Gases and the Kinetic Molecular Theory A Properties

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Gases and the Kinetic Molecular Theory A. Properties of Gases ¬ we use many

Gases and the Kinetic Molecular Theory A. Properties of Gases ¬ we use many that were designed technologies with the knowledge of the properties of gases eg) SCUBA equipment, hot air balloons, jackhammers

¬gases have several distinct macroscopic (visible) properties: Ø gases are compressible ie) pressure =

¬gases have several distinct macroscopic (visible) properties: Ø gases are compressible ie) pressure = volume Ø gases expand as temperature increases ie) temperature = volume (not confined) temperature = pressure (confined) Ø gases have low resistance to flow …allows them to escape quickly (viscosity) through small openings

Ø gases have low densities Ø gases mix evenly and completely, they all are

Ø gases have low densities Ø gases mix evenly and completely, they all are miscible Ø gases have they no shape or volume, of the container they are in fill the shape

B. Kinetic Molecular Theory ¬we need to describe how gases behave on models the

B. Kinetic Molecular Theory ¬we need to describe how gases behave on models the molecular level kinetic molecular theory ¬the says that all particles are in motion at all times

¬an is ideal gas (which is hypothetical) defined by the following characteristics: 1. the

¬an is ideal gas (which is hypothetical) defined by the following characteristics: 1. the gas molecules are in constant random move in a straight motion where they with a particle or collide line until they wall of the container 2. the gas molecules are “point masses” (they have but no ) mass volume 3. the only interaction between molecules of the gas and container are elastic collisions… collisions where is kinetic energy conserved

¬ do not have these perfect characteristics real gases however their behaviour is not

¬ do not have these perfect characteristics real gases however their behaviour is not that far off of ideal gases

Gases and Pressure A. Atmospheric Pressure ¬although gas molecules in the atmosphere have ,

Gases and Pressure A. Atmospheric Pressure ¬although gas molecules in the atmosphere have , the Earth’s very little mass gravitational pull keeps them near the surface of the planet ¬ pressure = force per unit area ¬pressure is exerted in all directions to the same extent

¬atmospheric pressure is the that a column of force particular area air exerts on

¬atmospheric pressure is the that a column of force particular area air exerts on a on the Earth’s surface ¬air is as altitude less compressed increases pressure is exerted less

B. Measuring Pressure ¬Pascal and Perier used to prove that Hg(l) atmospheric pressure decreases

B. Measuring Pressure ¬Pascal and Perier used to prove that Hg(l) atmospheric pressure decreases with altitude ¬the work of Pascal, Perier and Torricelli all led to the development of the mercury barometer

¬there are several different units used to measure pressure: Ø millimetres of mercury (mm.

¬there are several different units used to measure pressure: Ø millimetres of mercury (mm. Hg) Øthe Pascal (Pa) Øthe kilopascal (k. Pa) Øthe atmosphere (atm)

¬you will be using the standard unit of in gas k. Pa law calculations

¬you will be using the standard unit of in gas k. Pa law calculations and therefore you must be able to convert mm. Hg and atm to k. Pa ¬memorize the following standard atmospheric pressures: 760 mm. Hg = 1 atm = 101. 325 k. Pa ¬to convert other units of pressure to k. Pa, set up a ratio

Example 1 Convert 650 mm. Hg to k. Pa. 101. 325 k. Pa =

Example 1 Convert 650 mm. Hg to k. Pa. 101. 325 k. Pa = x 760 mm. Hg 650 mm. Hg x = 86. 6… k. Pa

Example 2 Convert 2. 5 atm to k. Pa. 101. 325 k. Pa =

Example 2 Convert 2. 5 atm to k. Pa. 101. 325 k. Pa = x 1 atm 2. 5 atm x = 253. 3… k. Pa

Try These: Convert the following pressures to k. Pa (unrounded): 1. 4. 0 atm

Try These: Convert the following pressures to k. Pa (unrounded): 1. 4. 0 atm 405. 3 k. Pa 2. 855 mm. Hg 113. 9…k. Pa 3. 0. 625 atm 63. 3…k. Pa 4. 150 mm. Hg 19. 9…k. Pa

C. Boyle’s Law Robert Boyle ¬Irish scientist studied the pressure volume relationship between the

C. Boyle’s Law Robert Boyle ¬Irish scientist studied the pressure volume relationship between the and of gases at constant temperatures ¬pressure on the is caused by the walls of a container collisions of the gas molecules with the walls ¬as you of a contained gas, reduce the volume there is less room for the gas particles so they collide more ¬ more collisions = higher pressure http: //michele. usc. edu/java/gassim. html

Volume vs. Pressure for a Gas Volume (L) Pressure (k. Pa)

Volume vs. Pressure for a Gas Volume (L) Pressure (k. Pa)

 volume of a gas ¬Boyle’s Law states that the varies inversely with the

volume of a gas ¬Boyle’s Law states that the varies inversely with the pressure at a constant temperature and mass eg) lungs – to inhale, we the volume increase decreases of our chest cavity which the pressure which makes the air move in

eg) breath-hold diving – all air containing spaces in increases shrink body as pressure

eg) breath-hold diving – all air containing spaces in increases shrink body as pressure with depth…this doesn’t happen with SCUBA gear

P 1 V 1 = P 2 V 2 where: P 1, P 2

P 1 V 1 = P 2 V 2 where: P 1, P 2 = pressures in k. Pa V 1, V 2 = volumes in L

Example 1 A balloon is filled with 30. 0 L of helium gas at

Example 1 A balloon is filled with 30. 0 L of helium gas at 100 k. Pa. What is the volume when the balloon rises to an altitude where the pressure is only 25. 0 k. Pa? (assume constant temperature) P 1 = 100 k. Pa P 2 = 25. 0 k. Pa V 1 = 30. 0 L P 1 V 1 = P 2 V 2 (100 k. Pa)(30. 0 L) = (25. 0 k. Pa) V 2 = 120 L

Example 2 The pressure on 2. 50 L of anesthetic gas is 100 k.

Example 2 The pressure on 2. 50 L of anesthetic gas is 100 k. Pa. If 6. 25 L of gas is the required volume, what pressure must it be under assuming constant temperature? P 1 = 100 k. Pa V 1 = 2. 50 L V 2 = 6. 25 L P 1 V 1 = P 2 V 2 (100 k. Pa)(2. 50 L) = P 2 (6. 25 L) P 2 = 40. 0 k. Pa

Gases and Pressure A. Volume vs. Temperature volume vs. temperature ¬when of a gas

Gases and Pressure A. Volume vs. Temperature volume vs. temperature ¬when of a gas are graphed, the plot is linear (as long as amount of gas and pressure were constant) ¬it was also noticed that when these linear plots were down to all the zero volume, extrapolated lines at one point converged

Volume vs. Temperature for a Gas Volume (L) -273. 15 C Temperature ( C)

Volume vs. Temperature for a Gas Volume (L) -273. 15 C Temperature ( C)

¬the temperature when the volume of a gas is zero is 273. 15 C

¬the temperature when the volume of a gas is zero is 273. 15 C ¬ , in 1848, suggested Lord Kelvin that this is the lowest possible temperature or absolute zero ¬he established a new temperature scale which is Kelvin called the scale in his honour

 C ¬ is used for temperature in t ¬ is used for temperature

C ¬ is used for temperature in t ¬ is used for temperature in T K ¬ = -273. 15 C absolute zero Kelvin ¬to go from C to K… you add 273. 15 eg) 0 C = 273. 15 K 25 C = 298. 15 K -30 C = 243. 15 K -273. 15 C = 0 K

B. Charles’ Law Jacques Charles ¬ (and Joseph Louis Gay-Lussac) noticed that there was

B. Charles’ Law Jacques Charles ¬ (and Joseph Louis Gay-Lussac) noticed that there was a relationship between volume the and temperature of a gas increases ¬ kinetic as temperature , so does the energy of the gas molecules ¬as the molecules move , they exert faster higher pressure ¬the volume of the gas will expand under this until it reaches atmospheric pressure

¬ volume of a gas Charles’ Law states that the varies directly with the

¬ volume of a gas Charles’ Law states that the varies directly with the temperature at a constant pressure and mass V 1 = V 2 T 1 T 2 where: T 1, T 2 = temperatures in K V 1, V 2 = volumes in L

Example 1 A balloon was inflated at 27 C and has a volume of

Example 1 A balloon was inflated at 27 C and has a volume of 4. 0 L. If it is heated to 57 C, what is the new volume? (assume constant pressure) T 1 = 27 C = 300. 15 K V 1 = 4. 0 L T 2 = 57 C = 330. 15 K V 1 = V 2 T 1 T 2 (4. 0 L) = V 2 (300. 15 K) 330. 15 K V 2 = 4. 39… L V 2 = 4. 4 L

Example 2 A sample of gas occupies 6. 8 L at 110 C. What

Example 2 A sample of gas occupies 6. 8 L at 110 C. What will the final temperature be in C when the volume is decreased to 5. 6 L? T 1 = 110 C = 383. 15 K V 1 = V 2 V 1 = 6. 8 L T 1 T 2 V 2 = 5. 6 L (6. 8 L) = 5. 6 L (383. 15 K) T 2 = 315. 5… K – 273. 15 T 2 = 42 C p. 1 in booklet Read p. 150 -151 in textbook DO! p. 152 #5 -10

Gases and the Kinetic Molecular Theory A. Combined Gas Law Calculations ¬ = standard

Gases and the Kinetic Molecular Theory A. Combined Gas Law Calculations ¬ = standard temperature and pressure STP = 273. 15 K (0 C) and 101. 325 k. Pa ¬ = standard ambient temperature and pressure SATP = 298. 15 K (25 C) and 100. 000 k. Pa ¬now we’ll combine Boyle’s Law and Charles’ Law P 1 V 1 = P 2 V 2 T 1 T 2 where: V 1, V 2 = volumes in L T 1, T 2 = temperatures in K P 1, P 2 = pressures in k. Pa

Example 1 A weather balloon is filled with H 2(g) at 20 C and

Example 1 A weather balloon is filled with H 2(g) at 20 C and 100 k. Pa. It has a volume of 7. 50 L. It rises to an altitude where the air temperature is -36 C and the pressure is 28 k. Pa. What is the new volume of the balloon? T 1 = 20 C = 293. 15 K V 1 = 7. 50 L P 1 = 100 k. Pa P 2 = 28 k. Pa T 2 = -36 C = 237. 15 K P 1 V 1 = P 2 V 2 T 1 T 2 (100 k. Pa)(7. 50 L) = (28 k. Pa)V 2 (293. 15 K) (237. 15 K) V 2 = 21. 6…L V 2 = 22 L

Example 2 A large syringe was filled with 50. 0 m. L of ammonia

Example 2 A large syringe was filled with 50. 0 m. L of ammonia gas at STP. If the gas was compressed to 25. 0 m. L with a pressure of 210 k. Pa, what was the final temperature in C? T 1 = 0 C = 273. 15 K V 1 = 0. 0500 L P 1 = 101. 325 k. Pa P 2 = 210 k. Pa P 1 V 1 = P 2 V 2 = 0. 0250 L T 1 T 2 (100 k. Pa)(7. 50 L) = (28 k. Pa)(0. 0250 L) (293. 15 K) T 2 = 283. 0…K – 273. 15 T 2 = 9. 91 C

B. Combining Volumes of Gases ¬ Gay-Lussac analyzed chemical reactions that involved gases ¬he

B. Combining Volumes of Gases ¬ Gay-Lussac analyzed chemical reactions that involved gases ¬he studied the of volumes the gaseous reactants and products and concluded that the gases combine in very simple proportions

¬the states that, Law of Combining Volumes volumes when gases react, the of the

¬the states that, Law of Combining Volumes volumes when gases react, the of the gaseous reactants and products, measured at constant temperature and pressure, are always in whole number ratios eg) 1 N 2(g) + 3 H 2(g) 2 NH 3(g) 1 : 3 : 2 is the volume ratio

Example What volume of nitrogen is used up if 100 m. L of ammonia

Example What volume of nitrogen is used up if 100 m. L of ammonia is formed in a composition reaction? N 2(g) + 3 H 2(g) 2 NH 3(g) x m. L 100 m. L x m. L = 100 m. L 1 2 x m. L = 50 m. L

Ideal Gas Law A. Ideal Gas Law Calculations ¬ this law combines all four

Ideal Gas Law A. Ideal Gas Law Calculations ¬ this law combines all four variables (P, V, n and T) into one equation: PV = n. RT where: V = volume in L T = temperature in K P = pressure in k. Pa n = number of moles in mol R = universal gas constant = 8. 314 k. Pa L/mol K

Example 1 What is the volume of 10. 8 mol of oxygen gas at

Example 1 What is the volume of 10. 8 mol of oxygen gas at 100. 00 k. Pa and 15. 5 C? n = 10. 8 mol T = 15. 5 C = 288. 65 K P = 100. 00 k. Pa R = 8. 314 k. Pa L/mol K PV = n. RT (100. 00 k. Pa)V = (10. 8 mol)(8. 314 k. Pa L/mol K)(288. 65 K) V = 259. 1…L = 259 L

Example 2 A rigid steel vessel with a volume of 20. 0 L is

Example 2 A rigid steel vessel with a volume of 20. 0 L is filled with nitrogen gas to a pressure of 20 000 k. Pa at 27. 0 C. What is the number of moles of nitrogen? T = 27 C = 300. 15 K V = 20. 0 L P = 20 000 k. Pa R = 8. 314 k. Pa L/mol K PV = n. RT (20 000 k. Pa)(20. 0 L) = n(8. 314 k. Pa L/mol K)(300. 15 K) n = 160. 2…mol = 160 mol

Example 3 What is the pressure exerted by 15. 5 g of methane, CH

Example 3 What is the pressure exerted by 15. 5 g of methane, CH 4(g), if it occupies a volume of 10. 0 L at 25 C? T = 25 C = 298. 15 K V = 10. 0 L R = 8. 314 k. Pa L/mol K m = 15. 5 g M = 16. 05 g/mol n = m M = 15. 5 g 16. 05 g/mol = 0. 965…mol PV = n. RT P(10. 0 L) = (0. 965…mol)(8. 314 k. Pa L/mol K)(298. 15 K) P = 239. 3…k. Pa = 2. 4 102 k. Pa

Example 4 What is the mass of hydrogen gas contained in a 4. 5

Example 4 What is the mass of hydrogen gas contained in a 4. 5 L weather balloon at 25 C and 102. 0 k. Pa? T = 25 C = 248. 15 K V = 4. 5 L P = 102. 0 k. Pa R = 8. 314 k. Pa L/mol K M = 2. 02 g/mol PV = n. RT (102. 0 k. Pa)(10. 0 L) =n(8. 314 k. Pa L/mol K)(298. 15 K) n = 0. 222…mol m = n. M = (0. 222…mol)(2. 02 g/mol) = 0. 449…g = 0. 45 g

B. Dalton’s Law of Partial Pressures ¬ Dalton’s law of partial pressures states that

B. Dalton’s Law of Partial Pressures ¬ Dalton’s law of partial pressures states that in a not react chemically, mixture of gases that do the sum of the partial total pressure is the pressures of each individual gas Ptotal = P 1 + P 2 + P 3

Example Two gases are pumped into a 32. 0 L reaction vessel at 25.

Example Two gases are pumped into a 32. 0 L reaction vessel at 25. 0 C one after another. 6. 20 mol of O 2(g) is pumped in first, then 8. 30 mol of H 2(g) is pumped in. What would the pressure gauge reading be after each gas is pumped in? O 2(g) n = 6. 20 mol T = 25. 0 C = 298. 15 K V = 32. 0 L R = 8. 314 k. Pa L/mol K PV = n. RT P(32. 0 L) = (6. 20 mol)(8. 314 k. Pa L/mol K)(298. 15 K) P = 480. 2…k. Pa 1 st reading = 480 k. Pa

O 2(g) + H 2(g) n =6. 20 mol + 8. 30 mol =

O 2(g) + H 2(g) n =6. 20 mol + 8. 30 mol = 14. 5 mol T = 25. 0 C = 298. 15 K V = 32. 0 L R = 8. 314 k. Pa L/mol K PV = n. RT P(32. 0 L) = (14. 5 mol)(8. 314 k. Pa L/mol K)(298. 15 K) P = 1123. 2…k. Pa 2 nd reading = 1. 12 x 103 k. Pa

C. Ideal Gases and Real Gases ¬ for ideal gases we assume that there

C. Ideal Gases and Real Gases ¬ for ideal gases we assume that there are no between the intermolecular attractions molecules of the gas ¬ in real gases, however, there are attractions between the molecules ¬ we don’t have to worry about considering this in our calculations because at standard P and T conditions, moving very far apart the molecules are and are don’t interact very quickly they much with each other

¬ real gases behave like ideal gases at high temperatures and low pressures ¬

¬ real gases behave like ideal gases at high temperatures and low pressures ¬ real gases deviate from ideal gas behaviour at very low temperatures (moving very slowly) (molecules close and very high pressures together)