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Chapter 5: Gases I.Kinetic Molecular Theory II.Pressure III.Effect of ____ on Volume a)Pressure (Boyles’ Law) b)Temperature (Charles’ Law) c)# Moles (Avogadro’s Law) IV.Ideal Gas Law V.STP/ Molar Volume

Chapter 5: Gases VI.Gas Mixtures (Dalton’s Law) VII.Diffusion of Gases VIII.Real Gases

Elements that exist as gases at 25 0 C and 1 atmosphere 4.1

4.2

Macroscopic vs. Microscopic Representation

Kinetic Molecular Theory of Gases 1.Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.

Kinetic Molecular Theory of Gases 2.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy 3.A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 4.Gas molecules exert neither attractive nor repulsive forces on one another. 5.Each gas molecule “behaves” as if it were alone in container (due to #3 and #4) 5.7

Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. 5.1 Physical Characteristics of Gases

Pressure – KMT Viewpoint Origin of Pressure – Gas molecules hitting container walls –  Temp,  KE,  # collisions,  P –  Volume,  # collisions,  P

تعريف الضغط : الضغط عند أي نقطة هو القوة المؤثرة على وحدة المساحة لسطح ما. ويعرف الضغط رياضيا الضغط = القوة / المساحة P= F/A تعريف الضغط الجوي : وزن عمود الهواء المؤثر على وحدة المساحة ويمتد رأسيا من السطح إلى نهاية الغلاف الجوي. الضغط الجوي يكون أكبر ما يمكن بالقرب من سطح الأرض في أي مكان ويقل مع الارتفاع رأسيا إلى أعلى

Pressure – Macroscopic Viewpoint  Temp,  KE,  Force,  P  Volume,  Area,  P Pressure = Force Area Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa

Atmospheric Pressure Barometer

Gases Move From a Region of High to Low Pressure Vacuum Cleaner Blow balloon up; then let it go

Effect of Air Pressure

Effect of Pressure on Volume Boyle’s Law atm atm atm 5

Which picture represents what the gas will look like when the pressure is doubled? (Assume constant n, T)

4.6 Manometers Used to Measure Gas Pressures

4.7 As P (h) increases V decreases Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

P  1/V P x V = constant P 1 x V 1 = P 2 x V Boyle’s Law Constant temperature Constant amount of gas

Kinetic Molecular theory of gases and … Boyle’s Law P  collision rate with wall Collision rate Increases with decreased volume P  1/V Increase P, decrease volume 5.7

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg 5.3

Effect of Temperature on Volume (Charles’ Law) Low Temperature High Temperature

The volume of a gas increases with and increase in temperature. Introductory Chemistry; 2 nd Ed; by Nivaldo Tro; Prentice Hall Publishing 2006, p356

Which picture represents what the gas will look like when the temperature is increased? (Assume constant n, P)

As T increases, V Increases 5.3 V   T V = kT V/T = k V 1 /T 1 = V 2 /T 2 Constant pressure Constant amount of gas

Kinetic theory of gases and … Charles’ Law -Average kinetic energy  T -Increase T, Gas Molecules hit walls with greater Force, this Increases the Pressure BUT since pressure must remain constant, and only volume can change -Volume Increase to reduce Pressure -Increase Temperature, Increase Volume 5.7

Determination of Absolute Zero

A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V L x K 3.20 L = = 192 K 5.3 V 1 /T 1 = V 2 /T 2

Avogadro’s Law V  number of moles (n) V = constant x n V 1 /n 1 = V 2 /n Constant temperature Constant pressure

Kinetic theory of gases and … Avogadro’s Law More moles of gas, more collisions with walls of container More collisions, higher pressure BUT since pressure must remain constant and only volume can change Volume increases to decrease pressure to original value 5.7

Which picture represents what the gas will look like when the moles of gas is doubled? (Assume constant P, T)

Ideal Gas Equation 5.4 Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT R = L atm / (mol K)

Obtaining Other Gas Law Relationship PV = nRT

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm 5.4

What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = K P = 1 atm n = 49.8 g x 1 mol HCl g HCl = 1.37 mol V = 1 atm 1.37 mol x x K Latm molK V = 30.6 L 5.4

Boyle’s Law

Charles Law

Avogadro’s Law

Types of Problems Make Substitution into PV = nRT Given initial conditions, determine final conditions; Cancel out what is constant

Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4

The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)( K) R = L atm / (mol K) 5.4 Experiments show that at STP, 1 mole of an ideal gas occupies L.

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? 5.4 dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x x K Latm molK M = 54.6 g/mol

Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO g C 6 H 12 O 6 1 mol C 6 H 12 O g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = mol CO 2 V = nRT P mol x x K Latm molK 1.00 atm = = 4.76 L 5.5

Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = Repulsive Forces Attractive Forces

Effect of intermolecular forces on the pressure exerted by a gas. 5.8

Van der Waals equation nonideal gas P + (V – nb) = nRT an 2 V2V2 () } corrected pressure } corrected volume

2KClO 3 (s) 2KCl (s) + 3O 2 (g) Bottle full of oxygen gas and water vapor P T = P O + P H O

Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2 5.6

Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T 5.6 mole fraction (X i ) = nini nTnT

A sample of natural gas contains 8.24 moles of CH 4, moles of C 2 H 6, and moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = P T = 1.37 atm = P propane = x 1.37 atm= atm 5.6

Apparatus for studying molecular speed distribution 5.7

The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature 5.7 u rms = 3RT M 

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. 5.7 NH 3 17 g/mol HCl 36 g/mol NH 4 Cl r1r1 r2r2 M2M2 M1M1  =

Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. 5.7 r1r1 r2r2 t2t2 t1t1 M2M2 M1M1  == Nickel forms a gaseous compound of the formula Ni(CO) x What is the value of x given that under the same conditions methane (CH 4 ) effuses 3.3 times faster than the compound? r 1 = 3.3 x r 2 M 1 = 16 g/mol M 2 = r1r1 r2r2 ( ) 2 x M 1 = (3.3) 2 x 16 = x 28 = x = 4.1 ~ 4

Chemistry in Action: Scuba Diving and the Gas Laws PV Depth (ft)Pressure (atm)

An increase in pressure decreases the volume of air in the lungs. Introductory Chemistry; 2 nd Ed; by Nivaldo Tro; Prentice Hall Publishing 2006, p351