. Relationship between Kelvin and centigrade.

Charles’ Law: The Volume-Temperature Relationship Mathematical form of Charles’ law.

Charles’ Law: The Volume-Temperature Relationship Example 12-2: A sample of hydrogen, H2, occupies 1.00 x 102 mL at 25.0oC and 1.00 atm. What volume would it occupy at 50.0oC under the same pressure? T1 = 25 + 273 = 298 T2 = 50 + 273 = 323

Standard Temperature and Pressure Standard temperature and pressure is given the symbol STP. It is a reference point for some gas calculations. Standard P  1.00000 atm or 101.3 kPa Standard T  273.15 K or 0.00oC

The Combined Gas Law Equation Boyle’s and Charles’ Laws combined into one statement is called the combined gas law equation. Useful when the V, T, and P of a gas are changing.

The Combined Gas Law Equation Example 12-3: A sample of nitrogen gas, N2, occupies 7.50 x 102 mL at 75.00C under a pressure of 8.10 x 102 torr. What volume would it occupy at STP?

The Combined Gas Law Equation Example 12-4 : A sample of methane, CH4, occupies 2.60 x 102 mL at 32oC under a pressure of 0.500 atm. At what temperature would it occupy 5.00 x 102 mL under a pressure of 1.20 x 103 torr? You do it!

The Combined Gas Law Equation

Avogadro’s Law and the Standard Molar Volume

Avogadro’s Law and the Standard Molar Volume Avogadro’s Law states that at the same temperature and pressure, equal volumes of two gases contain the same number of molecules (or moles) of gas. If we set the temperature and pressure for any gas to be STP, then one mole of that gas has a volume called the standard molar volume. The standard molar volume is 22.4 L at STP. This is another way to measure moles. For gases, the volume is proportional to the number of moles. 11.2 L of a gas at STP = 0.500 mole 44.8 L = ? moles

Avogadro’s Law and the Standard Molar Volume Example 12-5: One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a given temperature and pressure. (a) What is its molar mass? (b) What is its density at STP?

Summary of Gas Laws: The Ideal Gas Law Boyle’s Law - V  1/P (at constant T & n) Charles’ Law – V  T (at constant P & n) Avogadro’s Law – V  n (at constant T & P) Combine these three laws into one statement V  nT/P Convert the proportionality into an equality. V = nRT/P This provides the Ideal Gas Law. PV = nRT R is a proportionality constant called the universal gas constant.

Summary of Gas Laws: The Ideal Gas Law We must determine the value of R. Recognize that for one mole of a gas at 1.00 atm, and 273 K (STP), the volume is 22.4 L. Use these values in the ideal gas law.

Summary of Gas Laws: The Ideal Gas Law R has other values if the units are changed. R = 8.314 J/mol K Use this value in thermodynamics. R = 8.314 kg m2/s2 K mol Use this later in this chapter for gas velocities. R = 8.314 dm3 kPa/K mol This is R in all metric units. R = 1.987 cal/K mol This the value of R in calories rather than J.

Summary of Gas Laws: The Ideal Gas Law Example 12-6: What volume would 50.0 g of ethane, C2H6, occupy at 1.40 x 102 oC under a pressure of 1.82 x 103 torr? To use the ideal gas law correctly, it is very important that all of your values be in the correct units! T = 140 + 273 = 413 K P = 1820 torr (1 atm/760 torr) = 2.39 atm 50 g (1 mol/30 g) = 1.67 mol

Summary of Gas Laws: The Ideal Gas Law

Summary of Gas Laws: The Ideal Gas Law Example 12-7: Calculate the number of moles in, and the mass of, an 8.96 L sample of methane, CH4, measured at standard conditions. You do it!

Summary of Gas Laws: The Ideal Gas Law

Summary of Gas Laws: The Ideal Gas Law Example 12-8: Calculate the pressure exerted by 50.0 g of ethane, C2H6, in a 25.0 L container at 25.0oC. You do it!

Determination of Molecular Weights and Molecular Formulas of Gaseous Substances Example 12-9: A compound that contains only carbon and hydrogen is 80.0% carbon and 20.0% hydrogen by mass. At STP, 546 mL of the gas has a mass of 0.732 g . What is the molecular (true) formula for the compound? 100 g of compound contains 80 g of C and 20 g of H.

Determination of Molecular Weights and Molecular Formulas of Gaseous Substances

Determination of Molecular Weights and Molecular Formulas of Gaseous Substances

Determination of Molecular Weights and Molecular Formulas of Gaseous Substances Example 12-10: A 1.74 g sample of a compound that contains only carbon and hydrogen contains 1.44 g of carbon and 0.300 g of hydrogen. At STP 101 mL of the gas has a mass of 0.262 gram. What is its molecular formula? You do it!

Determination of Molecular Weights and Molecular Formulas of Gaseous Substances

Determination of Molecular Weights and Molecular Formulas of Gaseous Substances

Dalton’s Law of Partial Pressures Dalton’s law states that the pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases. Ptotal = PA + PB + PC + .....

Dalton’s Law of Partial Pressures Example 12-11: If 1.00 x 102 mL of hydrogen, measured at 25.0 oC and 3.00 atm pressure, and 1.00 x 102 mL of oxygen, measured at 25.0 oC and 2.00 atm pressure, were forced into one of the containers at 25.0 oC, what would be the pressure of the mixture of gases?

Dalton’s Law of Partial Pressures Vapor Pressure is the pressure exerted by a substance’s vapor over the substance’s liquid at equilibrium.

Dalton’s Law of Partial Pressures Example 12-12: A sample of hydrogen was collected by displacement of water at 25.0 oC. The atmospheric pressure was 748 torr. What pressure would the dry hydrogen exert in the same container?

Dalton’s Law of Partial Pressures Example 12-13: A sample of oxygen was collected by displacement of water. The oxygen occupied 742 mL at 27.0 oC. The barometric pressure was 753 torr. What volume would the dry oxygen occupy at STP? You do it!

Dalton’s Law of Partial Pressures

Mass-Volume Relationships in Reactions Involving Gases In this section we are looking at reaction stoichiometry, like in Chapter 3, just including gases in the calculations. 2 mol KClO3 yields 2 mol KCl and 3 mol O2 2(122.6g) yields 2 (74.6g) and 3 (32.0g) Those 3 moles of O2 can also be thought of as: 3(22.4L) or 67.2 L at STP

Mass-Volume Relationships in Reactions Involving Gases Example 12-14: What volume of oxygen measured at STP, can be produced by the thermal decomposition of 120.0 g of KClO3? You do it!

Mass-Volume Relationships in Reactions Involving Gases

The Kinetic-Molecular Theory The basic assumptions of kinetic-molecular theory are: Postulate 1 Gases consist of discrete molecules that are relatively far apart. Gases have few intermolecular attractions. The volume of individual molecules is very small compared to the gas’s volume. Proof - Gases are easily compressible.

The Kinetic-Molecular Theory Postulate 2 Gas molecules are in constant, random, straight line motion with varying velocities. Proof - Brownian motion displays molecular motion.

The Kinetic-Molecular Theory Postulate 3 Gas molecules have elastic collisions with themselves and the container. Total energy is conserved during a collision. Proof - A sealed, confined gas exhibits no pressure drop over time.

The Kinetic-Molecular Theory Postulate 4 The kinetic energy of the molecules is proportional to the absolute temperature. The average kinetic energies of molecules of different gases are equal at a given temperature. Proof - Brownian motion increases as temperature increases.

The Kinetic-Molecular Theory The kinetic energy of the molecules is proportional to the absolute temperature. The kinetic energy of the molecules is proportional to the absolute temperature. Displayed in a Maxwellian distribution.

The Kinetic-Molecular Theory The gas laws that we have looked at earlier in this chapter are proofs that kinetic-molecular theory is the basis of gaseous behavior. Boyle’s Law P  1/V As the V increases the molecular collisions with container walls decrease and the P decreases. Dalton’s Law Ptotal = PA + PB + PC + ..... Because gases have few intermolecular attractions, their pressures are independent of other gases in the container. Charles’ Law V  T An increase in temperature raises the molecular velocities, thus the V increases to keep the P constant.

The Kinetic-Molecular Theory

The Kinetic-Molecular Theory The root-mean square velocity of gases is a very close approximation to the average gas velocity. Calculating the root-mean square velocity is simple: To calculate this correctly: The value of R = 8.314 kg m2/s2 K mol And M must be in kg/mol.

The Kinetic-Molecular Theory Example 12-17: What is the root mean square velocity of N2 molecules at room T, 25.0oC?

The Kinetic-Molecular Theory Example 12-18: What is the root mean square velocity of He atoms at room T, 25.0oC? You do it!

The Kinetic-Molecular Theory Can you think of a physical situation that proves He molecules have a velocity that is so much greater than N2 molecules? What happens to your voice when you breathe He?

Diffusion and Effusion of Gases Diffusion is the intermingling of gases. Effusion is the escape of gases through tiny holes.

Diffusion and Effusion of Gases This is a demonstration of diffusion.

Diffusion and Effusion of Gases The rate of effusion is inversely proportional to the square roots of the molecular weights or densities.

Diffusion and Effusion of Gases Example 12-15: Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SO2, at the same temperature and pressure.

Diffusion and Effusion of Gases Example 12-16: A sample of hydrogen, H2, was found to effuse through a pinhole 5.2 times as rapidly as the same volume of unknown gas (at the same temperature and pressure). What is the molecular weight of the unknown gas? You do it!

Real Gases: Deviations from Ideality Real gases behave ideally at ordinary temperatures and pressures. At low temperatures and high pressures real gases do not behave ideally. The reasons for the deviations from ideality are: The molecules are very close to one another, thus their volume is important. The molecular interactions also become important.

Real Gases: Deviations from Ideality van der Waals’ equation accounts for the behavior of real gases at low temperatures and high pressures. The van der Waals constants a and b take into account two things: a accounts for intermolecular attraction b accounts for volume of gas molecules At large volumes a and b are relatively small and van der Waal’s equation reduces to ideal gas law at high temperatures and low pressures.

Real Gases: Deviations from Ideality What are the intermolecular forces in gases that cause them to deviate from ideality? For nonpolar gases the attractive forces are London Forces. For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds.

Real Gases: Deviations from Ideality Example 12-19: Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the ideal gas law. You do it!

Real Gases: Deviations from Ideality Example 12-20: Solve Example 12-19 using the van der Waal’s equation.

Real Gases: Deviations from Ideality

Gases and Kinetic Molecular Theory 12 Gases and Kinetic Molecular Theory