Chapter 9 Gases: Their Properties and Behavior

Chapter 9 Gases: Their Properties and Behavior 國防醫學院生化學科王明芳老師 2011-11-15 & 2011-11-22 Chapter 9/1

Gases and Gas Pressure Gas mixtures are homogeneous and compressible. Air-the mixture of gases. Molecular view of a gas Chapter 9/2

Gases and Gas Pressure Nitrogen and oxygen account for more than 99% by volume of dry air. Chapter 9/3

Gases and Gas Pressure Pressure System International unit (SI units) SI units = Newton/meter 2 = 1 Pascal (Pa) 1 standard atmosphere (atm) = 101,325 Pa 1 atm = 101 x 10 3 kg/m.s 2 = 101 x 10 3 Pa 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr Chapter 9/4

Gases and Gas Pressure Pressure (P) is defined as a force F exerted per unit area A. Pressure is equal to force/unit area Chapter 9/5

Gases and Gas Pressure Examples for System International unit (SI units) (1). 10.2 mg resting on an area of 1 cm 2. (2). 10300 kg on an area of 1 m 2 = 1 atm. (3). Density of mercury = 1.35951 x 10 4 kg/m 3 Chapter 9/6

Gases and Gas Pressure Force Pressure: Unit area Atmospheric pressure Chapter 9/7

Gases and Gas Pressure A mercury barometer Chapter 9/8

Gases and Gas Pressure Barometer Units Pa torr mm Hg atm bar Chapter 9/9

Gases and Gas Pressure Conversions 1 atm = 760 mm Hg 1 torr = 1 mm Hg 1 bar = 1 x 10 5 Pa (exact) (exact) (exact) 1 atm = 101 325 Pa Chapter 9/10

Example 9.1 Converting Between Different Units Of Pressure Typical atmospheric pressure on top of Mt. Everest (29,035 ft) is 265 mm Hg. Convert this value to pascals, atmospheres, and bars. Solution conversion factors 101,325 Pa/760 mm Hg, 1 atm/760 mm Hg, and 1 bar/10 5 Pa

Gases and Gas Pressure Open-end manometers for measuring pressure in a gas-filled bulb

Example 9.2 Using An Open-end Manometer Assume that you are using an open-end manometer (Figure 9.4) filled with mineral oil rather than mercury. What is the gas pressure in the bulb in millimeters of mercury if the level of mineral oil in the arm connected to the bulb is 237 mm higher than the level in the arm connected to the atmosphere and atmospheric pressure is 746 mm Hg? The density of mercury is 13.6 g/ml, and the density of mineral oil is 0.822 g/ml. Solution The gas pressure in the bulb equals the difference between the outside pressure and the manometer reading. The manometer reading indicates that the pressure of the gas in the bulb is less than atmospheric because the liquid level is higher on the side connected to the sample. Because mercury is more dense than mineral oil by a factor of 13.6/0.822, or 16.5, a given pressure will hold a column of mercury only 1/16.5 times the height of a column of mineral oil.

The Gas Laws The physical properties of a gas can be defined by four variables: P T V n pressure temperature volume number of moles A gas whose behavior follows the laws exactly is called an ideal gas. Chapter 9/14

The Gas Laws V a 1 P Boyle s Law (constant n and T)

The Gas Laws V a Boyle s Law plot 1 (constant n and T) P

The Gas Laws Chapter 9/17

The Gas Laws V a 1 P Boyle s Law (constant n and T) PV = k P initial V initial = P final V final Chapter 9/18

The Gas Laws Charles s Law V a T (constant n and P)

The Gas Laws

The Gas Laws Charles s Law plot V a T (constant n and P)

The Gas Laws Charles s Law V a T (constant n and P) V T = k V initial T initial = V final T final Chapter 9/22

The Gas Laws Avogadro s Law V a n (constant T and P)

The Gas Laws Avogadro s Law V a n (constant T and P) V n V initial n initial = k V = final n final Chapter 9/24

Example 9.3 Visual Representations of Gas Laws Show the approximate level of the movable piston in drawings (a) and (b) after the indicated changes have been made to the initial gas sample.

Example 9.3 Visual Representations of Gas Laws Continued (a) The temperature T has increased by a factor of 375/250 = 1.5 while the molar amount n and the pressure P are unchanged. According to Charles s law, the volume will increase by a factor of 1.5. (b) The temperature T is unchanged, while both the molar amount n and the pressure P are halved. According to Avogadro s law, halving the molar amount will halve the volume, and according to Boyle s law, halving the pressure will double the volume. The two changes cancel, so the volume is unchanged.

The Ideal Gas Law Summary Boyle s Law: P initial V initial = P final V final Charles Law: V initial T initial V = final T final Avogadro s Law: V initial n initial = V final n final Chapter 9/27

The Ideal Gas Law Is there a mathematical relationship between P, V, n, and T for an ideal gas? Chapter 9/28

The Ideal Gas Law Ideal Gas Law: PV = nrt R is the gas constant and is the same for all gases. L atm R = 0.08206 K mol Standard Temperature and Pressure (STP) for Gases T = 0 o C (273.15 K) P = 1 atm Chapter 9/29

The Ideal Gas Law What is the volume of 1 mol of gas at STP? V = nrt = P L atm (1 mol) 0.08206 (273.15 K) K mol (1 atm) = 22.41 L Chapter 9/30

The Ideal Gas Law Chapter 9/31

Example 9.4 Gas Law Calculations How many moles of gas (air) are in the lungs of an average adult with a lung capacity of 3.8 L? Assume that the lungs are at 1.00 atm pressure and a normal body temperature of 37 C. Solution The lungs of an average adult hold 0.15 mol of air.

Example 9.5 Gas Law Calculations In a typical automobile engine, the mixture of gasoline and air in a cylinder is compressed from 1.0 atm to 9.5 atm prior to ignition. If the uncompressed volume of the cylinder is 410 ml, what is the volume in milliliters when the mixture is fully compressed? Solution This is a Boyle s law problem because only P and V are changing, while n and T remain fixed. (PV) initial = (PV) final = nrt

Stoichiometric Relationships with Gases The reaction used in the deployment of automobile air bags is the high-temperature decomposition of sodium azide, NaN 3, to produce N 2 gas. How many liters of N 2 at 1.15 atm and 30.0 o C are produced by decomposition of 45.0 g NaN 3? 2NaN 3 (s) 2Na(s) + 3N 2 (g) Chapter 9/34

Stoichiometric Relationships with Gases 2NaN 3 (s) 2Na(s) + 3N 2 (g) Moles of N 2 produced: 45.0 g NaN 3 x 1 mol NaN 3 65.0 g NaN 3 x 3 mol N 2 2 mol NaN 3 = 1.04 mol N 2 Volume of N 2 produced: V = nrt = P L atm (1.04 mol) 0.08206 (303.2 K) K mol (1.15 atm) = 22.5 L Chapter 9/35

Example 9.6 Finding A Mass Using Gas Law Calculations A typical high-pressure tire on a bicycle might have a volume of 365 ml and a pressure of 7.80 atm at 25 C. Suppose the rider filled the tire with helium to minimize weight. What is the mass of the helium in the tire? Solution

Stoichiometric Relationships with Gases Determining the density of an unknown gas Chapter 9/37

Stoichiometric Relationships with Gases d mass volume n M V P M R T (M: molecular weight = Molar Mass = g/mol) 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 9/38

Example 9.7 Finding A Density Using Gas Law Calculations What is the density in g/l of ammonia at STP if the gas in a 1.000 L bulb weighs 0.672 g at 25 C and 733.4 mm Hg pressure? Strategy For the ammonia sample, the mass is 0.672 g but the volume of the gas is given under nonstandard conditions and must first be converted to STP. Because the amount of sample is constant, we can set the quantity PV/RT measured under nonstandard conditions equal to PV/RT at STP and then solve for V at STP. Solution The amount of gas in the 1.000 L bulb under the measured nonstandard conditions would have a volume of only 0.884 L at STP. Dividing the given mass by this volume gives the density of ammonia at STP:

Example 9.8 Identifying An Unknown Using Gas Law Calculations To identify the contents of an unlabeled cylinder of gas, a sample was collected and found to have a density of 5.380 g/l at 15 C and 736 mm Hg pressure. What is the molar mass of the gas? Strategy Let s assume we have a 1.000 L sample of the gas, which weighs 5.380 g. We know the temperature, volume, and pressure of the gas and can therefore use the ideal gas law to find n, the number of moles in the sample. Dividing the mass by the number of moles then gives the molar mass. Solution The gas is probably xenon (atomic mass = 131.3 amu).

Stoichiometric Relationships with Gases Summary Standard Temperature and Pressure(STP) STP P = 1 atmosphere T = 0 The molar volume of an ideal gas is 22.414 liters at STP Molar Mass of a Gas(Molecular weight) Density = mass / volume Molar mass = drt / p Chapter 9/41

Partial Pressure and Dalton s Law Dalton s Law of Partial Pressures: The total pressure exerted by a mixture of gases in a container at constant V and T is equal to the sum of the pressures of each individual gas in the container. P total = P 1 + P 2 + + P N Mole Fraction (X) = Moles of component Total moles in mixture n X i = i ntotal or P X i = i Ptotal Chapter 9/42

Partial Pressure and Dalton s Law In a mixture of gases the total pressure, P tot, is the sum of the partial pressures of the gases: P total Dalton s law allows us to work with mixtures of gases. RT V n For a two-component system, the moles of components A and B can be represented by the mole fractions (X A and X B ). na X A n n X X B A n X A A B n B n 1 B B Chapter 9/43

Partial Pressure and Dalton s Law The partial pressure of each gas in a mixture of gases in a container depends on the number of moles of that gas. Similarly, X a +X b +X c + =1 for all components. Mole fraction is related to the total pressure by: P i X i P tot Chapter 9/44

Example 9.9 Using Partial Pressures And Mole Fractions At an underwater depth of 250 ft, the pressure is 8.38 atm. What should the mole percent of oxygen in the diving gas be for the partial pressure of oxygen in the gas to be 0.21 atm, the same as in air at 1.0 atm? Solution The diving gas should contain 2.5% O 2 for the partial pressure of O 2 to be the same at 8.38 atm as it is in air at 1.0 atm.

Think about it 1 On a beautiful autumn day in NDMC, you are holding two balloons with your lover, each with 10 mole of gas. Which balloon is larger? The one filled with neon (Ne) gas. The one filled with argon (Ar) gas. Chapter 9/46

Think about it 2 Consider the following container of helium (He). Initially the valve is closed. The total pressure in the container after the valve is opened is: <5.0 atm 5.0 atm >5.0 atm 2.00 atm 3.00 atm 9.00 L 3.00 L Chapter 9/47

The Kinetic-Molecular Theory of Gases So far we have considered what happens, but not why. In science, what always comes before why. Chapter 9/48

The Kinetic-Molecular Theory of Gases This theory presents physical properties of gases in terms of the motion of individual molecules. Assumptions: 1. A gas consists of tiny particles, either atoms or molecules, moving about at random. 2. The volume of the particles themselves is negligible compared with the total volume of the gas. Most of the volume of a gas is empty space. 3. The gas particles act independently of one another; there are no attractive or repulsive forces between particles. 4. Collisions of the gas particles, either with other particles or with the walls of a container, are elastic (constant temperature). 5. The average kinetic energy of the gas particles is proportional to the Kelvin temperature of the sample. Chapter 9/49

A kinetic-molecular view of the gas laws

The Kinetic-Molecular Theory of Gases Average Kinetic Energy (KE) is given by: The total kinetic energy of a mole of gas particles equals 3RT / 2. The average kinetic energy per particle is 3RT / 2N A. (N A is Avogadro s number) Ek 1 mu 2 3RT u M 2 = 3RT / 2 ; M Mwt = molar mass Mean free path: At room temperature and 1 atm pressure, the average distance between collisions. Chapter 9/51

The Kinetic-Molecular Theory of Gases molar mass average speed Chapter 9/52

The Kinetic-Molecular Theory of Gases The distribution of speeds for helium atoms at different temperatures. Chapter 9/53

Graham s Law: Diffusion and Effusion of Gases Chapter 9/54

Root Mean Square Velocity (u rms ): Molecular Speeds 1. Gas molecules do not all move at the same speed, they have a wide distribution of speeds. 2. The root-mean-square velocity, u rms, is the square root of the average of the squares of the molecular speeds. Rate a u rms 3RT M 3. At a fixed temperature, molecules of higher mass (M) move more slowly than molecules of lower mass. Chapter 9/55

Graham s Law: Diffusion and Graham s Law Rate a 1 M Effusion of Gases Graham s Law: Rate of effusion is proportional to its rms speed, u rms. Rate a u rms 3RT M For two gases at same temperature and pressure: Rate Rate 1 2 M M 2 1 M M 2 1 Chapter 9/56

Example 9.10 Using Graham s Law To Calculate Diffusion Rates Assume that you have a sample of hydrogen gas containing H 2, H D, and D 2 that you want to separate into pure components (H = 1 H and D = 2 H). What are the relative rates of diffusion of the three molecules according to Graham s law? (for H 2, m = 2.016 amu; for HD, m = 3.022 amu; for D 2, m = 4.028 amu.) Solution Because D 2 is the heaviest of the three molecules, it will diffuse most slowly. If we call its relative rate 1.000, we can then compare HD and H 2 with D 2 : Thus, the relative rates of diffusion are H 2 (1.414) > HD (1.154) > D 2 (1.000).

The Behavior of Real Gases An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law, although many gases come very close at low pressure and/or high temperature. In fact, we often learn more about nature from the failures of our models than from their successes.

The Behavior of Real Gases The volume of a real gas is larger than predicted by the ideal gas law. The volume of a real gas Chapter 9/59

The Behavior of Real Gases Attractive forces between particles become more important at higher pressures. Molecules attract one another at distances up to about 10 molecular diameters. The result is a decrease in the actual volume of most real gases when compared with ideal gases at pressures up to 300 atm. Chapter 9/60

The Behavior of Real Gases For an ideal gas, PV/nRT = 1 under all conditions, but notice that for real gases, PV/nRT approaches 1 only at very low pressure (typically below 1 atm). Plots of PV/nRT versus P for Several Gases (200K)

The Behavior of Real Gases Note that the behavior of the gas appears to become more nearly ideal as the temperature is increased. Plots of PV/nRT versus P for Nitrogen Gas at Three Temperatures

The Behavior of Real Gases (a) Gas at Low Concentration--Relatively Few Interactions Between Particles (b) Gas at High Concentration Many More Interactions Between Particles

The Behavior of Real Gases Since the 1 2 pair is the same as the 2 1 pair, this analysis counts each pair twice. Thus, for N particles, there are N(N-1)/2 pairs. If N is a very large number, N-1 approximately equals N, giving N 2 /2 possible pairs. Illustration of Pairwise Interactions Among Gas Particles

How can we modify the assumptions of the KMT to fit the behavior of real gases? nrt P V nb n: number of mole b: constant (from experimental results) nrt Pobs ( P correction factor) correction factor V nb The size of the correction factor depends on the concentrations of gas molecules defined in terms of moles of gas particles per liter (n/v). For large numbers of particles, the number of interacting pairs of particles depends on the square of the number of particles and thus on the square of the concentration, or (n/v) 2

The Behavior of Real Gases 2 2 n nrt n obs = P P a a V V nb V Van der waals equation: 2 n P a V nb V obs = nrt

The Behavior of Real Gases van der Waals equation Correction for intermolecular attractions. P a + V - n b n 2 V 2 = nrt Correction for molecular volume. Chapter 9/67

The Behavior of Real Gases a is a constant which includes the factor of ½ from N 2 /2 Values of the van der Waals Constants for Some Common Gases

The Earth s Atmosphere The change of P and T at different altitude. Variations of atmospheric pressure and average temperature with altitude Chapter 9/69

The Earth s Atmosphere Annual concentration of atmospheric CO 2 since 1850 Chapter 9/70