Gases Chapter 5

The Gaseous State Ideal Gas Concept Ideal gas - a model of the way that particles of a gas behave at the microscopic level. We can measure the following of a gas: temperature, volume, pressure and quantity (mass) We can systematically change one of the properties and see the effect on each of the others.

The most important gas laws involve the relationship between Measurement of Gases The most important gas laws involve the relationship between number of moles (n) of gas volume (V) temperature (T) pressure (P) Pressure - force per unit area. Gas pressure is a result of force exerted by the collision of particles with the walls of the container.

Pressure P = Force/unit area Force exerted per unit area of surface by molecules in motion. P = Force/unit area 1 atmosphere = 14.7 psi 1 atmosphere = 760 mm Hg 1 atmosphere = 101,325 Pascals 1 Pascal = 1 kg/m.s2 2

Barometer - measures atmospheric pressure. Invented by Evangelista Torricelli A commonly used unit of pressure is the atmosphere (atm). 1 atm is equal to: 760 mmHg 760 torr 76 cmHg

Worked Example 5.1

Elements that exist as gases at 250C and 1 atmosphere

Gases assume the volume and shape of their containers. 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.

Boyle’s Law Boyle’s Law - volume of a gas is inversely proportional to pressure if the temperature and number of moles is held constant. PV = k1 or PiVi = PfVf

A Problem to Consider A sample of chlorine gas has a volume of 1.8 L at 1.0 atm. If the pressure increases to 4.0 atm (at constant temperature), what would be the new volume? 7

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg

Examples using Boyle's Law 2 1. A 5.0 L sample of a gas at 25oC and 3.0 atm is compressed at constant temperature to a volume of 1.0 L. What is the new pressure? 2. A 3.5 L sample of a gas at 1.0 atm is expanded at constant temperature until the pressure is 0.10 atm. What is the volume of the gas?

Charles’ Law Charles’ Law - volume of a gas varies directly with the absolute temperature (K) if pressure and number of moles of gas are constant. or

A Problem to Consider A sample of methane gas that has a volume of 3.8 L at 5.0 oC is heated to 86.0 oC at constant pressure. Calculate its new volume. 7

A sample of carbon monoxide gas occupies 3. 20 L at 125 0C A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 /T1 = V2 /T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T1 = 125 (0C) + 273.15 (K) = 398.15 K V2 x T1 V1 1.54 L x 398.15 K 3.20 L = T2 = = 192 K

Examples Using Charles' Law 2 1. A 2.5 L sample of gas at 25oC is heated to 50oC at constant pressure. Will the volume double? 2. What would be the volume in question 1? 3. What temperature would be required to double the volume in question 1?

The Empirical Gas Laws Gay-Lussac’s Law: The pressure exerted by a gas at constant volume is directly proportional to its absolute temperature. P a Tabs (constant moles and V) or 3

A Problem to Consider An aerosol can has a pressure of 1.4 atm at 25 oC. What pressure would it attain at 1200 oC, assuming the volume remained constant? 7

Combined Gas Law 1 This law is used when a sample of gas undergoes change involving volume, pressure, and temperature simultaneously.

A Problem to Consider A sample of carbon dioxide occupies 4.5 L at 30 oC and 650 mm Hg. What volume would it occupy at 800 mm Hg and 200 oC? 7

Example Using the Combined 2 Example Using the Combined Law Calculate the temperature when a 0.50 L sample of gas at 1.0 atm and 25oC is compressed to 0.05 L of gas at 5.0 atm.

Avogadro’s Law 1 Avogadro’s Law - equal volumes of an ideal gas contain the same number of moles if measured under the same conditions of temperature and pressure. or

Example Using Avogaro's Law 2 Assuming no change in temperature and pressure, how many moles of gas would be needed to double the volume occupied by 0.50 moles of gas?

Molar Volume - the volume occupied by 1 mol of any gas. Molar Volume of a Gas Molar Volume - the volume occupied by 1 mol of any gas. STP - Standard Temperature and Pressure T = 273 K (or 0oC) P = 1 atm At STP the molar volume of a gas is 22.4 L/mol We will learn to calculate the volume later.

The molar volume of a gas. Photo courtesy of James Scherer.

Figure 5.11

Gas Densities We know: density = mass/volume Let’s calculate the density of H2. What is the mass of 1 mol of H2? 2.0 g What is the volume of 1 mol of H2? 22.4 L Density = 2.0 g/22.4 L = 0.089 g/L

The Ideal Gas Law Combining Boyle’s Law, Charles’ Law and Avogadro’s Law gives the Ideal Gas Law. PV=nRT R (ideal gas constant) = 0.08206 L.Atm/mol.K

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

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K)

Let’s calculate the molar volume at STP using the ideal gas law: PV = nRT What would be the pressure? 1 atm What would be the temperature? 273 K What would be the number of moles? 1 mol 22.4 L

A Problem to Consider An experiment calls for 3.50 moles of chlorine, Cl2. What volume would this be if the gas volume is measured at 34 oC and 2.45 atm? 5

What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = nRT P V = 1 atm 1.37 mol x 0.0821 x 273.15 K L•atm mol•K V = 30.6 L

Worked Example 5.3

Worked Example 5.4

Examples using The Ideal Gas Law 1. What is the volume of gas occupied by 5.0 g CH4 at 25oC and 1 atm? 2. What is the mass of N2 required to occupy 3.0 L at 100oC and 700 mmHg?

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K P1 T1 P2 T2 = P2 = P1 x T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm

Worked Example 5.5

Worked Example 5.6

Worked Example 5.7

Molecular Weight Determination The relationship between moles and mass. or 12

Molecular Weight Determination If we substitute this in the ideal gas equation, we obtain If we solve this equation for the molecular mass, we obtain 12

A Problem to Consider A 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25 oC and a pressure of 1.08 atm. Calculate its molecular mass. 5

A 2. 10-L vessel contains 4. 65 g of a gas at 1. 00 atm and 27. 0 0C A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L 2.21 g L 1 atm x 0.0821 x 300.15 K L•atm mol•K M = M = 54.6 g/mol

Density Determination If we look again at our derivation of the molecular mass equation, we can solve for m/V, which represents density. 12

A Problem to Consider Calculate the density of ozone, O3 (Mm = 48.0g/mol), at 50 oC and 1.75 atm of pressure. 5

Density (d) Calculations m is the mass of the gas in g m V = PM RT d = M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L

Worked Example 5.8

Gas Stoichiometry g C6H12O6 mol C6H12O6 mol CO2 V CO2 What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) g C6H12O6 mol C6H12O6 mol CO2 V CO2 1 mol C6H12O6 180 g C6H12O6 x 6 mol CO2 1 mol C6H12O6 x 5.60 g C6H12O6 = 0.187 mol CO2 0.187 mol x 0.0821 x 310.15 K L•atm mol•K 1.00 atm = nRT P V = = 4.76 L

Stoichiometry Problems Involving Gas Volumes Consider the following reaction, which is often used to generate small quantities of oxygen. Suppose you heat 0.0100 mol of potassium chlorate, KClO3, in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm? 13

Stoichiometry Problems Involving Gas Volumes First we must determine the number of moles of oxygen produced by the reaction. 13

Stoichiometry Problems Involving Gas Volumes Now we can use the ideal gas equation to calculate the volume of oxygen under the conditions given. 13

1 Dalton’s Law of Partial Pressures Dalton’s Law - a mixture of gases exerts a pressure that is the sum of the pressures that each gas would exert if it were present alone under the same conditions. Pt=p1+p2+p3+... For example, the total pressure of our atmosphere is equal to the sum of the pressures of N2 and O2.

Partial Pressures of Gas Mixtures The partial pressure of a component gas, “A”, is then defined as Applying this concept to the ideal gas equation, we find that each gas can be treated independently. 13

Partial Pressures of Gas Mixtures The composition of a gas mixture is often described in terms of its mole fraction. The mole fraction, c , of a component gas is the fraction of moles of that component in the total moles of gas mixture. 13

A sample of natural gas contains 8. 24 moles of CH4, 0 A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane = = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

A Problem to Consider Given a mixture of gases in the atmosphere at 760 torr, what is the partial pressure of N2 (c = 0 .7808) at 25 oC? 5

Kinetic Molecular Theory of Gases 3 Kinetic Molecular Theory of Gases Provides an explanation of the behavior of gases that we have studied in this chapter. Summary follows: 1. Gases are made up of small atoms or molecules that are in constant and random motion. 2. The distance of separation is very large compared to the size of the atoms or molecules. the gas is mostly empty space.

3. All gas particles behave independently. No attractive or repulsive forces exist between them. 4. Gas particles collide with each other and with the walls of the container without losing energy. The energy is transferred from one atom or molecule to another. 5. The average kinetic energy of the atoms or molecules is proportional to absolute temperature. K.E. = 1/2mv2 so as temperature goes up, the speed of the particles goes up.

4 How does the Kinetic Molecular Theory of Gases explain the following statements? Gases are easily compressible. Gases will expand to fill any available volume. Gases have low density.

Remember: pressure is a force per unit area resulting from collision of gas particles with the walls of the container. If pressure remains constant why does volume increase with temperature?

 The distribution of speeds of three different gases at the same temperature The distribution of speeds for nitrogen gas molecules at three different temperatures urms = 3RT M 

Worked Example 5.16

Gases behave most ideally at low pressure and high temperatures. Ideal Gases Vs. Real Gases In reality there is no such thing as an ideal gas. Instead this is a useful model to explain gas behavior. Non-polar gases behave more ideally than polar gases because attractive forces are present in polar gases.

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. M2 M1  r1 r2 = NH4Cl NH3 17 g/mol HCl 36 g/mol

Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. M2 M1  r1 r2 t2 t1 = =

Real Gases a corrects for interaction between atoms. Real gases do not follow PV = nRT perfectly. The van der Waals equation corrects for the nonideal nature of real gases. a corrects for interaction between atoms. b corrects for volume occupied by atoms. 29

Effect of intermolecular forces on the pressure exerted by a gas.

Deviations from Ideal Behavior 1 mole of ideal gas Repulsive Forces PV = nRT n = PV RT = 1.0 Attractive Forces

( ) } } Van der Waals equation nonideal gas an2 P + (V – nb) = nRT V2 corrected pressure } corrected volume

A Problem to Consider The following values for SO2 If sulfur dioxide were an “ideal” gas, the pressure at 0 oC exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure. The following values for SO2 a = 6.865 L2.atm/mol2 b = 0.05679 L/mol 29

A Problem to Consider First, let’s rearrange the van der Waals equation to solve for pressure. R= 0.0821 L. atm/mol. K T = 273.2 K V = 22.41 L a = 6.865 L2.atm/mol2 b = 0.05679 L/mol 29

A Problem to Consider The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the “ideal” pressure. 29

Solubility of Gases: Henry’s Law Henry’s Law - the number of moles of a gas dissolved in a liquid is proportional to the partial pressure of the gas above the liquid. (At constant temperature) Carbonated Beverages Respiration

Pressure and Solubility of Gases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). concentration = Ch P concentration is (M) of the dissolved gas P is the pressure of the gas over the solution Ch is a constant for each gas (mol/L•atm) that depends only on temperature low P high P low c high c

Pressure and Solubility of Gases Amount of Gas dissolved Unit Volume of solvent Pure gas: = Ch X gas pressure Gas Mixture: = Ch X gas pressure Amount of Gas dissolved Unit Volume of solvent mL gas mL solvent x atm Ch =