kinetic molecular theory teacher: h. michael hayes

Kinetic Molecular Theory Teacher: H. Michael Hayes

Kinetic Theory of Gases Seeks to explain the similarities observed in the behaviour of gases based on the movement of particles that compose them.

Particle model of matter All matter is composed of particles with more or less space between them depending on their phase. Particles attract or repel each other and that force depends on the distance between the particles. Particles are always moving.

Particle model of matter Particle behaviour in solids Weak vibration. Particle behaviour in liquids Vibration, rotation and weak translation. Particle behaviour in gases Vibration, rotation and translation.

Particle model of matter Summary Table

Kinetic Energy of Gas Particles and Temperature E k varies directly with mass and velocity. What variable has the greater effect on E k and why?

Kinetic Energy of Gas Particles and Temperature In a sample of any gas, not all particles have the same kinetic energy. Therefore not all particles have the same velocity. This is why it is appropriate to consider mean velocity to determine the mean kinetic energy of a gas.

Kinetic Energy of Gas Particles and Temperature

Hypotheses of the kinetic theory of gases H YPOTHESIS 1 T HE PARTICLES OF A GAS ARE INFINITELY SMALL AND THE SIZE OF A PARTICLE IS NEGLIGIBLE COMPARED TO THE VOLUME OF THE CONTAINER THAT HOLDS THE GAS. H YPOTHESIS 2 T HE PARTICLES OF A GAS ARE IN CONSTANT MOTION AND MOVE IN A STRAIGHT LINE IN ALL DIRECTIONS. H YPOTHESIS 3 T HE PARTICLES OF A GAS DO NOT EXERT ANY FORCE OF ATTRACTION OR REPULSION ON EACH OTHER. H YPOTHESIS 4 T HE MEAN KINETIC ENERGY OF A GAS IS DIRECTLY PROPORTIONAL TO THE ABSOLUTE TEMPERATURE.

Kinetic theory of gases Questions Numbers 1 - 11 Page 62 of Quantum Be able to answer and understand all! Computer simulation of particle motion and mean kinetic energy. http://youtu.be/A0May2m6cIc

Behavior of gases. Compressibility Supported by Hypothesis 1. Due to the large inter particle distance gases can be stored in small spaces under high pressure. Expansion Supported by Hypothesis 2. Gases dilate indefinitely by filling any accessible space. Greater dilation leads to greater inter particle distance. Expansion varies directly with atmospheric pressure.

Behavior of gases. Diffusion The random dispersion of aimlessly colliding particles.

Behavior of gases. Diffusion Not all gases diffuse at the same rate. The rate depends on the velocity of particles. Therefore: At the SAME TEMPERATURE particles with a small mass move more quickly than particles with a large mass.

Behavior of gases. Effusion When gas flows across a barrier through a small opening, it is termed EFFUSION. Example is a deflating Helium balloon In this picture two balloons were inflated. One with Nitrogen and the other with Helium. Due to the small radius of He it effuses more readily through the balloon (membrane).

Behavior of gases. Diffusion & Effusion of Two Different Gases at the Same Temperature and Pressure. Hypothesis 4 tells us: At the same temperature and pressure, two different gases will have the same Kinetic Energy If we use the following expression for the kinetic energy of the two gases: Given the statement above (hypothesis 4)

Behavior of gases. Diffusion & Effusion of Two Different Gases at the Same Temperature and Pressure. If we use the following expression for the kinetic energy of the two gases: We can express the relationship like this: Simplify to this:(cancel out ) Simplify to this:(divide both sides m 1 v 2 2 ) Simplify to this:GRAHAMS LAW

Behavior of gases. Diffusion & Effusion of Two Different Gases at the Same Temperature and Pressure. Grahams Law Simplify to this:(cancel out ) Simplify to this:(divide both sides m 1 v 2 2 ) Simplify to this:GRAHAMS LAW Under identical conditions of temperature and pressure, the relative rates of diffusion and effusion of two gases are inversely proportional to the square roots of their molar masses.

Grahams Law Problem. A 3.00 L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane in 25 hours. A 3.00 L sample of oxygen was placed in an identical container. How many hours will it take for half of the oxygen to effuse though the membrane?: Given: Applying Grahams Law: How long for half of O 2 to effuse?

Grahams Law Problem. A 3.00 L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane in 25 hours. A 3.00 L sample of oxygen was placed in an identical container. How many hours will it take for half of the oxygen to effuse though the membrane?: Given: Applying Grahams Law: How long for half of O 2 to effuse? See more problems in Quantum text page 67

Pressure of Gases The pressure of a gas corresponds to the force they exert on a surface. Supported by Hypothesis 2 and 4 (particles are in constant motion & the greater their kinetic energy the faster they move.

Pressure of Gases From Hypothesis 4 the mean kinetic energy of the particles of 2 gases is the same Therefore: a light gas will collide more frequently with less force and a heavy gas colliding less frequently with more force given the same pressure.

Atmospheric Pressure Atmospheric pressure is measured using a barometer invented by Evangelista Torricelli in 1643.

Atmospheric Pressure - measurement Covert the lowest ever pressure recorded on earth in kPa to mm of Hg.

Measuring Gas Pressure Barometers are exclusively used to measure atmospheric pressure. Gas pressure is measured using a manometer or pressure gauge. The pictures below illustrate DIAL Manometers.

Measuring Gas Pressure Manometers U-Tube 2 types 1.Closed-end 2.Open-end ClosedOpen

Measuring Gas Pressure in a close-end manometer Real pressure of a gas using a closed-end manometer. P gas = h P gas = Pressure of the gas in the container expressed in mm of Hg. h = Height of the column of Hg expressed in mm of Hg.

Measuring Gas Pressure in a close-end manometer What is the real pressure of the gas in this manometer? F IND THE DIFFERENCE IN HEIGHTS. P gas = h h = 88 cm Hg 32 cm Hg h= 56 cm Hg = 560 mm Hg P gas = 560 mm of HG

Measuring Gas Pressure in a open-end manometer If P gas > P atm Then P gas = P atm + h If P gas

Measuring Gas Pressure in a open-end manometer What is the gas pressure? F IND THE DIFFERENCE IN HEIGHTS. h= 75 cm of Hg 45 cm Hg h=30 cm = 300 mm Since P gas > P atm then P gas = P atm + 300 mm of Hg P gas = 760 mm + 300 mm P gas = 1060 mm of Hg Ex #1

Measuring Gas Pressure in a open-end manometer What is the gas pressure? F IND THE DIFFERENCE IN HEIGHTS. h= 62 cm of Hg 25 cm Hg h=37 cm = 370 mm Since P gas < P atm then P gas = P atm - 370 mm of Hg P gas = 769.8 mm - 370 mm P gas = 399.8 mm of Hg Ex #2 In this example you first need to convert 102.6 kPa to mm of Hg

Measuring Gas Pressure in a open-end manometer What is the pressure of Ne? F IND THE DIFFERENCE IN HEIGHTS. h = 47 cm Hg 39 cm Hg h = 8 cm = 80 mm Hg Since P Ne < P atm then P Ne = P atm - 80 mm of Hg P Ne = 650 mm - 80 mm P Ne = 570 mm of Hg Ex #3

Simple Gas Laws The simple gas laws help to solve problems that establish a relationships between TWO of the FOUR variables that describe gases. Pressure (P) Volume (V) Absolute Temperature (T) Quantity of gas number of moles (n) .while the remaining two variables remain constant These laws are the results of several independent scientific investigations that began in the 17 th century. At the time no International System of Units existed and laboratory conditions varied greatly. (Temperature and pressure varied from lab to lab). This brought about standards so data could be compared. These standard conditions are now known as STP StandardTemperaturePressure Standard Temperature and Pressure (STP) 0 o C / 273 o K101.3 kPa Standard Ambient Temperature and Pressure (SATP) 25 o C / 298 o K101.3 kPa

Simple Gas Laws By combining the simple gas laws, a more general gas law evolved: Simple Gas Laws The Ideal Gas Law The General Gas Law This will be our focus over the coming weeks.

Relationship between pressure and volume

When the pressure on a gas increases the volume of a gas decreases proportionately In the expression to the left we have introduced a proportionality constant. The constant permits us to create a simple equation. The equation tells us that Pressure times Volume is constant.

Relationship between pressure and volume If we calculate the slope of this line we get the value for the constant. For any pair of co- ordinates on the line the following is true. If we now plot pressure versus the inverse of volume (1/v) we get the graph to the right.

Relationship between pressure and volume

Relationship between pressure and volume sample problem.

Relationship between volume and absolute temperature. Jacques Charles observed that the volume of a gas increased by 1/273 of its original volume when heated by 1 o C. He concluded that when a gas is heated from 0 o C to 273 o C its volume doubles.

Gas Volume versus Absolute Temperature It is Charles extrapolated data (dotted lines) that leads Lord Kelvin to the discovery of the Kelvin scale for temperature and the concept of absolute zero.

Charles Law

Relationship between pressure and temperature.

Gay-Lussacs Law provides us with a means to predict/compare pressures and temperatures for a given sample of gas at a CONSTANT volume

Relationship between pressure and temperature. Connection with the Kinetic Molecular Theory of gases. Increase in temperature translates to more rapid movement of particles Hypothesis 4 Increased temperature leads to more rapid rate of translation, leading to more collisions with the wall of the container causing a direct increase in pressure.

Relationship between volume and quantity of gas in moles. Gay-Lussacs fundamental law of chemistry: When gases react, the volumes of the reactants and the products are ALWAYS present in the form of simple whole number ratios if they are measured at constant temperatures and pressures.

Relationship between volume and quantity of gas in moles. Avogadros Hypothesis: Given the same conditions of temperature and pressure, equal volumes of different gases contain the same number of particles. Avogadros Law: Given the same temperature and pressure conditions, the volume of a gas is directly proportional to its quantity expressed in moles.

Relationship between volume and quantity of gas in moles. - Sample problem A helium balloon occupies a volume of 15L and contains 0.50 moles of helium at SATP. What will the new volume of the balloon be if 0.20 mol of helium is added to the balloon be if 0.20 mol of helium is added to the balloon under the same conditions? V 1 = 15L n 1 = 0.50 mol V 2 = ? n 2 = 0.50 mol + 0.20 mol What we know.

Relationship between volume and quantity of gas in moles. Connection with the Kinetic Molecular Theory of gases. as the number of particles of a gas increases, they collide more frequently with each other and the walls of the container which increases the pressure. At a constant temperature the additional pressure increases the volume of the container until the internal pressure of the gas equals the external pressure on the container. - Hypothesis 2

Molar Volume of Gases Definition: The volume occupied by ONE mole of any gas, under fixed conditions of temperature and pressure. The molar volume (V m ) is expressed in L/mol. Avogadros Law states that under the same conditions of temperature and pressure, equal volumes of any gas contain the same number of particles. Therefore it follows that 1 mole (6.02 x 10 23 ) particles of a gas under the same conditions will occupy the same volume. The size of the gas particles and the molar mass of the gas particles have NO INFLUENCE on the volume occupied by the gas.

Molar Volume of Gases Connection to the Kinetic Molecular Theory. The size of the particles of gases is negligible in relation to the volume of the particles occupy. Hypothesis 1

Molar Volume of Gases Molar Volume of any gas at STP = 22.4 L/mol. Molar Volume of any gas at SATP = 24.5 L/mol. Sample Problem: How many moles are there in a container holding 69.2 L of methane (CH4) gas at STP? What do we know? V = 69.2L n = ? V m = 22.4 L/mol. (at STP)

Relationship between pressure and the quantity of gas in moles. Under the same conditions of TEMPERATURE and VOLUME, the pressure of a gas is directly proportional to the number of moles. If the number of moles is doubled the pressure doubles.

Relationship between pressure and the quantity of gas in moles. Under the same conditions of TEMPERATURE and VOLUME, the pressure of a gas is directly proportional to the number of moles. If the number of moles is doubled the pressure doubles. Connection with Kinetic Theory of Gases Hypothesis 2 explains the relationship between gas pressure and the number of moles. As the number of particles (moles) of gas increase, they collide with each other and the walls of the container more frequently. Therefore the number of collisions per unit of surface area increases which increases the pressure.

Relationship between pressure and the quantity of gas in moles. Under the same conditions of TEMPERATURE and VOLUME, the pressure of a gas is directly proportional to the number of moles. Sample Problem

Ideal Gas Law It describes the interrelationship between the Four (4) variables that characterize a gas at a given moment in time. The variables are: (P) pressure, (T) absolute temperature, (V) volume, (n) quantity of gas in moles. An ideal gas does not exist in reality. It is an hypothetical gas that adheres to all the simple gas laws and conforms to Kinetic Molecular Theory. Ideal gases do not condense at low temperatures or high pressures. In contrast real gases do not resemble ideal gases at extreme temperatures and pressures.

Ideal Gases vs Real Gases

Ideal Gases Law Formula The Ideal Gas law is a combination of the 4 simple gas laws. R is the proportionality constant. It replaces the proportionality symbol (). We determine the value of R by finding its value for 1 mol of gas at STP.

Ideal Gases Law Formula R is the proportionality constant. It replaces the proportionality symbol (). We determine the value of R by finding its value for 1 mol of gas at STP. To determine the value for R:

Ideal Gases Law Problem When inflated to their maximum capacity, human lungs contain approximately 4.09 L of air at 37 degrees C. How many moles of air do lungs contain if the air pressure is 100kPa? Known Data P= 100kPa n= ? R= 8.31 (kPaL)/(molK) T= 37 o C V= 4.09L 1)Convert Temperature to K T = 37 o C + 273 = 310K 2)Calculate the number of moles

Ideal Gases Law Problem When inflated to their maximum capacity, human lungs contain approximately 4.09 L of air at 37 degrees C. How many moles of air do lungs contain if the air pressure is 100kPa? 1)Convert Temperature to K T = 37 o C + 273 = 310K 2)Calculate the number of moles Known Data P= 100kPa n= ? R= 8.31 (kPaL)/(molK) T= 37 o C V= 4.09L

Molar Mass of a Gas The molar mass of a gas (M) can be determined by dividing the mass of the gas (m) by its number of moles: We can use the Ideal Gas Law to determine the molar mass of a gas using the simple transformation on the left. Formula for Molar Mass derived from the Ideal Gas Law

Molar Mass of a Gas Sample Problem What is the molar mass of a sample of an unknown gas if, at a temperature of 0 o C and under a pressure of 102 kPa, a volume of 2.30 L of the gas weighs 4.23 grams? Known Data M=? T= 0 o C P=102 kPa V=2.30L m=4.23 g R=8.31 (kPaL)/(molK) 1)Convert Temperature to K T = 0 o C + 273 = 273K 2)Calculate the Molar Mass

General Gas Law The General Gas Law establishes a relationship between the 4 variables that describe P, V, T and n. It is used to predict the final conditions of a gas once its initial conditions have been modified. From this final equation for the General Gas Law we can deduce all formulas for the simple gas laws.

General Gas Law sample problem At SATP, 0.150 mol of water vapour occupies a volume of 55.0 mL. What is the new temperature in degrees Celsius if 0.030 mol of water vapour is removed while increasing the pressure to 115.0 kPa and decreasing the volume to 40.0 ml? Known Data P 1 = 101.3 kpa V 1 = 55.0 mL n 1 = 0.150 mol T 1 = 25 C P 2 = 115.0 kPa V 2 = 40.0mL n 2 = ? T 2 = ? 1 Convert Temperature to K 2 - Calculate the final number of moles 3- Use General Gas Law to solve problem

General Gas Law sample problem At SATP, 0.150 mol of water vapour occupies a volume of 55.0 mL. What is the new temperature in degrees Celsius if 0.030 mol of water vapour is removed while increasing the pressure to 115.0 kPa and decreasing the volume to 40.0 ml? Known Data P 1 = 101.3 kpa V 1 = 55.0 mL n 1 = 0.150 mol T 1 = 25 C P 2 = 115.0 kPa V 2 = 40.0mL n 2 = ? T 2 = ?

Stoichiometry of Gases..is a calculation method based on the ratios between the quantities of gas involved in a chemical reaction. This method is used to predict the quantity of a reactant or product involved in a chemical reaction in which at least one of the components is a gas. Sample problem: Ethanol vapour (C 2 H 5 OH) burns in air according to the following equation: 2 C 2 H 5 OH (g) + 6O 2 - 4 CO 2(g) + 6 H 2 O (l) a)If 2.5 L of ethanol burn at STP, what volume of oxygen (O2 ) is required? b)What volume of carbon dioxide (CO2) will be produced?

Stoichiometry of Gases Sample problem: Ethanol vapour (C 2 H 5 OH) burns in air according to the following equation: 2 C 2 H 5 OH (g) + 6O 2 - 4 CO 2(g) + 6 H 2 O (l) a)If 2.5 L of ethanol burn at STP, what volume of oxygen (O2 ) is required? b)What volume of carbon dioxide (CO2) will be produced?

Daltons Law Law of Partial Pressures At a given temperature, the total pressure of a mixture of gases equals the sum of the pressures of each of the gases.

Significant figures - Rules

Physical Properties of Gases Covers up to page 113 of Quantum Text

Homework Problems - solutions.

kinetic molecular theory teacher: h. michael hayes

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