Gases Pisgah High School M. Jones Revision history: 10/5/01 2/20/03 5/21/03 6/24/04 12/27/06 12/29/06 01/05/10 01/09/10 050212 Pisgah High School M. Jones

The Properties of Gases Part I The Properties of Gases

Properties of Gases Gases expand to fill the container. Gases take on the shape of the container. Gases are highly compressible. (Can be liquefied at high pressures). Gases have low densities. Gases mix uniformly.

The Kinetic Molecular Theory The kinetic molecular theory describes the behavior of ideal gases. An ideal gas is one that conforms to the KMT.

1. Molecules are in constant random motion Temperature is proportional to the average kinetic energy of the molecules. KE = ½ mv2 KE = ½ mass times speed squared The speed is proportional to the absolute temperature (Kelvin).

2. A gas is mostly empty space Molecules are far apart from each other. This accounts for the low density and high compressibility. The volume of the individual molecules is negligible compared to the volume of the gas.

3. No intermolecular forces There are no attractive or repulsive forces between gas molecules. Adjacent molecules do not attract or repel each other.

4. Collisions are elastic When gas molecules collide with each other they may speed up or slow down, BUT … The net (total) energy of the gas molecules does not change.

Kinetic Molecular Theory Gases in constant motion, speed depends on temperature. Molecules have negligible volume. No intermolecular forces. Elastic collisions. No change in energy.

Temperature reminder Use Kelvin degrees! When doing calculations, temperature must always be in an absolute temperature scale … … where the lowest possible temperature is zero degrees. Use Kelvin degrees!

Temperature conversion K = C + 273

Pressure 1. Pressure is the force per unit area exerted by the gas molecules. 2. Pressure is proportional to the number of collisions between the gas molecules and the walls of the container.

Pressure force P = area pascal (Pa) = N/m2 1. Pressure is a measure of the force per unit area. force P = Pressure can be in pounds per square inch (PSI), or … area … newtons per square meter (N/m2) pascal (Pa) = N/m2

Pressure Torricelli force P = area 1. Pressure is a measure of the force per unit area. force P = area Glass tube with Hg At sea level, air pressure holds up a column of mercury 760 mm high. Torricelli Bowl of Hg

Pressure Measurements Standard sea level pressure is… 1.00 atmospheres (atm) 760 mm Hg 760 torr (from Torricelli) 101.3 kilopascals (kPa) 14.7 lb/in2

Pressure Measurements Standard sea level pressure is… 1.00 atmospheres (atm) 760 mm Hg 760 torr (from Torricelli) 101.3 kilopascals (kPa) 14.7 lb/in2 Exact

Pressure 2. Pressure is proportional to the number of collisions between the gas molecules and the walls of the container. If you change the number of collisions, you change the pressure.

Part II The Gas Laws

The Gas Laws Boyle’s Law Amonton’s Law Charles’s Law Combined Gas Law Gay-Lussac’s Law Avogadro’s Law Dalton’s Law

Boyle’s Law Pressure Volume At a constant temperature, pressure is inversely proportional to volume.

Boyle’s Law 1/Pressure Volume At a constant temperature, pressure is inversely proportional to volume.

Boyle’s Law 1 P µ V PV = k P1V1 = P2V2 At a constant temperature, pressure is inversely proportional to volume. P µ 1 V PV = k P1V1 = P2V2 Year: 1662

Amonton’s Law P µ T P1 T1 = P2 T2 “Air thermometer”, 1695. This is not Gay-Lussac’s law. Diagram from http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html

Amonton’s Law P µ T P1 T1 = P2 T2 This is why you measure your tire pressure when the tire is cold. Tire pressures vary with temperature. P µ T P1 T1 = P2 T2

Amonton’s Law Amonton’s air thermometer was used to find the value of absolute zero. Measure the pressures of a gas at various temperatures at a constant volume.

Finding Absolute Zero -273 C Pressure Temperature (C) Extrapolate to the x-axis -300 -150 0 100 200 300 Temperature (C)

Charles’s Law Volume Temperature At constant pressure, volume is directly proportional to temp.

Charles’s Law V µ T V T = k V1 T1 = V2 T2 At constant pressure, volume is directly proportional to temperature. V µ T V T = k V1 T1 = V2 T2

Charles’s Law Studied gases during 1780’s. Hydrogen balloon assents, 3000 m, in 1783. Collaborated with the Montgolfier brothers on hot air balloons, 1783. Charles’s gas studies published by Gay-Lussac in1802.

Charles’s Law Hydrogen balloon assent, 3000 m, 1783.

Combined Gas Law P µ 1 V T gives P µ T V

Combined Gas Law T P µ V kT P = V Next we convert to an equation by adding a constant. P = kT V

Combined Gas Law Rearranging the equation gives: = k PV T P = kT V

Combined Gas Law Combining the laws of Boyle, Amonton and Charles produces the combined gas law. = k PV T

Combined Gas Law P2V2 P1V1 k k = = T2 T1 Consider a confined gas at two sets of conditions. Since the number of molecules is the same and the values of k are the same, then we can combine the two equations.

Combined Gas Law P1V1 T1 = k P2V2 T2 = k P1V1 T1 = P2V2 T2

Combined Gas Law P1V1 T1 = P2V2 T2 Use the combined gas law whenever you are asked to find a new P, V or T after changes to a confined gas. P1V1 T1 = P2V2 T2

Sample Combined Gas Law Problem Consider a confined gas in a cylinder with a movable piston. The pressure is 0.950 atm. Find the new pressure when the volume is reduced from 100.0 mL to 65.0 mL, while the temperature remains constant? 100 mL 65 mL

Sample Combined Gas Law Problem Start with the equation for the combined gas law. 100 mL 65 mL

Sample Combined Gas Law Problem P1V1 T1 = P2V2 T2 100 mL Since the temperature is constant, we can cancel out T1 and T2. 65 mL

Sample Combined Gas Law Problem P1V1 = P2V2 This becomes Boyle’s Law 100 mL P1V1 = V2 P2 Next, solve for P2. 65 mL

Sample Combined Gas Law Problem P1V1 = V2 P2 100 mL (0.950 atm)(100.0 mL) = 65.0 mL P2 = P2 1.46 atm 65 mL

Combined Gas Law Problems 1. A sample of neon gas has a volume of 2.00 L at 20.0 C and 0.900 atm. What is the new pressure when the volume is reduced to 0.750 L and the temperature increases to 24.0 C? The answer is 2.43 atm

Combined Gas Law Problems 2. Some “left over” propane gas in a rigid steel cylinder has a pressure of 24.6 atm at a temperature of 20.C. When thrown into a campfire the temperature in the cylinder rises to 313C. What will be the pressure of the propane? The answer is 49.2 atm

Combined Gas Law Problems 3. Consider the fuel mixture in the cylinder of a diesel engine. At its maximum, the volume is 816 cc. The mixture comes in at 0.988 atm and 31 C. What will be the temperature (in C) when the gas is compressed to 132 cc and 42.4 atm? The answer is 1837 C

Gay-Lussac’s Law Also known as the law of combining volumes. At a given temperature and pressure, the volumes of reacting gases are in a ratio of small, whole numbers. Year: 1802

2:1 Gay-Lussac’s Law 2 H2(g) + O2(g)  2 H2O(g) Gay-Lussac found that the volumes of gases in a reaction were in ratios of small, whole numbers. 2 H2(g) + O2(g)  2 H2O(g) 200 mL of hydrogen reacts with 100 mL of oxygen. 2:1

Gay-Lussac’s Law 2 H2(g) + O2(g)  2 H2O(g) The ratio of volumes of gases come from the ratios of the coefficients in the balanced equation.

Avogadro’s Law V µ n V n = k V1 n1 = V2 n2 Avogadro’s law followed Dalton’s atomic theory and Gay-Lussac’s law. Year: 1811. V µ n V n = k Equal volumes of gases at the same temperature and pressure have equal numbers of molecules. V1 n1 = V2 n2

Dalton’s Law Ptotal = P1 + P2 + P3 … Dalton’s law of partial pressures deals with mixtures of gases. The total pressure is the sum of the partial pressures: Ptotal = P1 + P2 + P3 … Use when dealing with the pressure of H2O(g) when collecting a gas over water.

Dalton’s Law “Collecting a gas over water” Inverted gas collecting bottle Rubber tubing carrying H2 gas Flask with metal and HC l Water in trough

Dalton’s Law “Collecting a gas over water” Inverted gas collecting bottle Rubber tubing carrying H2 gas Flask with metal and HC l Water in trough

Dalton’s Law Water is displaced through the mouth of the bottle as H2 gas bubbles in. H2 gas Flask with metal and HC l Water in trough

Dalton’s Law A mixture of hydrogen gas and water vapor is in the collection bottle. H2 gas Flask with metal and HC l Water in trough

Dalton’s Law The pressure of the mixture is the sum of the pressures of H2O gas and H2 gas. H2 gas Flask with metal and HC l Water in trough

Ptotal = PH2O + PH2 Dalton’s Law Flask with metal and HC l H2 gas Flask with metal and HC l Water in trough

Sample Problem The total pressure in the bottle is 712.7 torr. The temperature is 19 C. What is the H2 pressure? H2 gas Flask with metal and HC l Water in trough

Sample Problem At 19C, the vapor pressure of water is 16.5 torr. The H2 pressure is … H2 gas Flask with metal and HC l Water in trough

Sample Problem PH2 = Ptotal - PH2O PH2 = 712.7 torr – 16.5 torr = 696.2 torr H2 gas Flask with metal and HC l Water in trough

Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C.

Ptotal = P1 + P2 Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. Ptotal = P1 + P2

Ptotal = Pair+ PHg Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. Ptotal = Pair+ PHg

PHg = Ptotal - Pair Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. PHg = Ptotal - Pair

PHg = 693 torr – 684 torr Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. PHg = 693 torr – 684 torr

PHg = 9 torr Another Problem A pressure sensor is attached to a sealed flask containing 80.0 mL of mercury at 50.0 C. The pressure sensor indicates 693 torr. The pressure of the air inside the empty flask at 50.0 C is 684 torr. Determine the vapor pressure of mercury at 50.0 C. PHg = 9 torr

More on the gas laws: http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html

Part III Gases and Moles The Ideal Gas Equation

What factors affect the pressure of a confined gas? Think in terms of the number of collisions. Number of molecules Temperature Volume of the container

P µ n Number of molecules Increasing the number of molecules increases the number of collisions … … which increases the pressure. Where n is the number of moles of molecules P µ n

P µ T Temperature … which increases the pressure. Increasing the temperature makes the molecules move faster, increasing the number of collisions … … which increases the pressure. P µ T Where T is the absolute temperature

Volume Increasing the volume of the container decreases the number of collisions … … which decreases the pressure. 1 V P µ Where V is the volume

Sooooo… P µ n P µ 1 V P µ T n T P µ V

Make it into an equation P µ V n R T P = V

P V = n R T n R T P = V The Ideal Gas Equation … is usually written as … P V = n R T

P = V n R T R = 0.0821 L atm mol K The Ideal Gas Equation R is the “gas constant” L atm mol K R = 0.0821

P = V n R T L atm ? mol K The Ideal Gas Equation Can R be in units other than L atm ? mol K

P = V n R T The Ideal Gas Equation R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

P = V n R T The Ideal Gas Equation R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

P = V n R T The Ideal Gas Equation R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

P = V n R T The Ideal Gas Equation R = 0.0821 L atm/mol K R = 8.314 L kPa/mol K R = 62.4 L torr/mol K

PV = nRT Ideal Gas Equation The ideal gas equation relates pressure, volume, temperature and the number of moles of a quantity of gas. PV = nRT

PV = nRT Ideal Gas Equation Use the ideal gas equation whenever the problem gives you mass or moles, or asks for a mass or a number of moles. PV = nRT

Ideal gas equation problem: Some ammonia gas (NH3) is contained in a 2.50 L flask at a temperature of 20.0 C. If there are 0.0931 moles of the gas, what is its pressure?

Solution P = 0.896 atm PV = nRT P = (nRT)/V (0.0821 )(293 K) L atm mol K )(293 K) = (0.0931 mol) 2.50 L P = 0.896 atm

Here’s another one Find the volume of 1.00 mole of nitrogen gas (N2) at 0.0 C and 1.00 atm of pressure.

Solution V = 22.4 L PV = nRT V = (nRT)/P (0.0821 )(273 K) L atm mol K )(273 K) V = (1.00 mol) 1.00 atm V = 22.4 L

Ideal gas equation problem: How many grams of sulfur trioxide are in an 855 mL container at a pressure of 1585 torr and a temperature of 434 C? The answer is 2.46 g SO3

The Ideal Gas Equation can be used to derive the Combined Gas Law

The Combined Gas Law Start with the ideal gas equation: PV = nRT

The Combined Gas Law P1V1 = nRT1 P2V2 = nRT2 Suppose the volume, pressure and temperature change to give a new pressure, volume and temperature. P1V1 = nRT1 P2V2 = nRT2 and

The Combined Gas Law P1V1 = nRT1 P2V2 = nRT2 Now, solve for what doesn’t change, the constants n and R: P1V1 = nRT1 P2V2 = nRT2 and

The Combined Gas Law P1V1 = nR P2V2 T1 = nR T2 Now, solve for what doesn’t change, the constants n and R: P1V1 = nR T1 P2V2 = nR T2 and

The Combined Gas Law P1V1 = nR P2V2 T1 = nR T2 Since both are equal to nR, we can make a new equation. P1V1 = nR T1 P2V2 = nR T2 and

The Combined Gas Law P1V1 T1 P2V2 T2 Since both are equal to nR, we can make a new equation. P1V1 T1 P2V2 T2 =

This is the Combined Gas Law P1V1 T1 P2V2 T2 =

The Combined Gas Law P1V1 T1 P2V2 T2 = It can be derived from the laws of Boyle, Amonton and Charles, or the Ideal Gas Equation

The Ideal Gas Equation and Density

Density calculations m D = V m n = M Start with the equation for density: m D = V And an equation for “moles”: n = m M Where m = mass and M = molar mass

Density calculations m n = M PV = nRT mRT PV = M Now substitute into the ideal gas equation … PV = nRT PV = mRT M and get

Density calculations PV = mRT M Now rearrange to get P = mRT VM

Density calculations D = m V Recall that P = mRT VM P = DRT M

Density calculations Solving for density, P = DRT M becomes: D = PM RT

Density calculations PM D = RT The density of a gas depends on the molar mass and the pressure and temperature. D = PM RT

Density Problem 1. Determine the density of nitrogen, N2, gas (a) at STP (b) at a pressure of 695 torr and a temperature of 40.0 C. The answers are 1.25 g/L, and 0.996 g/L.

Another Problem 2. Determine the molar mass of a gas which has a density of 8.53 g/L at a pressure of 2.50 atm and a temperature of 500.0 K? The answer is 140. g/mol

Solutions of Gases

Solutions of Gases Many gases are soluble in water. Were it not for the solubility of oxygen, fish would have wear scuba tanks. Besides the characteristics of the gas itself, what are the two external factors which affect the solubility of a gas?

Solutions of Gases Two factors affect the solubility of a gas: The temperature of the solution. The pressure of the gas above the solution. See “Henry’s Law”.

Solutions of Gases The temperature of the solution. Why is the champagne chilled?

Solutions of Gases More CO2 is released from the champagne as its temperature rises.

Solutions of Gases The solubility of gases decreases with increasing temperature.

Solutions of Gases Thermal pollution This 1988 thermal image of the Hudson River highlights temperature changes caused by discharge of 2.5 billion gallons of water each day from the Indian Point power plant. The plant sits in the upper right of the photo — hot water in the discharge canal is visible in yellow and red, spreading and cooling across the entire width of the river. Two additional outflows from the Lovett coal-fired power plant are also clearly visible against the natural temperature of the water, in green and blue. http://www.switchstudio.com/waterkeeper/issues/Winter%2007/fight_power_plants.html

Solutions of Gases Thermal pollution The result is a decrease in dissolved oxygen which is needed by the plants and animals living in the heated water.

Solutions of Gases Why do we find trout in cold mountain streams, often near riffles and waterfalls? High dissolved oxygen.

Solutions of Gases The second factor is the pressure of the dissolved gas above the solution. CO2 is dissolved in a can of Pepsi. Above the liquid in the can is CO2 gas. The high pressure of the CO2 in the can keeps the can rigid and the CO2 in solution.

Solutions of Gases Henry’s Law: At a constant temperature, the amount of a given gas dissolved in a solution is directly proportional to the partial pressure of that gas in equilibrium with the solution.

Solutions of Gases p = kHc Henry’s Law: Where p is the partial pressure of the gas, kH is the Henry’s law constant, and c is the concentration of the gas in moles per liter. Some Henry’s law constants at 298K O2 769.2 Latm/mol CO2 29.4 Latm/mol H2 1282.1 Latm/mol

Solutions of Gases p = kHc Henry’s Law: The pressure of the CO2 in the headspace of can of Pepsi is 4.50 atm at 298K. What is the concentration of the CO2? Some Henry’s law constants at 298K O2 769.2 Latm/mol CO2 29.4 Latm/mol H2 1282.1 Latm/mol Ans: 0.153 M

Solutions of Gases p = kHc Henry’s Law: How do we predict the relative solubilities of gases based on the values of kH? Which gas below is the most soluble in water? Some Henry’s law constants at 298K O2 769.2 Latm/mol CO2 29.4 Latm/mol H2 1282.1 Latm/mol Answer: CO2

Graham’s Law

Graham’s Law Molecules at the same temperature have the same average kinetic energy. (T µ KE) KE is proportional to the speed of the molecules and the mass of the molecules. KE = ½ mv2

Graham’s Law Diffusion – “a gas spreading out” Effusion – a gas moving through a small hole Graham’s Law deals with the effusion of two gases into each other.

Graham’s Law Gas A Gas B Consider two gases at the same temperature and pressure in a box with two partitions, separated by a wall with a hole that can be opened.

…eventually both gases will be evenly distributed in the box. Graham’s Law Gas A Gas A effuses into Gas B, and … Gas B Gas B effuses into Gas A, so that … …eventually both gases will be evenly distributed in the box.

Graham’s Law Gas A Gas B The ratio of the rates at which Gas A and Gas B effuse into each other is … … inversely proportional to the square roots of their molar masses.

Graham’s Law Gas A Gas B The bottom line: the lighter the gas, the faster it moves!

Graham’s Law Gas A Gas B Suppose nitrogen and an unknown gas, both at the same temperature and pressure, are in the box. The rates of effusion are …

Graham’s Law Gas A Gas B … 0.0160 moles/L/min for the nitrogen and 0.0205 mol/L/min for the unknown. What is the molecular weight of the unknown?

Graham’s Law Gas A Gas B Start with the equation for Graham’s Law

Graham’s Law Gas A Gas B Square both sides, then solve for the molecular mass of compound x

Graham’s Law Gas A Gas B

Graham’s Law Gas A Gas B Mx = 17.0g/mol

Graham’s Law Gas A Gas B What is the unknown gas? The gas is “smelly”, dissolves in water to make a basic solution, and has a molar mass of 17 g/mol.

Ammonia gas, NH3, has a strong smell, and ... Graham’s Law Gas A Gas B Ammonia gas, NH3, has a strong smell, and ...

Graham’s Law Gas A Gas B … it reacts with water … NH3(g) + HOH(l)  NH4+ + OH- OH- … which is a basic solution,

Graham’s Law Gas A Gas B … and ammonia, NH3, has a molar mass of 17.0 g/mol. 1 x 14.0 + 3 x 1.0 = 17.0 g/mol