Gases Properties of Gases Gas Laws pressure volume

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Gases • Properties of Gases • Gas Laws (pressure, volume, temperature, moles) • Gases

Gases • Properties of Gases • Gas Laws (pressure, volume, temperature, moles) • Gases in Chemical Reactions • The Kinetic Model of Gases

Gaseous elements

Gaseous elements

Pressure = Force/Area

Pressure = Force/Area

Boyle’s Law V ~ 1/P If a fixed amount of gas is released into

Boyle’s Law V ~ 1/P If a fixed amount of gas is released into a larger container, how much force does it exert?

If the volume of a gas is constrained to a smaller container, how much

If the volume of a gas is constrained to a smaller container, how much force does it exert? P ~ 1/V Boyle’s Law: P 1 V 1 = P 2 V 2

Relationship between volume and pressure. V 1 P 1 V 1 = P 2

Relationship between volume and pressure. V 1 P 1 V 1 = P 2 V 2 Decrease volume more collisions V 2

 • Ex. 1 A sample of gas occupies 21 liters at a pressure

• Ex. 1 A sample of gas occupies 21 liters at a pressure of 2. 2 atm. What would be the volume if the pressure was increased to 6. 2 atm? • Ex. 2 A sample of O 2 occupies 10. 0 L at 785 torr. At what pressure would it occupy 14. 5 L?

If the temperature of gas is increased, how will its volume respond (if the

If the temperature of gas is increased, how will its volume respond (if the pressure is kept constant)? Charle’s Law: V 1/T 1 = V 2 / T 2

Gas 1 Gas 2 Gas 3 Gas 4 The point where a gas would

Gas 1 Gas 2 Gas 3 Gas 4 The point where a gas would have zero volume!

If a gas is heated, how much force does it exert (if the volume

If a gas is heated, how much force does it exert (if the volume is kept constant)? P~T

Combined Gas Law (Boyle and Charle’s) P 1 V 1 T 1 = P

Combined Gas Law (Boyle and Charle’s) P 1 V 1 T 1 = P 2 V 2 T 2

Re-cap of Gases: Boyle’s Law: P 1 V 1 = P 2 V 2

Re-cap of Gases: Boyle’s Law: P 1 V 1 = P 2 V 2 Charles’s Law: V 1/T 1 = V 2 / T 2 P 1 V 1 = P 2 V 2 T 1 T 2 1) Gas equations should make sense: volume, pressure, temperature 2) Watch your units! (K, L, atm)

One standardized set of conditions = “STP” • 0 o C (273 K) •

One standardized set of conditions = “STP” • 0 o C (273 K) • 1 atm (760 Torr)

 • Ex. 3 A balloon filled with He occupies 413 m. L at

• Ex. 3 A balloon filled with He occupies 413 m. L at 100. o. C. At what temperature would it occupy 577 m. L if the pressure was constant? • Ex. 4 A sample of hydrogen sulfide (H 2 S) occupies 210 L at 27 o. C at 1200 T. What volume would it occupy at STP?

How much volume does 1 mole of gas occupy at 0 o C, 1

How much volume does 1 mole of gas occupy at 0 o C, 1 atm? If one mole of Argon occupies 22. 1 L, how much volume will be occupied by: one mole CO 2 one mole N 2 one mole O 2 one mole H 2

How much volume does 1 mole of gas occupy at 0 o C, 1

How much volume does 1 mole of gas occupy at 0 o C, 1 atm? Liters Ideal gas: 22. 41 L per mole

Ex. 5 What is the density of a gas that has a molar mass

Ex. 5 What is the density of a gas that has a molar mass of 44. 01 g/mol at STP?

Ideal Gas Law: PV = n. RT (where n = moles) What is “R”?

Ideal Gas Law: PV = n. RT (where n = moles) What is “R”?

“STP” = 0 o C and 1 atm

“STP” = 0 o C and 1 atm

 • Ex. 6 What is the volume of a balloon filled with 32.

• Ex. 6 What is the volume of a balloon filled with 32. 02 grams of Helium when the atmospheric pressure is 722 torr and the temperature is 40 o C? • Ex. 7 The Goodyear blimp must be inflated with Helium prior to a football game. Its volume is 7601 ft 3. How many grams of He are needed for a pressure of 740 torr at 22 o C? (1 ft 3 = 28. 3 L)

 • Ex. 9 A 0. 723 g sample of a gas occupies 176

• Ex. 9 A 0. 723 g sample of a gas occupies 176 m. L at 100. o C and 750. torr. What is its molar mass?

Stoichiometric Calculations Involving Gases C 3 H 8 + 5 O 2 3 CO

Stoichiometric Calculations Involving Gases C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O 25. 0 g of propane produces how many moles of CO 2? 25. 0 g of propane produces how many liters of CO 2 at STP? How many grams of propane are need to react with 50. 0 L of O 2 at 25 o C and 1. 0 atm?

Key Concepts: • Ideal Gas Law PV = n. RT • STP (0 o

Key Concepts: • Ideal Gas Law PV = n. RT • STP (0 o C and 1 atm) • Standard molar volume: 22. 4 L/mole at STP • Gases in stoichiometric calculations

An airbag inflates in less than 50 msec by the reaction of Na. N

An airbag inflates in less than 50 msec by the reaction of Na. N 3 to produce Na and nitrogen gas: Na. N 3 Na + N 2 The volume of the airbag is about 30 L when inflated, and it is filled to a pressure of 1. 4 atm. How many grams of Na. N 3 must be used for each air bag? The molar mass of Na. N 3 is 65. 1 g/mole. Assume the process occurs at room temperature.

Gas mixtures • Dalton’s Law of partial pressures The total pressure of a mixture

Gas mixtures • Dalton’s Law of partial pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone PT=P 1+P 2+P 3+…. Pn Exercise: A gaseous mixture is made from 6. 00 g oxygen and 9. 00 g methane placed in a 15 L vessel at 0 o. C. What is the partial pressure of each gas and the total pressure in the vessel?

Dalton’s law of partial pressures PT = P a + P b + P

Dalton’s law of partial pressures PT = P a + P b + P c + P d = … Remember that each pressure is also equal to: n. XRT/V

Mole Fractions • The ratio n 1/n. T is called the mole fraction (denoted

Mole Fractions • The ratio n 1/n. T is called the mole fraction (denoted x 1), a dimensionless number between 0 and 1. Mole fraction of N 2 in air is 0. 78, therefore if the total barometric pressure is 760 torr, the partial pressure of N 2 is (0. 78)(760) = 590 torr.

Kinetic –Molecular Theory describing why gas laws are obeyed (explains both pressure and temperature

Kinetic –Molecular Theory describing why gas laws are obeyed (explains both pressure and temperature of gases on a molecular level). • Complete form of theory, developed over 100 years or so, published by Clausius in 1857. Ø Gases consist of large numbers of molecules that are in continuous, random motion Ø Volume of all molecules of the gas is negligible, as are attractive/repulsive interactions Ø Interactions are brief, through elastic collisions (average kinetic energy does not change) Ø Average kinetic energy of molecules is proportional to T, and all gases have the same average kinetic energy at any given T. Because each molecule of gas will have an individual kinetic energy, and thus individual speed, the speed of molecules in the gas phase is usually characterised by the root-mean-squared (rms) speed, u, (not the same though similar to the average speed). Average kinetic energy є = ½mu 2

Application to Gas Laws • Increasing V at constant T: Constant T means that

Application to Gas Laws • Increasing V at constant T: Constant T means that u is unchanged. But if V is increased the likelihood of collision with the walls decreases, thus the pressure decreases (Boyle’s Law) • Increasing T at constant V: Increasing T increases u, increasing collisional frequency with the walls, thus the pressure increases (Ideal Gas Equation).

 • Ex. 10 A 20. 5 L bulb contains 0. 200 moles of

• Ex. 10 A 20. 5 L bulb contains 0. 200 moles of methane, 0. 300 moles of hydrogen, and 0. 400 moles of nitrogen at 20. 0 o C. It is stinky and explosive. What is the pressure inside the bulb? How much pressure is contributed by each of the three gases? • Ex. 12 One tank of gas contains 5. 00 L of N 2 at 32. 0 atm. A second tank contains 3. 00 L of O 2 at 24. 0 atm. What pressure is attained when the valve between the tanks is opened?

Mole fraction: the portion of a specific substance within a mixture What is the

Mole fraction: the portion of a specific substance within a mixture What is the portion (mole fraction) of red spheres?

Kinetic model of gases each dot is one gas molecule

Kinetic model of gases each dot is one gas molecule

The Kinetic-Molecular Theory • The basic assumptions of kinetic-molecular theory are: • Postulate 1

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

The Kinetic-Molecular Theory • Postulate 2 – Gas molecules are in constant, random, straight

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

The Kinetic-Molecular Theory • Postulate 3 – Gas molecules have elastic collisions with themselves

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

The Kinetic-Molecular Theory • Postulate 4 – The kinetic energy of the molecules is

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

The Kinetic-Molecular Theory • The kinetic energy of the molecules is proportional to the

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

The Kinetic-Molecular Theory • The gas laws that we have looked at earlier in

The Kinetic-Molecular Theory • The gas laws that we have looked at earlier in this chapter are proofs that kinetic-molecular theory is the basis of gaseous behavior. • Boyle’s Law – P 1/V – As the V increases the molecular collisions with container walls decrease and the P decreases. • Dalton’s Law – Ptotal = PA + PB + PC +. . . – Because gases have few intermolecular attractions, their pressures are independent of other gases in the container.

 • Charles’ Law – V T – An increase in temperature raises the

• Charles’ Law – V T – An increase in temperature raises the molecular velocities, thus the V increases to keep the P constant.

The Kinetic-Molecular Theory • The root-mean square velocity of gases is a very close

The Kinetic-Molecular Theory • The root-mean square velocity of gases is a very close approximation to the average gas velocity. • Calculating the root-mean square velocity is simple: • To calculate this correctly: – The value of R = 8. 314 kg m 2/s 2 K mol – And M must be in kg/mol.

The Kinetic-Molecular Theory • Example 12 -17: What is the root mean square velocity

The Kinetic-Molecular Theory • Example 12 -17: What is the root mean square velocity of N 2 molecules at room T, 25. 0 o. C?

The Kinetic-Molecular Theory • What is the root mean square velocity of He atoms

The Kinetic-Molecular Theory • What is the root mean square velocity of He atoms at room T, 25. 0 o. C?

The Kinetic-Molecular Theory • Can you think of a physical situation that proves He

The Kinetic-Molecular Theory • Can you think of a physical situation that proves He molecules have a velocity that is so much greater than N 2 molecules? • What happens to your voice when you breathe He?

Diffusion and Effusion of Gases • Diffusion is the intermingling of gases. • Effusion

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

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

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

Diffusion and Effusion of Gases • The rate of effusion is inversely proportional to

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

Diffusion and Effusion of Gases • Calculate the ratio of the rate of effusion

Diffusion and Effusion of Gases • Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SO 2, at the same temperature and pressure.

Diffusion and Effusion of Gases • Example 12 -16: A sample of hydrogen, H

Diffusion and Effusion of Gases • Example 12 -16: A sample of hydrogen, H 2, was found to effuse through a pinhole 5. 2 times as rapidly as the same volume of unknown gas (at the same temperature and pressure). What is the molecular weight of the unknown gas?

Real Gases: Deviations from Ideality • • • Real gases behave ideally at ordinary

Real Gases: Deviations from Ideality • • • Real gases behave ideally at ordinary temperatures and pressures. At low temperatures and high pressures real gases do not behave ideally. The reasons for the deviations from ideality are: 1. The molecules are very close to one another, thus their volume is important. 2. The molecular interactions also become important.

Real Gases: Deviations from Ideality • van der Waals’ equation accounts for the behavior

Real Gases: Deviations from Ideality • van der Waals’ equation accounts for the behavior of real gases at low temperatures and high pressures. • The van der Waals constants a and b take into account two things: 1. a accounts for intermolecular attraction 2. b accounts for volume of gas molecules • At large volumes a and b are relatively small and van der Waal’s equation reduces to ideal gas law at high temperatures and low pressures.

Real Gases: Deviations from Ideality • What are the intermolecular forces in gases that

Real Gases: Deviations from Ideality • What are the intermolecular forces in gases that cause them to deviate from ideality? 1. For nonpolar gases the attractive forces are London Forces 2. For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds.

Real Gases: Deviations from Ideality • Example 12 -19: Calculate the pressure exerted by

Real Gases: Deviations from Ideality • Example 12 -19: Calculate the pressure exerted by 84. 0 g of ammonia, NH 3, in a 5. 00 L container at 200. o. C using the ideal gas law.

Real Gases: Deviations from Ideality • Solve Example 12 -19 using the van der

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

Real Gases: Deviations from Ideality

Real Gases: Deviations from Ideality

“Real” gases • Deviations from ideal behavior as gases are exposed to: – high

“Real” gases • Deviations from ideal behavior as gases are exposed to: – high pressures – low temperatures (What happens at high pressures and low temperatures? ? ? )

 • The molar volume is not constant as is expected for ideal gases.

• The molar volume is not constant as is expected for ideal gases. • These deviations due to an attraction between some molecules. • Finite molar molecular volume. • For compounds that deviate from ideality the van der Waals equation is used: where a and b are constants that are characteristic of the gas. • Applicable at high pressures and low temperatures.

Ideal: PV = n. RT Real: (P + a n 2/V 2) (V –

Ideal: PV = n. RT Real: (P + a n 2/V 2) (V – nb) = n. RT • a is a molecular attractivity factor • b is a molecular volume factor

Ideal: P = n. RT/V Real: P = n. RT/(V – nb) - a

Ideal: P = n. RT/V Real: P = n. RT/(V – nb) - a 2 2 n /V

“Real” gases

“Real” gases

Ex. 13 Calculate the pressure exerted by 5. 00 moles of NH 3 in

Ex. 13 Calculate the pressure exerted by 5. 00 moles of NH 3 in a 1. 00 L vessel at 25. 0 o. C assuming ideal and non-ideal behavior.

Compression Factor Variation of Z with Pressure At 0 C Low pressures, Z=1, all

Compression Factor Variation of Z with Pressure At 0 C Low pressures, Z=1, all gases ideal At high pressures, Vreal >Videal ; Z>1 repulsive forces dominate Low pressure, Vreal <Videal, Z< 1 attractive forces dominate Low temperatures -> molecules moving less rapidly more influenced by attractive forces

Van der Waals Equation (1873) a reflects how strongly molecules attract each other b

Van der Waals Equation (1873) a reflects how strongly molecules attract each other b corrects for the molecule’s size

“Derivation” of vdw Eq. State Repulsive interactions cause molecules to behave as impenetrable spheres

“Derivation” of vdw Eq. State Repulsive interactions cause molecules to behave as impenetrable spheres Molecules restricted to smaller volume V-nb, where nb is volume molecules take up Pressure depends on frequency of collisions with walls and force of each collision – both reduced by attractive forces proportional to molar concentration (n/V) Pressure is then reduced according to a(n/V)2

Phase Transition: Condensation Imagine you are compressing gas with piston At C, piston slides

Phase Transition: Condensation Imagine you are compressing gas with piston At C, piston slides without further change in pressure Liquid begins to appear – Two phases coexist Going to E – amount of liquid increases Pressure along CDE is vapour pressure At E, sample is liquid – now the compressibility changes …. Similar behave happens in lipidic or polymeric systems, which exhibit a rich variety of phase transitions … Experimental isotherms of CO 2 at several temperatures.

Critical Constants The `critical isotherm', the isotherm at the critical temperature Tc, is at

Critical Constants The `critical isotherm', the isotherm at the critical temperature Tc, is at 31. 04 C. The critical point is marked with a star. Critical pressure Pc Critical molar volume Vc Tc, Pc, Vc critical constants If you compress along Tc, liquid does not appear – these highly dense materials are called “supercritical fluids” and their properties are a subject of intense current research

Virial Equation of State Most fundamental and theoretically sound Polynomial expansion – Viral Expansion

Virial Equation of State Most fundamental and theoretically sound Polynomial expansion – Viral Expansion = 0 at Boyle temperature Used to summarize P, V, T data Also allow derivation of exact correspondence between virial coefficients and intermolecular interactions