Chapter 13 Kinetic Theory (Kinetikos- “Moving”) Based on the idea that particles of matter are always in motion The motion has consequences Behavior of Gases Physical Properties of Gases Ideal Gas – an imaginary gas that conforms perfectly to all assumptions

Five Assumptions of the KMT Gases consist of large numbers of tiny particles The Particles are in Constant Motion, moving in straight lines. The collisions between particles & w/ the container wall are elastic. There are no forces of attraction or repulsion between the particles of a gas. The average K.E. of the particles is directly proportional to the Kelvin Temperature. KE = ½ mv2

Measuring Gases Four factors that can affect the behavior of a gas. Amount of gas (n) = moles Volume (V), 1000 cm3 = 1000mL = 1L Temperature (T), Celsius and Kelvins Kelvins = oC + 273 Pressure(P), atmospheres(atm), mmHg, or kPa

Atmospheric Pressure Pressure exerted by the column of air in the atmosphere. Result of the earth’s gravity attracting the air downward. Barometer – device used to measure the atmospheric pressure on earth. Manometer – device used to measure the pressure of a gas in an enclosed container.

Nature of Gases 1 mole of any gas at STP equals 22.4L of volume. STP is defined at sea level. Standard Temperature = 0oC = 273K Standard Pressure = 1 atm = 101.3 kPa = 760mmHg = 760 torrs Normal boiling point of water is 100oC at sea level. Higher elevation lower boiling points. Less Pressure above the surface of water.

Physical Properties of Gases Gases have mass Easily compressed Fill their containers completely Different gases move easily through each other. Diffusion – more mass = slower gas Gases exert pressure Pressure of a gas depends on temperature Volumes of gas particles themselves are assumed to be zero and exert no force on each other.

13.2 - Summary Pressure = Force / Area P = F/A Reduce the area - Increase the Pressure Increase the force - Increase the Pressure S.I Unit for Force - N (Newton) S.I Unit for Area - m2 S.I Unit for Pressure - Pa (Pascal) = 1 N/ m2

Pressure and Volume at Constant Temperature Pressure is exerted by gas particles colliding with walls of its container. What would happen when a gas in a 1-liter container is placed into a 1/4 liter container? Less space = More collisions More collisions = Greater Pressure Conclusion Volume Decreases - Pressure Increases. Inverse Relationship Boyle’s Law

13.3- Boyle’s Law: Pressure-Volume Relationship 2 Conditions P1V1 = k P2V2 = k Then P1V1 = P2V2 If you know 3 you can find the 4th Boyle’s Law - the volume of a fixed gas varies inversely with the pressure at constant temperature.

Algebraic Equations for Boyle P1V1 = P2V2

Sample Problem A sample of gas collected occupies a volume of 150mL when its pressure is 720 mmHg. What volume will it occupy if its pressure is changed to 750 mmHg? Given V1 = 150 mL V2 = ? P1 = 720 mmHg P2 = 750 mmHg Equation P1V1 = P2V2 (720)(150) = (750) X X = (720 x 150) / (750) = 144 mL

Charles’ Law: Temperature-Volume Relationship The volume of a fixed amount of gas varies directly with the Kelvin temperature at constant pressure. V1 / T1 = V2 / T2 V1 T2 = V2 T1

Charles’ Law Temperature must be in Kelvin! Absolute Zero - lowest possible temperature, all kinetic energy ceases. -273.15 °C

Sample Problem A sample of neon gas occupies a volume of 752 mL at 25 °C. What volume will it occupy at 50 °C. P, n are constant. V1= 752 mL V2= ? T1 = 25 °C +273 = 298 K T2 = 50 °C + 273 = 323 K V1 T2 = V2 T1 X = (752)(323) / 298 = 815.1 mL

Gay-Lussac’s Law The pressure of a fixed gas varies directly with the temperature at constant volume. Mathematically P = k T or P / T = k P1T2 = P2T1

Sample Problem (3.0)( 325) = X (298) X = (325)(3.0) / 298 = 3.27 atm The gaseous contents in an aerosol can are under a pressure of 3.00 atm at 25 °C. If the temperature is increased to 52 °C, what would the pressure of the can be? P1= 3.00 atm T1 = 25 + 273 = 298 K P2= ? T2 = 52 + 273 = 325 K P1T2 = P2T1 (3.0)( 325) = X (298) X = (325)(3.0) / 298 = 3.27 atm

Avogadro’s Law Equal volumes of gases at the same temperature and pressure contain equal number of gas particles. At STP, 22.4L = 1 mol V1n2 = V2n1

Sample Problem Determine the number of moles of helium that are held in a 250mL container. Consider that 2.0 moles can be held in a 3L container. V1 = 250mL V2 = 3000mL n1 = ? n2 = 2.0moles

The Combined Gas Law Expresses the relationship between P,T, & V of a fixed amount of gas. Mathematically PV/T = k P1V1T2 = P2V2T1

Sample Problem A helium-filled balloon has a volume of 50.0 L at 25°C and 820 mmHg. What volume will it occupy at 650 mmHg and 10 °C? P1 = 820 mmHg P2 = 650 mmHg V1 = 50 L V2 = ? T1 = 298 K T2 = 283 K V2 = (820)(50)(283) (650)(298) V2= 59.9 L

Dalton’s Law of Partial Pressure The total pressure of a mixture of gases is equal to the sum of all the partial pressures. Partial pressure - pressure of one gas in a mixture of gases PT = P1 + P2 + P3 + …

Sample Problem Determine the pressure of oxygen gas in a container that is under 1 atm of pressure and contains carbon dioxide and nitrogen. Note: PCO2 = .285mmHg, PN2 = 594mmHg 760 mmHg = PO2 + .285mmHg + 594mmHg PO2 = 165.715 mmHg

13-4 : Ideal Gas Law Describes the physical behavior of an ideal gas in terms of pressure, volume, temperature and number of moles. The combination of all 4 gas laws from the previous section.

Derived Equation for the Ideal Gas Law Needed an Ideal Gas Law Constant (R). The second conditions were set at STP to equal the ideal behavior.

Ideal Gas Constant

Practice Problem A camping stove uses a propane tank that holds 4.0 moles of liquid C3H8. How many liters will be needed to hold the same amount of propane at 25oC and 3atm? V = ? n = 4 mol T = 25oC = 298K P = 3 atm R = .0821

Gas Density at STP The density of a gas at STP is constant, due to the standard molar volume of a gas. Note: Molar Volume = 22.4L/mol

Gas Density Problems Determine the density of CO2 at STP. What is the molar mass of gas that has a density of 1.28g/L at STP?

Molar Mass and Ideal Gas Law Considering that moles are in the Ideal Gas Law equation, we can substitute the equivalent of moles(n) into the equation.

Density and the Ideal Gas Law Now that mass(m) is in the equation we can substitute density(d) into the equation.

Molar Mass not at STP Using the previous equations : Example: A 1.25g sample of gas was found to have a volume of 350mL at 20oC and 750mmHg. What is the molar mass of this gas?

Classwork What is the molar mass of a gas that has a density of 2.08g/L at STP? What is the density at STP of NO2? What is the molar mass of a gas, if a 1.39g sample of gas has a volume of 375mL at 22oC and 755mmHg? 46.6g/mol 2.05g/L 90.4 g/mol

Ratm = .0821 RmmHg = 62.4 RkPa = 8.314

Corrected Vapor Pressure When a gas is collected through water displacement, there is always a trace of water vapor in the container. To correctly use the gas laws you must subtract the water vapor pressure from the atmospheric pressure. Pgas = Patm – PH2O

Water Displacement A sample of methane gas that was collected through water displacement had a volume of 350mL at 27.0oC and 720mmHg. What is the volume at 2.0oC and 600.2mmHg? T1 = 300 K T2 = 275K P1 = 720mmHg P2 = 600.2mmHg V1 = 350mL V2= ? V2 = P1V1T2 T1 P2

Solution P1=720mmHg –26.7 mmHg = 693.3mmHg

Graham’s Law Diffusion – Tendency of gas particles to travel toward areas of lower concentration. Effusion – Gas escapes a tiny opening in a container. (one way diffusion) Graham’s Law Rate of effusion of a gas is inversely proportional to the square root of its molar mass. Less mass = faster gas

Graham’s Law Problems Which gas will diffuse into a container faster? CO2 or NH3? Why? Compare the rates of effusion for F2 and Cl2. NH3, has less mass.

At a certain temperature and pressure, Cl2 has a velocity of. 038m/s At a certain temperature and pressure, Cl2 has a velocity of .038m/s. What is the velocity of SO2 at the same condition?

Determining the Molar Mass An unknown gas was placed into a container with N2 gas. The nitrogen was found to travel 1.2 times faster than the unknown gas. What is the molar mass of this unknown gas?