Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Elements that exist as gases at 25 0 C and 1 atmosphere 5.1

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

Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 5.2 Barometer Pressure = Force Area

Sea level1 atm 6.4 km 4 miles 0.5 atm 16 km 10 miles 0.2 atm 5.2

When gas pressure is less than atmospheric When gas pressure is more than atmospheric

5.3 As P (h) increases V decreases P h = 1 atm

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

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

As T increasesV increases 5.3

Variation of gas volume with temperature at constant pressure. 5.3 V  TV  T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) Charles’ & Gay-Lussac’s Law Temperature must be in Kelvin

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 = 398 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V L x 398 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

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of O 2 are needed to react with one volume of ammonia at the same temperature and pressure? 4NH 3 + 5O 2 4NO + 6H 2 O 4 mole NH 3 5 mole O 2 At constant T and P 4 volume NH 3 5 volume O volume NH volume O 2

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

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.

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

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

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 310 K Latm molK 1.00 atm = = 4.76 L 5.5

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 fractions partial pressures

A sample of natural gas contains 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

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

Kinetic Molecular Theory of Gases 1.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. 2.Gas molecules exert neither attractive nor repulsive forces on one another. 3. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. 4. 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 & 2. are the definition of an ideal gas

Kinetic theory of gases and … Compressibility of Gases Low Density 5.7 Assume shape and volume of container Diffusion - the gas particles are in continual random (thermal) motion. - there is enormous empty space between the particles of a gas, compared to their size

Charles’ Law collision rate  average kinetic energy of gas molecules  T collision rate with wall  P V increases until gas P drops V  T Boyle’s Law: collision rate with wall  P collision rate  density density  1/V  P  1/V

Avagadro’s Law collision rate  number of particles n collision rate with wall  P V increases until gas P drops n  T P vs. T Law collision rate  average kinetic energy of gas molecules  T collision rate with wall  P P  T

Kinetic theory of gases and … Dalton’s Law of Partial Pressures - Molecules do not attract or repel one another - P exerted by one type of molecule is unaffected by the presence of another gas - P total =  P i 5.7

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

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 ( P + )( V – bn ) = nRT an 2 V2V2 Modifies the Ideal Gas Law to describe real gases accounts for real forces between molecules accounts for size of real molecules