Pressure
Dr. L. Dawe
Chemistry 110/120
Lecture 4 and 5 Chapter 2: The Behaviour of Gases • Section 2.1 (Pages 62-63): Pressure – Units of Pressure – Suggested end-of-chapter exercises: 2.5, 2.6 • Section 2.2 (Pages 64-69): Describing Gases – Variations in Gas Volume – The Ideal Gas Equation – Suggested end-of-chapter exercises: 2.7, 2.8, 2.11, 2.12
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Lecture 4 and 5 Learning Outcomes Students should be able to:
2.1 understand gas pressure and express pressure in various units 2.2 explain the relationships between pressure, volume, temperature and amount of gas 2.3 express the relationship between pressure, volume, temperature and amount of gas as the ideal gas law (PV = nRT)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
General Gas Characteristics • Gases – Fill and assume the shape of their containers – Diffuse into each other – Mix in all proportions
• Gas properties are determined by – Quantity of gas – Volume – Temperature – Pressure Liquid bromine with brown bromine vapor above
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Definition of pressure • Pressure: force per unit area • P = F/A P = pressure (Pa = kg m-1 s-2) F = force (N = kg m s-2) A = area (m2)
• Note that the SI unit for pressure is the Pascal (Pa), and for force, it is the Newton (N). • It is difficult to measure total force exerted by gas molecules!
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Origin of Pressure: Pressure is a measure of the force a gas exerts on a surface. How do individual gas particles create this force? A: When a moving object collides with any surface it exerts a force and therefore pressure. When a particle collides with the container wall, it exerts a force. All of the collisions occurring at any instant make up the observed pressure. The more particles there are the more frequent are the collisions with the wall and therefore the greater the pressure.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Liquid pressure • Gas pressure is normally measured indirectly by comparison to liquid pressure. • Pliquid = g ·h ·d P = pressure (Pa = kg m-1 s-2) g = acceleration due to gravity (9.80665 m s-2) h = height (m) d = density (kg m-3)
All the interconnected vessels fill to the same height. As a result the liquid pressures are the same, despite the different shapes and volumes of the containers.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Barometric pressure • Barometer: device used to measure the pressure of the atmosphere (ie. the barometric pressure). • Normally, a barometer is a tube ~1m long, closed at one end, filled with mercury and inverted into a dish containing more mercury. • At sea level and under normal atmospheric conditions the mercury stops flowing out when the level of mercury in the tube and above the pool of mercury is 760 mm.
• Barometric pressure varies with atmospheric conditions and altitude.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Barometric pressure Problem: Why is it practical to construct barometers using mercury, given that it is rare, expensive and poisonous? (dHg = 13.6 g/cm3, dH2O = 1.00 g/cm3)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Barometric pressure and common pressure units
• Manometer: similar to a barometer, but gas pressure is exerted on both liquid surfaces, and the difference in pressures in measured. • Atmospheric pressure can be expressed in various units:
Dr. L. Dawe
Chemistry 110/120
Boyle's law • For a fixed amount of gas at a constant temperature, the gas volume is inversely proportional to the gas pressure. P a 1/V
or
PV = constant Note that the constant is dependent on the amount of gas and on temperature.
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Boyle's law Problem: 1.50 L of gas at 2.25 atm pressure is connected to a flask of volume 8.10 L. What is the final gas pressure expressed in mmHg? (Hint: What is the total volume?) V = 1.50 L
V = 8.10 L
P=?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Boyles Law: A change in gas pressure in one direction causes a change in gas volume in the other. What happens to the particles when external pressure compresses the gas volume? Why aren’t liquids and solids compressible? A: A gas is composed of very tiny particles of negligible volume and consists of mostly empty space. When an external pressure is applied to the gas, the molecules get closer together, decreasing the volume of the sample. The pressure exerted by the gas increases simultaneously because a smaller volume has shorter distances between gas molecules and the walls so there are more frequent collisions. Solids and liquids are different in that the molecules are already very close together.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Kelvin temperature scale • Kelvin temperature: an absolute temperature. The lowest attainable temperature is 0 K = -273.15oC (ie. the temperature at which molecular motion ceases). • Kelvin and Celsius temperatures are related by: T (K) = t (oC) + 273.15
Dr. L. Dawe
Chemistry 110/120
Barometric pressure • Barometer: device used to measure the pressure of the atmosphere (ie. the barometric pressure). • Normally, a barometer is a tube ~1m long, closed at one end, filled with mercury and inverted into a dish containing more mercury.
• At sea level and under normal atmospheric conditions the mercury stops flowing out when the level of mercury in the tube and above the pool of mercury is 760 mm.
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Charles's law • The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin (absolute) temperature. VaT or V = (constant)(T)
Dr. L. Dawe
Chemistry 110/120
Charles's law
Fall 2014
V = (constant)(T)
• Note: The value of the constant depends on the amount of gas and the pressure. • Note: The VT relationship is linear for both the Celcius and Kelvin scales, but volume is directly proportional only to the absolute (Kelvin) scale.
Dr. L. Dawe
Chemistry 110/120
Charles's law
Fall 2014
V = (constant)(T)
• William Thompson (Lord Kelvin) found that if you extend or extrapolate any V vs T plot back to 0 L they all intersect at -273.15 oC.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Charles’s law and the Kelvin temperature scale Problem: A gas at 25 oC and 0.987 atm pressure is confined in a cylinder by a piston. When the cylinder is heated, the gas volume expands from 0.250 L to 1.65 L. What is the new temperature of the gas, assuming the pressure remains constant?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Charles Law: A change in temperature is accompanied by a corresponding change in volume. What effect does higher temperature have on gas particles that increase the volume or increases the pressure if volume is fixed?
A: As the temperature increases, the most probable molecular speed and average kinetic energy increases. Therefore the molecules hit the container walls more frequently and more energetically which causes a greater gas pressure.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Standard conditions of temperature and pressure (STP) • Standard conditions of temperature and pressure (STP) refers to a gas maintained at a temperature of exactly 0 oC (273.15 K) and 105 Pa (1 bar). • Other systems of measurement were used in the past, in particular, STP pressure used to be defined as 1 atm • The above definition agrees with The International Union of Pure and Applied Chemistry.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Avogadro's Hypothesis 1.
2.
Equal volumes of different gases compared at the same temperature and pressure contain equal numbers of molecules Equal numbers of molecules of different gases compared at the same temperature and pressure occupy equal volumes.
If equal volumes of gases contain equal numbers of molecules, this means the volume of O2(g) is one half that of H2(g) .
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Avogadro's Law Avogadro’s Law: At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas.
Dr. L. Dawe
Chemistry 110/120
Avogadro's Law 1 mol gas = 6.022 x 1023 molecules gas = 22.7 L gas (at STP) or = 22.4 L gas at 1 atm and 273.15 K
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Avagadro’s Law: Gas volume (or pressure) depends on the number of moles present, not on the nature of the particular gas. But shouldn’t one mole of larger molecules occupy more space than one mole of smaller ones? And why doesn’t one mole of heavier molecules exert more pressure than one mole of lighter ones? A: Adding more molecules to a container increases the total number of collisions with the walls and, therefore, the internal pressure. As a result, the volume expands to yield the same pressure as before.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Avogadro's Law Avogadro’s Law: At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas. V a n (where n is the number of moles) or V = (constant) (n)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Avogadro's Law Problem: A 128 g piece of solid carbon dioxide sublimes into CO2(g). How many litres are formed at STP? Recall: STP refers to a gas maintained at a temperature of exactly 0 oC (273.15 K) and 105 Pa (1 bar).
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Combining the gas laws • Boyle’s Law: P a 1/V • Charles’s Law: V a T • Avogadro’s Law: V a n • These simple laws can be combined to give:
nT V P
or
RnT V P
where R is the gas constant
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The Ideal Gas Equation and R • The ideal gas equation normally takes the form: PV nRT
• And R can be determined by: PV 1atm 22.4140 L R 0.082057 L atm mol 1 K 1 nT 1mol 273.15 K
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The gas constant R • Depending on the choice of units, R can have different common values 0.082057 L atm mol-1K-1 62.364 L Torr mol-1K-1 8.3145 m3 Pa mol-1K-1 8.3145 J mol-1 K-1
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The Ideal Gas Equation Problem: At what temperature will a 13.7 g Cl2 sample exert a pressure of 745 mmHg when confined in a 7.50 L container? Problem: How many moles of He(g) are in a 5.00 L storage tank filled with He at 10.5 atm pressure at 30.0 oC?
Dr. L. Dawe
Chemistry 110/120
Lecture 6 and 7 Chapter 2: The Behaviour of Gases • Section 2.2 (Pages 64-69): Describing Gases – Variations on The Ideal Gas Equation
• Section 2.6 (Pages 85-90): Additional Gas Properties – Determination of Molar Mass – Gas Density – Suggested end-of-chapter exercises: 2.45-2.48
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Lecture 6 and 7 Chapter 2: The Behaviour of Gases • Section 2.4 (Pages 74-78): Gas Stoichiometry – Summary of Mole Conversions – Suggested end-of-chapter exercises: 2.24, 2.26, 2.27, • Section 2.3 (Pages 70-73): Gas Mixtures – Dalton’s Law of Partial Pressures – Describing Gas Mixtures – Suggested end-of-chapter exercises: 2.17, 2.18, 2.21
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Lecture 6 and 7 Learning Outcomes Students should be able to:
2.4 use the concept of partial pressures in gas mixtures 2.5 use stoichiometry to solve problems involving gas-phase chemical reactions 2.7 calculate gas densities and molar masses from pressurevolume-temperature data
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Combining the Gas laws: The general gas equation
• These can be combined to give the general gas equation:
PiVi P f V f ni Ti n f Tf
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Combining the Gas laws: The general gas equation
Problem: A 1.00 mL sample of N2(g) at 36.2oC and 2.14 atm is heated to 37.8oC, and the pressure changed to 1.02 atm. What volume does the gas occupy at this final temperature and pressure?
Dr. L. Dawe
Chemistry 110/120
Applications of the Ideal Gas Equation
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Molar mass determination • The ideal gas equation can be rearranged to solve for molar mass directly. • We know: n
m MM
and
PV nRT
therefore by substitution we arrive at: mRT PV MM
or
mRT MM PV
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Molar mass determination • Problem: A 2.650 g sample of a gaseous compound occupies 428 mL at 24.3 oC and 742 mmHg. The compound consists of 15.5% C, 23.0% Cl and 61.5% F by mass. What is the molecular formula? • Problem: A 132.10 mL glass vessel weighs 56.1035 g when evacuated and 56.2445 g when filled with the gaseous hydrocarbon acetylene at 749.3 mmHg and 20.02 oC. What is the molecular mass of acetylene?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas densities • Gas density differs from the density of liquids and solids in that: – Gas densities depend on both temperature and pressure, while for solids and liquids, density depends slightly on temperature, and far less on pressure (think about the compressibility of a gas vs. that of the same compound in the solid phase!)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas densities • Gas density differs from the density of liquids and solids in that: – The density of a gas is directly proportional to its molar mass. This is not the case for solids or liquids.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas densities • Gas densities are normally much smaller than the densities of liquids or solids. • Gas densities are normally given in g/L, while solids and liquids are expressed in g/mL.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas densities Problem: The density of phosphorus vapor at 310oC and 775 mmHg is 2.64 g/L. What is the molecular formula of the phosphorus under these conditions?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gases in Chemical Reactions Airbags:
2 NaN3(s) 2 Na(l) + 3 N2(g)
The ideal gas law may now be applied to reactions that involve gaseous reactants or products. (1) Use stoichiometric factors to relate the amount (mols) of gas to other reactants or products (2) Use the ideal gas law to relate the amount of gas to volume, temperature and pressure.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Stoichiometric calculations involving gases Problem: Calculate the volume of H2(g), measured at 26oC and 751 mmHg, required to react with 28.5 L of CO(g), measured at 0oC and 760 mmHg, in the following reaction: 3CO(g) + 7H2(g) C3H8(g) + 3H2O(l)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Mixtures of Gases • Simple gas laws and the ideal gas law were originally derived for air (a mixture of gases). • When dealing with a mixture of non-reactive gases, the simplest approach is to use the number of total moles of gas for n.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Mixtures of Gases Problem: 2.0 L of O2(g) and 8.0 L of N2(g), each at STP, are mixed together. The non-reactive gaseous mixture is compressed to occupy 2.0 L at 298 K. What is the pressure exerted by this mixture?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Dalton's law of partial pressures • Dalton’s law of partial pressures states that in a mixture of gases, the total pressure is the sum of the partial pressures of the gases present. That is: Ptot PA PB
Note that since gases expand to fill their containers, here:
Vtot V A VB
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Dalton’s Law: The pressure of a gas mixture is the sum of the pressures of the individual gases. Why does each gas contribute to the total pressure in proportion to its mole fraction?
A: Adding another component to a gas simply increases the number of particles which increases the frequency of collisions and therefore the pressure. Each gas exerts a fraction of the total pressure based on the fraction of molecules (or moles) of that gas in the mixture.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Mole fraction • Mole fraction (c) describes a mixture in terms of the fraction of all the molecules that are of a particular type. It is the amount of one component, in moles, divided by the total amount of all the substances in the mixture. VA PA nA cA Vtot Ptot ntot
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Mole fraction Problem: A mixture of 0.197 mol CO2(g) and 0.00278 mol H2O(g) is held at 30.0oC and 2.50 atm. What is the partial pressure of each gas?
Dr. L. Dawe
Chemistry 110/120
Lecture 7 and 8 Chapter 2: The Behaviour of Gases • Section 2.8 (Pages 97): Chemistry of the Earth’s Atmosphere – Vapour Pressure
• Section 2.5 (Pages 79-84): Molecular View of Gases – Molecular Speeds – Speed and Energy – Average Kinetic Energy – Ideal Gases – Suggested end-of-chapter exercises: 2.51, 2.52, 2.33-2.36
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Lecture 7 and 8 Learning Outcomes Students should be able to:
2.10 calculate water vapour pressure and relative humidity 2.6 explain the basic concepts of kinetic molecular theory: molecular speed, energy, and the effects of temperature and volume on gas pressure
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Collecting a gas and downward displacement of a liquid • A pneumatic trough is a device that was commonly used to collect gas from a reaction or other gas generating device, for further analysis. • The gas that is collected displaces water, and is actually a mixture of gases; that which we want to analyze, and water vapour.
Dr. L. Dawe
Chemistry 110/120
Collecting a gas and downward displacement of a liquid
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Collecting a gas and downward displacement of a liquid Gas pressure can be adjusted to atmospheric pressure by adjusting the height of the bottle, to give: Ptot = Pbar = Pgas + PH2O Pgas = Pbar – PH2O Values of water vapour pressure can be obtained from tables:
Dr. L. Dawe
Chemistry 110/120
Equilibrium Vapor Pressure Problem: In the following reaction, if 35.5 mL of H2(g) is collected over water at 26 oC and 755 mmHg barometric pressure, how many moles of HCl must be consumed? (The vapor pressure of water at 26 oC is 25.2 mmHg). 2 Al(s) + 6 HCl(aq) 2 AlCl3(aq) + 3 H2(g)
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Postulates of the Kinetic Molecular theory Postulate 1. Particle Volume. • A gas consists of a large collection of individual particles. • The volume of an individual particle is extremely small compared to the volume of the container. • Most of the space occupied by a gas is empty.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Postulates of the Kinetic Molecular theory Postulate 2. Particle Motion. • Gas particles are in constant motion. • Their motion is random and in straight lines. • At any one time there are gas particles moving in all directions. • When they collide with the walls of the container or one another they change their direction.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Postulates of the Kinetic Molecular theory Postulate 3. Particle Collisions. • The forces of attraction between molecules in a gas are negligible. Molecules do not influence one another unless they collide. • Collisions between molecules or with container walls are fleeting and most of the time molecules are not involved in collisions. • When molecules collide they can transfer kinetic energy.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Postulates of the Kinetic Molecular theory Postulate 4. Temperature. • In a collection of molecules at a certain temperature the total energy remains constant although each individual molecule may have a certain kinetic energy and can transfer energy in a collision.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Molecular speed • The average kinetic energy of a gas depends on its temperature and its mass. • The average speed of molecules of a gas is greater at higher temperature. • Species with different molecular masses have the same average kinetic energies at any given temperature, but the smaller molecules have a greater average speed.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Molecular speed • Molecules have a distribution of speeds. For example, they could be moving very fast one second, then collide with another molecule and transfer energy to the other molecule and be sitting almost still at another instant. um = modal speed uav = average speed urms = root mean square speed
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Average and root-mean-square speed Problem: Five molecules are traveling at different speeds: 400, 450, 525, 585 and 600 m/s. Calculate the average speed and the root-mean-square speed for this collection of molecules.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Translational kinetic energy • Translational kinetic energy (ek): energy possessed by objects (ie. gas molecules) traveling through space.
1 E k mu 2 2 m = mass of the molecule (g/molecule) u = velocity 1 2 E k mu rms 2
urms = root-mean square speed. A molecule moving with this speed has the average kinetic energy.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Pressure • Translational kinetic energy, • Frequency of collisions, • Impulse or momentum transfer, • Pressure proportional to impulse times frequency
• Three dimensional systems lead to:
1 e k mu 2 2
N vu V I mu N P mu 2 V 1N P m u2 3V
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Pressure • Translational kinetic energy, • Frequency of collisions, • Impulse or momentum transfer, • Pressure proportional to impulse times frequency
• Three dimensional systems lead to:
1 e k mu 2 2
N vu V I mu N P mu 2 V 1N P m u2 3V
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Root-mean-square speed Assume one mole:
1 2 PV N A m u 3
PV=RT so:
3RT N A m u
NAm = M:
3RT M u
Rearrange:
u rms
2
3RT M
2
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Root-mean-square speed Assume one mole:
1 2 PV N A m u 3
PV=RT so:
3RT N A m u
NAm = M:
3RT M u
2
2 8.3145 J mol-1K-1
Rearrange:
u rms
3RT M
J has units kg m2 s-2
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Root-mean-square speed In a qualitative sense, this tells us that the higher the molar mass of a molecule, the Assume lower its root-mean-square speed. one mole:
PV=RT so:
1 PV N A m u 2 3 3RT N A m u 2
NAm = M:
Rearrange:
3RT M u 2 u rms
3RT M
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Root-mean-square speed A member of the cat family, the cheetah leads the way as the world's fastest mammal. Cheetahs can attains speeds of 114 km/hr, and have been recorded accelerating from 0-60mph in just 3 seconds - faster than most sports cars. Problem: Which has the greatest speed: a cheetah running at its maximum speed, or the root-mean-square speed of H2 molecules at 25oC?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Distribution of molecular speeds and effect of temperature
The average kinetic energy of a gas particle is proportional to the Kelvin temperature of the sample. That is, on average, for particles of the same gas, those at a higher temperature will possess more kinetic energy, reflected in a higher average speed.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The meaning of temperature and absolute zero Kelvin • The Kelvin temperature (T) of a gas is directly proportional to the average translational kinetic energy (Ek) of its molecules. • The absolute zero of temperature is the temperature at which translational molecular motions should cease.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas Properties Related to the Kinetic Molecular Theory: Diffusion • Diffusion: the migration of molecules as a result of random molecular motion.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas Properties Related to the Kinetic Molecular Theory: Diffusion NH3(g) + HCl(g) NH4Cl(s)
1. The gases diffuse toward each other and where they meet a white cloud of ammonium chloride forms. 2. Because of their greater average speed, NH3(g) molecules diffuse faster than HCl(g) molecules, so the cloud forms closer to the drop of HCl(aq).
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas Properties Related to the Kinetic Molecular Theory: Effusion • Effusion: the escape of gas molecules from their container through a tiny orifice or pinhole.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Effusion • The rate of effusion is directly proportional to molecular speed. • The rate of effusion for two different gases at the same temperature can be compared by: EffusionRateA ( urms ) A EffusionRateB ( urms ) B
3 RT / M A 3 RT / M B
MB MA
• At a given temperature and pressure, the gas with the lower molar mass effuses faster because the most probable speed of its molecules is higher, therefore more molecules escape through the tiny hole per unit time.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Effusion • Graham’s Law: The rates of effusion of two different gases are inversely proportional to the square roots of their molar masses. • Graham’s law only applies to gases at very low pressures (ie. molecules escape through the pinhole, one at a time) and when the orifice is very tiny, so that no collisions take place as molecules are passing through.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Effusion Problem: 2.2 x 10-4 mol N2(g) effuses through a tiny hole in 105 s. How much O2(g) would effuse through the same orifice in 105 s? Problem: A sample of Kr(g) escapes through a tiny hole in 87.3 s. An unknown gas requires 131.3 s to escape under the same conditions. What is the molar mass of this unknown gas?
Dr. L. Dawe
Chemistry 110/120
Lecture 8 Chapter 2: The Behaviour of Gases • Section 2.7 (Pages 93-95): Non-Ideal (Real) Gases • Suggested end-of-chapter exercises: 2.55-2.57, 2.59
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Lecture 8 Learning Outcomes Students should be able to:
2.8 explain the two principle ways in which ideal gases and real gases differ 2.9 calculate the pressure of a gas under non-ideal conditions
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Qualitative comparison of nonideal gases to ideal gases • Up to now, our gas laws have applied to ideal gases. That is, it was assumed that individual gas molecules acted as points, and occupied zero volume, and that intermolecular forces between these points was negligible. This is not the case for real gases. • Actually: – Gases tend to behave ideally at high temperatures and low pressures. – Gases tend to behave nonideally at low temperatures and high pressures. » WHY?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The van der Waals equation • Compressibility factor of a gas is defined as PV/nRT For ideal gases: PV/nRT = 1 For real gases: PV/nRT < 1 intermolecular forces are significant. PV/nRT > 1 molecular volume is significant • The van der Waals equation corrects for the volume associated with the molecules themselves and for intermolecular forces:
n2a P 2 (V nb) nRT V
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The van der Waals equation n2a P 2 (V nb) nRT V
Dutch physicist Johannes van der Waals reasoned that the decrease in pressure due to intermolecular attractions should be proportional to the square of the concentration; “a” provides a measure of how strongly the molecules attract each other.
“b” is called the “excluded volume per mole” and is roughly equal to the volume that one mole of gas occupies when condensed to the liquid phase.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The van der Waals equation • Compressibility factor of a gas is defined as PV/nRT For ideal gases: PV
nRT
1
For real gases:
PV 1 nRT
PV 1 nRT
intermolecular forces are molecular volume is significant significant • The van der Waals equation corrects for the volume associated with the molecules themselves and for intermolecular forces:
n2a P 2 (V nb) nRT V
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The correction for intermolecular forces, n2a/V2 • The effect of attractive forces between molecules is to decrease the pressure that a mole of gas (at standard temperature in a 22.4 L container) from that which it would exert if it were ideal. • That is, PV/nRT < 1
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The excluded volume term, nb
• The effect of molecular volume is to make the volume occupied by a mole of gas larger than predicted by the ideal gas law. • That is, PV/nRT > 1
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Application of the van der Waals Equation
• Problem: Use the van der Waals equation to calculate the pressure exerted by 1.00 mol Cl2 (g) confined to a volume of 2.00 L at 273 K. What is the pressure exerted for CO2 (g), under the same conditions? Which gas shows the greater departure from ideal gas behaviour? What are the main causes of the departures from ideality for each gas?
The constants a and b must be given! For Cl2(g): a = 6.49 L2 atm mol-2 and b = 0.0562 L mol-1 For CO2(g): a = 3.59 L2 atm mol-2 and b = 0.0427 L mol-1
Chemistry 110/120
Lecture 4 and 5 Chapter 2: The Behaviour of Gases • Section 2.1 (Pages 62-63): Pressure – Units of Pressure – Suggested end-of-chapter exercises: 2.5, 2.6 • Section 2.2 (Pages 64-69): Describing Gases – Variations in Gas Volume – The Ideal Gas Equation – Suggested end-of-chapter exercises: 2.7, 2.8, 2.11, 2.12
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Lecture 4 and 5 Learning Outcomes Students should be able to:
2.1 understand gas pressure and express pressure in various units 2.2 explain the relationships between pressure, volume, temperature and amount of gas 2.3 express the relationship between pressure, volume, temperature and amount of gas as the ideal gas law (PV = nRT)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
General Gas Characteristics • Gases – Fill and assume the shape of their containers – Diffuse into each other – Mix in all proportions
• Gas properties are determined by – Quantity of gas – Volume – Temperature – Pressure Liquid bromine with brown bromine vapor above
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Definition of pressure • Pressure: force per unit area • P = F/A P = pressure (Pa = kg m-1 s-2) F = force (N = kg m s-2) A = area (m2)
• Note that the SI unit for pressure is the Pascal (Pa), and for force, it is the Newton (N). • It is difficult to measure total force exerted by gas molecules!
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Origin of Pressure: Pressure is a measure of the force a gas exerts on a surface. How do individual gas particles create this force? A: When a moving object collides with any surface it exerts a force and therefore pressure. When a particle collides with the container wall, it exerts a force. All of the collisions occurring at any instant make up the observed pressure. The more particles there are the more frequent are the collisions with the wall and therefore the greater the pressure.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Liquid pressure • Gas pressure is normally measured indirectly by comparison to liquid pressure. • Pliquid = g ·h ·d P = pressure (Pa = kg m-1 s-2) g = acceleration due to gravity (9.80665 m s-2) h = height (m) d = density (kg m-3)
All the interconnected vessels fill to the same height. As a result the liquid pressures are the same, despite the different shapes and volumes of the containers.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Barometric pressure • Barometer: device used to measure the pressure of the atmosphere (ie. the barometric pressure). • Normally, a barometer is a tube ~1m long, closed at one end, filled with mercury and inverted into a dish containing more mercury. • At sea level and under normal atmospheric conditions the mercury stops flowing out when the level of mercury in the tube and above the pool of mercury is 760 mm.
• Barometric pressure varies with atmospheric conditions and altitude.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Barometric pressure Problem: Why is it practical to construct barometers using mercury, given that it is rare, expensive and poisonous? (dHg = 13.6 g/cm3, dH2O = 1.00 g/cm3)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Barometric pressure and common pressure units
• Manometer: similar to a barometer, but gas pressure is exerted on both liquid surfaces, and the difference in pressures in measured. • Atmospheric pressure can be expressed in various units:
Dr. L. Dawe
Chemistry 110/120
Boyle's law • For a fixed amount of gas at a constant temperature, the gas volume is inversely proportional to the gas pressure. P a 1/V
or
PV = constant Note that the constant is dependent on the amount of gas and on temperature.
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Boyle's law Problem: 1.50 L of gas at 2.25 atm pressure is connected to a flask of volume 8.10 L. What is the final gas pressure expressed in mmHg? (Hint: What is the total volume?) V = 1.50 L
V = 8.10 L
P=?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Boyles Law: A change in gas pressure in one direction causes a change in gas volume in the other. What happens to the particles when external pressure compresses the gas volume? Why aren’t liquids and solids compressible? A: A gas is composed of very tiny particles of negligible volume and consists of mostly empty space. When an external pressure is applied to the gas, the molecules get closer together, decreasing the volume of the sample. The pressure exerted by the gas increases simultaneously because a smaller volume has shorter distances between gas molecules and the walls so there are more frequent collisions. Solids and liquids are different in that the molecules are already very close together.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Kelvin temperature scale • Kelvin temperature: an absolute temperature. The lowest attainable temperature is 0 K = -273.15oC (ie. the temperature at which molecular motion ceases). • Kelvin and Celsius temperatures are related by: T (K) = t (oC) + 273.15
Dr. L. Dawe
Chemistry 110/120
Barometric pressure • Barometer: device used to measure the pressure of the atmosphere (ie. the barometric pressure). • Normally, a barometer is a tube ~1m long, closed at one end, filled with mercury and inverted into a dish containing more mercury.
• At sea level and under normal atmospheric conditions the mercury stops flowing out when the level of mercury in the tube and above the pool of mercury is 760 mm.
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Charles's law • The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin (absolute) temperature. VaT or V = (constant)(T)
Dr. L. Dawe
Chemistry 110/120
Charles's law
Fall 2014
V = (constant)(T)
• Note: The value of the constant depends on the amount of gas and the pressure. • Note: The VT relationship is linear for both the Celcius and Kelvin scales, but volume is directly proportional only to the absolute (Kelvin) scale.
Dr. L. Dawe
Chemistry 110/120
Charles's law
Fall 2014
V = (constant)(T)
• William Thompson (Lord Kelvin) found that if you extend or extrapolate any V vs T plot back to 0 L they all intersect at -273.15 oC.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Charles’s law and the Kelvin temperature scale Problem: A gas at 25 oC and 0.987 atm pressure is confined in a cylinder by a piston. When the cylinder is heated, the gas volume expands from 0.250 L to 1.65 L. What is the new temperature of the gas, assuming the pressure remains constant?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Charles Law: A change in temperature is accompanied by a corresponding change in volume. What effect does higher temperature have on gas particles that increase the volume or increases the pressure if volume is fixed?
A: As the temperature increases, the most probable molecular speed and average kinetic energy increases. Therefore the molecules hit the container walls more frequently and more energetically which causes a greater gas pressure.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Standard conditions of temperature and pressure (STP) • Standard conditions of temperature and pressure (STP) refers to a gas maintained at a temperature of exactly 0 oC (273.15 K) and 105 Pa (1 bar). • Other systems of measurement were used in the past, in particular, STP pressure used to be defined as 1 atm • The above definition agrees with The International Union of Pure and Applied Chemistry.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Avogadro's Hypothesis 1.
2.
Equal volumes of different gases compared at the same temperature and pressure contain equal numbers of molecules Equal numbers of molecules of different gases compared at the same temperature and pressure occupy equal volumes.
If equal volumes of gases contain equal numbers of molecules, this means the volume of O2(g) is one half that of H2(g) .
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Avogadro's Law Avogadro’s Law: At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas.
Dr. L. Dawe
Chemistry 110/120
Avogadro's Law 1 mol gas = 6.022 x 1023 molecules gas = 22.7 L gas (at STP) or = 22.4 L gas at 1 atm and 273.15 K
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Avagadro’s Law: Gas volume (or pressure) depends on the number of moles present, not on the nature of the particular gas. But shouldn’t one mole of larger molecules occupy more space than one mole of smaller ones? And why doesn’t one mole of heavier molecules exert more pressure than one mole of lighter ones? A: Adding more molecules to a container increases the total number of collisions with the walls and, therefore, the internal pressure. As a result, the volume expands to yield the same pressure as before.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Avogadro's Law Avogadro’s Law: At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas. V a n (where n is the number of moles) or V = (constant) (n)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Avogadro's Law Problem: A 128 g piece of solid carbon dioxide sublimes into CO2(g). How many litres are formed at STP? Recall: STP refers to a gas maintained at a temperature of exactly 0 oC (273.15 K) and 105 Pa (1 bar).
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Combining the gas laws • Boyle’s Law: P a 1/V • Charles’s Law: V a T • Avogadro’s Law: V a n • These simple laws can be combined to give:
nT V P
or
RnT V P
where R is the gas constant
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The Ideal Gas Equation and R • The ideal gas equation normally takes the form: PV nRT
• And R can be determined by: PV 1atm 22.4140 L R 0.082057 L atm mol 1 K 1 nT 1mol 273.15 K
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The gas constant R • Depending on the choice of units, R can have different common values 0.082057 L atm mol-1K-1 62.364 L Torr mol-1K-1 8.3145 m3 Pa mol-1K-1 8.3145 J mol-1 K-1
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The Ideal Gas Equation Problem: At what temperature will a 13.7 g Cl2 sample exert a pressure of 745 mmHg when confined in a 7.50 L container? Problem: How many moles of He(g) are in a 5.00 L storage tank filled with He at 10.5 atm pressure at 30.0 oC?
Dr. L. Dawe
Chemistry 110/120
Lecture 6 and 7 Chapter 2: The Behaviour of Gases • Section 2.2 (Pages 64-69): Describing Gases – Variations on The Ideal Gas Equation
• Section 2.6 (Pages 85-90): Additional Gas Properties – Determination of Molar Mass – Gas Density – Suggested end-of-chapter exercises: 2.45-2.48
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Lecture 6 and 7 Chapter 2: The Behaviour of Gases • Section 2.4 (Pages 74-78): Gas Stoichiometry – Summary of Mole Conversions – Suggested end-of-chapter exercises: 2.24, 2.26, 2.27, • Section 2.3 (Pages 70-73): Gas Mixtures – Dalton’s Law of Partial Pressures – Describing Gas Mixtures – Suggested end-of-chapter exercises: 2.17, 2.18, 2.21
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Lecture 6 and 7 Learning Outcomes Students should be able to:
2.4 use the concept of partial pressures in gas mixtures 2.5 use stoichiometry to solve problems involving gas-phase chemical reactions 2.7 calculate gas densities and molar masses from pressurevolume-temperature data
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Combining the Gas laws: The general gas equation
• These can be combined to give the general gas equation:
PiVi P f V f ni Ti n f Tf
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Combining the Gas laws: The general gas equation
Problem: A 1.00 mL sample of N2(g) at 36.2oC and 2.14 atm is heated to 37.8oC, and the pressure changed to 1.02 atm. What volume does the gas occupy at this final temperature and pressure?
Dr. L. Dawe
Chemistry 110/120
Applications of the Ideal Gas Equation
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Molar mass determination • The ideal gas equation can be rearranged to solve for molar mass directly. • We know: n
m MM
and
PV nRT
therefore by substitution we arrive at: mRT PV MM
or
mRT MM PV
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Molar mass determination • Problem: A 2.650 g sample of a gaseous compound occupies 428 mL at 24.3 oC and 742 mmHg. The compound consists of 15.5% C, 23.0% Cl and 61.5% F by mass. What is the molecular formula? • Problem: A 132.10 mL glass vessel weighs 56.1035 g when evacuated and 56.2445 g when filled with the gaseous hydrocarbon acetylene at 749.3 mmHg and 20.02 oC. What is the molecular mass of acetylene?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas densities • Gas density differs from the density of liquids and solids in that: – Gas densities depend on both temperature and pressure, while for solids and liquids, density depends slightly on temperature, and far less on pressure (think about the compressibility of a gas vs. that of the same compound in the solid phase!)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas densities • Gas density differs from the density of liquids and solids in that: – The density of a gas is directly proportional to its molar mass. This is not the case for solids or liquids.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas densities • Gas densities are normally much smaller than the densities of liquids or solids. • Gas densities are normally given in g/L, while solids and liquids are expressed in g/mL.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas densities Problem: The density of phosphorus vapor at 310oC and 775 mmHg is 2.64 g/L. What is the molecular formula of the phosphorus under these conditions?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gases in Chemical Reactions Airbags:
2 NaN3(s) 2 Na(l) + 3 N2(g)
The ideal gas law may now be applied to reactions that involve gaseous reactants or products. (1) Use stoichiometric factors to relate the amount (mols) of gas to other reactants or products (2) Use the ideal gas law to relate the amount of gas to volume, temperature and pressure.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Stoichiometric calculations involving gases Problem: Calculate the volume of H2(g), measured at 26oC and 751 mmHg, required to react with 28.5 L of CO(g), measured at 0oC and 760 mmHg, in the following reaction: 3CO(g) + 7H2(g) C3H8(g) + 3H2O(l)
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Mixtures of Gases • Simple gas laws and the ideal gas law were originally derived for air (a mixture of gases). • When dealing with a mixture of non-reactive gases, the simplest approach is to use the number of total moles of gas for n.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Mixtures of Gases Problem: 2.0 L of O2(g) and 8.0 L of N2(g), each at STP, are mixed together. The non-reactive gaseous mixture is compressed to occupy 2.0 L at 298 K. What is the pressure exerted by this mixture?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Dalton's law of partial pressures • Dalton’s law of partial pressures states that in a mixture of gases, the total pressure is the sum of the partial pressures of the gases present. That is: Ptot PA PB
Note that since gases expand to fill their containers, here:
Vtot V A VB
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Some questions and answers! Q: Dalton’s Law: The pressure of a gas mixture is the sum of the pressures of the individual gases. Why does each gas contribute to the total pressure in proportion to its mole fraction?
A: Adding another component to a gas simply increases the number of particles which increases the frequency of collisions and therefore the pressure. Each gas exerts a fraction of the total pressure based on the fraction of molecules (or moles) of that gas in the mixture.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Mole fraction • Mole fraction (c) describes a mixture in terms of the fraction of all the molecules that are of a particular type. It is the amount of one component, in moles, divided by the total amount of all the substances in the mixture. VA PA nA cA Vtot Ptot ntot
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Mole fraction Problem: A mixture of 0.197 mol CO2(g) and 0.00278 mol H2O(g) is held at 30.0oC and 2.50 atm. What is the partial pressure of each gas?
Dr. L. Dawe
Chemistry 110/120
Lecture 7 and 8 Chapter 2: The Behaviour of Gases • Section 2.8 (Pages 97): Chemistry of the Earth’s Atmosphere – Vapour Pressure
• Section 2.5 (Pages 79-84): Molecular View of Gases – Molecular Speeds – Speed and Energy – Average Kinetic Energy – Ideal Gases – Suggested end-of-chapter exercises: 2.51, 2.52, 2.33-2.36
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Lecture 7 and 8 Learning Outcomes Students should be able to:
2.10 calculate water vapour pressure and relative humidity 2.6 explain the basic concepts of kinetic molecular theory: molecular speed, energy, and the effects of temperature and volume on gas pressure
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Collecting a gas and downward displacement of a liquid • A pneumatic trough is a device that was commonly used to collect gas from a reaction or other gas generating device, for further analysis. • The gas that is collected displaces water, and is actually a mixture of gases; that which we want to analyze, and water vapour.
Dr. L. Dawe
Chemistry 110/120
Collecting a gas and downward displacement of a liquid
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Collecting a gas and downward displacement of a liquid Gas pressure can be adjusted to atmospheric pressure by adjusting the height of the bottle, to give: Ptot = Pbar = Pgas + PH2O Pgas = Pbar – PH2O Values of water vapour pressure can be obtained from tables:
Dr. L. Dawe
Chemistry 110/120
Equilibrium Vapor Pressure Problem: In the following reaction, if 35.5 mL of H2(g) is collected over water at 26 oC and 755 mmHg barometric pressure, how many moles of HCl must be consumed? (The vapor pressure of water at 26 oC is 25.2 mmHg). 2 Al(s) + 6 HCl(aq) 2 AlCl3(aq) + 3 H2(g)
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Postulates of the Kinetic Molecular theory Postulate 1. Particle Volume. • A gas consists of a large collection of individual particles. • The volume of an individual particle is extremely small compared to the volume of the container. • Most of the space occupied by a gas is empty.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Postulates of the Kinetic Molecular theory Postulate 2. Particle Motion. • Gas particles are in constant motion. • Their motion is random and in straight lines. • At any one time there are gas particles moving in all directions. • When they collide with the walls of the container or one another they change their direction.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Postulates of the Kinetic Molecular theory Postulate 3. Particle Collisions. • The forces of attraction between molecules in a gas are negligible. Molecules do not influence one another unless they collide. • Collisions between molecules or with container walls are fleeting and most of the time molecules are not involved in collisions. • When molecules collide they can transfer kinetic energy.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Postulates of the Kinetic Molecular theory Postulate 4. Temperature. • In a collection of molecules at a certain temperature the total energy remains constant although each individual molecule may have a certain kinetic energy and can transfer energy in a collision.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Molecular speed • The average kinetic energy of a gas depends on its temperature and its mass. • The average speed of molecules of a gas is greater at higher temperature. • Species with different molecular masses have the same average kinetic energies at any given temperature, but the smaller molecules have a greater average speed.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Molecular speed • Molecules have a distribution of speeds. For example, they could be moving very fast one second, then collide with another molecule and transfer energy to the other molecule and be sitting almost still at another instant. um = modal speed uav = average speed urms = root mean square speed
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Average and root-mean-square speed Problem: Five molecules are traveling at different speeds: 400, 450, 525, 585 and 600 m/s. Calculate the average speed and the root-mean-square speed for this collection of molecules.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Translational kinetic energy • Translational kinetic energy (ek): energy possessed by objects (ie. gas molecules) traveling through space.
1 E k mu 2 2 m = mass of the molecule (g/molecule) u = velocity 1 2 E k mu rms 2
urms = root-mean square speed. A molecule moving with this speed has the average kinetic energy.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Pressure • Translational kinetic energy, • Frequency of collisions, • Impulse or momentum transfer, • Pressure proportional to impulse times frequency
• Three dimensional systems lead to:
1 e k mu 2 2
N vu V I mu N P mu 2 V 1N P m u2 3V
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Pressure • Translational kinetic energy, • Frequency of collisions, • Impulse or momentum transfer, • Pressure proportional to impulse times frequency
• Three dimensional systems lead to:
1 e k mu 2 2
N vu V I mu N P mu 2 V 1N P m u2 3V
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Root-mean-square speed Assume one mole:
1 2 PV N A m u 3
PV=RT so:
3RT N A m u
NAm = M:
3RT M u
Rearrange:
u rms
2
3RT M
2
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Root-mean-square speed Assume one mole:
1 2 PV N A m u 3
PV=RT so:
3RT N A m u
NAm = M:
3RT M u
2
2 8.3145 J mol-1K-1
Rearrange:
u rms
3RT M
J has units kg m2 s-2
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Root-mean-square speed In a qualitative sense, this tells us that the higher the molar mass of a molecule, the Assume lower its root-mean-square speed. one mole:
PV=RT so:
1 PV N A m u 2 3 3RT N A m u 2
NAm = M:
Rearrange:
3RT M u 2 u rms
3RT M
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Root-mean-square speed A member of the cat family, the cheetah leads the way as the world's fastest mammal. Cheetahs can attains speeds of 114 km/hr, and have been recorded accelerating from 0-60mph in just 3 seconds - faster than most sports cars. Problem: Which has the greatest speed: a cheetah running at its maximum speed, or the root-mean-square speed of H2 molecules at 25oC?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Distribution of molecular speeds and effect of temperature
The average kinetic energy of a gas particle is proportional to the Kelvin temperature of the sample. That is, on average, for particles of the same gas, those at a higher temperature will possess more kinetic energy, reflected in a higher average speed.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The meaning of temperature and absolute zero Kelvin • The Kelvin temperature (T) of a gas is directly proportional to the average translational kinetic energy (Ek) of its molecules. • The absolute zero of temperature is the temperature at which translational molecular motions should cease.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas Properties Related to the Kinetic Molecular Theory: Diffusion • Diffusion: the migration of molecules as a result of random molecular motion.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas Properties Related to the Kinetic Molecular Theory: Diffusion NH3(g) + HCl(g) NH4Cl(s)
1. The gases diffuse toward each other and where they meet a white cloud of ammonium chloride forms. 2. Because of their greater average speed, NH3(g) molecules diffuse faster than HCl(g) molecules, so the cloud forms closer to the drop of HCl(aq).
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Gas Properties Related to the Kinetic Molecular Theory: Effusion • Effusion: the escape of gas molecules from their container through a tiny orifice or pinhole.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Effusion • The rate of effusion is directly proportional to molecular speed. • The rate of effusion for two different gases at the same temperature can be compared by: EffusionRateA ( urms ) A EffusionRateB ( urms ) B
3 RT / M A 3 RT / M B
MB MA
• At a given temperature and pressure, the gas with the lower molar mass effuses faster because the most probable speed of its molecules is higher, therefore more molecules escape through the tiny hole per unit time.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Effusion • Graham’s Law: The rates of effusion of two different gases are inversely proportional to the square roots of their molar masses. • Graham’s law only applies to gases at very low pressures (ie. molecules escape through the pinhole, one at a time) and when the orifice is very tiny, so that no collisions take place as molecules are passing through.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Effusion Problem: 2.2 x 10-4 mol N2(g) effuses through a tiny hole in 105 s. How much O2(g) would effuse through the same orifice in 105 s? Problem: A sample of Kr(g) escapes through a tiny hole in 87.3 s. An unknown gas requires 131.3 s to escape under the same conditions. What is the molar mass of this unknown gas?
Dr. L. Dawe
Chemistry 110/120
Lecture 8 Chapter 2: The Behaviour of Gases • Section 2.7 (Pages 93-95): Non-Ideal (Real) Gases • Suggested end-of-chapter exercises: 2.55-2.57, 2.59
Fall 2014
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Lecture 8 Learning Outcomes Students should be able to:
2.8 explain the two principle ways in which ideal gases and real gases differ 2.9 calculate the pressure of a gas under non-ideal conditions
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Qualitative comparison of nonideal gases to ideal gases • Up to now, our gas laws have applied to ideal gases. That is, it was assumed that individual gas molecules acted as points, and occupied zero volume, and that intermolecular forces between these points was negligible. This is not the case for real gases. • Actually: – Gases tend to behave ideally at high temperatures and low pressures. – Gases tend to behave nonideally at low temperatures and high pressures. » WHY?
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The van der Waals equation • Compressibility factor of a gas is defined as PV/nRT For ideal gases: PV/nRT = 1 For real gases: PV/nRT < 1 intermolecular forces are significant. PV/nRT > 1 molecular volume is significant • The van der Waals equation corrects for the volume associated with the molecules themselves and for intermolecular forces:
n2a P 2 (V nb) nRT V
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The van der Waals equation n2a P 2 (V nb) nRT V
Dutch physicist Johannes van der Waals reasoned that the decrease in pressure due to intermolecular attractions should be proportional to the square of the concentration; “a” provides a measure of how strongly the molecules attract each other.
“b” is called the “excluded volume per mole” and is roughly equal to the volume that one mole of gas occupies when condensed to the liquid phase.
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The van der Waals equation • Compressibility factor of a gas is defined as PV/nRT For ideal gases: PV
nRT
1
For real gases:
PV 1 nRT
PV 1 nRT
intermolecular forces are molecular volume is significant significant • The van der Waals equation corrects for the volume associated with the molecules themselves and for intermolecular forces:
n2a P 2 (V nb) nRT V
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The correction for intermolecular forces, n2a/V2 • The effect of attractive forces between molecules is to decrease the pressure that a mole of gas (at standard temperature in a 22.4 L container) from that which it would exert if it were ideal. • That is, PV/nRT < 1
Dr. L. Dawe
Chemistry 110/120
Fall 2014
The excluded volume term, nb
• The effect of molecular volume is to make the volume occupied by a mole of gas larger than predicted by the ideal gas law. • That is, PV/nRT > 1
Dr. L. Dawe
Chemistry 110/120
Fall 2014
Application of the van der Waals Equation
• Problem: Use the van der Waals equation to calculate the pressure exerted by 1.00 mol Cl2 (g) confined to a volume of 2.00 L at 273 K. What is the pressure exerted for CO2 (g), under the same conditions? Which gas shows the greater departure from ideal gas behaviour? What are the main causes of the departures from ideality for each gas?
The constants a and b must be given! For Cl2(g): a = 6.49 L2 atm mol-2 and b = 0.0562 L mol-1 For CO2(g): a = 3.59 L2 atm mol-2 and b = 0.0427 L mol-1
