Chapter 3 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 2. The Particles are in Constant Motion, moving in straight lines. 3. The collisions between particles & w/ the container wall are elastic. 4. There are no forces of attraction or repulsion between the particles of a gas. 5. The average K.E. of the particles is directly proportional to the Kelvin Temperature. KE ½ mv 2
Measuring Gases Four factors that can affect the behavior of a gas. Amount of gas (n) moles Volume (V), 000 cm 3 000mL L Temperature (T), Celsius and Kelvins Kelvins o C + 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 mole of any gas at STP equals 22.4L of volume. STP is defined at sea level. Standard Temperature 0 o C 273K Standard Pressure atm 0.3 kpa 760mmHg 760 torrs Normal boiling point of water is 00 o C 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.
3.2 - 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 - m 2 S.I Unit for Pressure - Pa (Pascal) N/ m 2
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 -liter container is placed into a /4 liter container? Less space More collisions More collisions Greater Pressure Conclusion Volume Decreases - Pressure Increases. Inverse Relationship Boyle s Law
3.3- Boyle s Law: Pressure- Volume Relationship Boyle s Law - the volume of a fixed gas varies inversely with the pressure at constant temperature. 2 Conditions P V k P 2 V 2 k Then P V P 2 V 2 If you know 3 you can find the 4th
Algebraic Equations for Boyle P V P 2 V 2 P 2 P V V 2 P P2V V 2 V 2 P V P 2 V P2V P 2
Sample Problem A sample of gas collected occupies a volume of 50mL when its pressure is 720 mmhg. What volume will it occupy if its pressure is changed to 750 mmhg? Given V 50 ml V 2? P 720 mmhg P 2 750 mmhg Equation P V P 2 V 2 (720)(50) (750) X X (720 x 50) / (750) 44 ml
Charles Law: Temperature- Volume Relationship The volume of a fixed amount of gas varies directly with the Kelvin temperature at constant pressure. V / T V 2 / T 2 V T 2 V 2 T
Charles Law Temperature must be in Kelvin! Absolute Zero - lowest possible temperature, all kinetic energy ceases. -273.5 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. V 752 ml V 2? T 25 C +273 298 K T 2 50 C + 273 323 K V T 2 V 2 T X (752)(323) / 298 85. 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 P T 2 P 2 T
Sample Problem 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? P 3.00 atm P 2? P T 2 P 2 T T 25 + 273 298 K T 2 52 + 273 325 K (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 mol V n 2 V 2 n
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. V 250mL V 2 3000mL n? n 2 2.0moles Vn 2 V2n 250mL 2.0mol 3000mL X ( )( ) ( ) X 250mL 2.0mol ( )( ) 3000mL ( ).67moles
The Combined Gas Law Expresses the relationship between P,T, & V of a fixed amount of gas. Mathematically PV/T k P V T P2V T P V T 2 P 2 V 2 T 2 2
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 0 C? P 820 mmhg P 2 650 mmhg V 50 L V 2? T 298 K T 2 283 K V 2 (820)(50)(283) (650)(298) V 2 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 P T P + P 2 + P 3 +
Sample Problem Determine the pressure of oxygen gas in a container that is under atm of pressure and contains carbon dioxide and nitrogen. Note: P CO2.285mmHg, P N2 594mmHg 760 mmhg P O2 +.285mmHg + 594mmHg P O2 65.75 mmhg
3-4 : 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 P V n T P V n 2 2 T 2 2 Gas Law Needed an Ideal Gas Law Constant (R). The second conditions were set at STP to equal the ideal behavior. P V n T R PV nrt R R R atm kpa mmhg.082 8.34 62.4
Ideal Gas Constant PV n T R atm ( atm)( 22.4L) PV n T R.082 atm L ( mol)( 273K) mol K R mmhg ( 760mmHg )( 22.4L) R 62. 4 ( mol)( 273K ) mol K mmhg L R kpa PV n T R ( 0.3 kpa)( 22.4L) kpa L 8. ( mol)( 273K ) mol K 34
Practice Problem A camping stove uses a propane tank that holds 4.0 moles of liquid C 3 H 8. How many liters will be needed to hold the same amount of propane at 25 o C and 3atm? V? n 4 mol T 25 o C 298K P 3 atm R.082 nrt V ( 4)(.082)( 298) P V 32.6L 3
Gas Density at STP The density of a gas at STP is constant, due to the standard molar volume of a gas. density (STP ) molar mass molar volume Note: Molar Volume 22.4L/mol
Gas Density Problems Determine the density of CO 2 at STP. molar mass 44g/mol density (STP).96g/L molar volume 22.4L/mol What is the molar mass of gas that has a density of.28g/l at STP? molar mass density Molar Volume (STP) molar mass.28g/l 22.4L/mol 28.7g/mol
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. m M PV nrt, n, PV mrt M MPV mrt
Density and the Ideal Gas Law Now that mass(m) is in the equation we can substitute density(d) into the equation. M mrt PV MP drt
Molar Mass not at STP Using the previous equations : Example: A.25g sample of gas was found to have a volume of 350mL at 20 o C and 750mmHg. What is the molar mass of this gas? MPV mrt M(750)(.35) (.25)(62.4)(293) M (.25)(62.4)(293) (750)(.35) 87.g/mol
Classwork. What is the molar mass of a gas that has a density of 2.08g/L at STP? 46.6g/mol 2. What is the density at STP of NO 2? 3. What is the molar mass of a gas, if a.39g sample of gas has a volume of 375mL at 22 o C and 755mmHg? 90.4 g/mol 2.05g/L
M T P drt P MPV mr mrt MV MPV d M V mrt MP RT nrt P mrt PV MP T P MP dr m nrt V drt P MPV RT T PV PV nr drt M V nrt n mrt MP PV RT R atm.082 R mmhg 62.4 R kpa 8.34
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. P gas P atm P H2O
Water Displacement A sample of methane gas that was collected through water displacement had a volume of 350mL at 27.0 o C and 720mmHg. What is the volume at 2.0 o C and 600.2mmHg? T 300 K T 2 275K P 720mmHg P 2 600.2mmHg V 350mL V 2? V 2 P V T 2 T P 2
Solution P 720mmHg 26.7 mmhg 693.3mmHg V 2 ( 693.3)( 350)( 275) ( 600.2)( 300) 370.6mL
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 Rate molar mass A B Rate molar mass RateB molar massa
Graham s Law Problems Which gas will diffuse into a container faster? CO 2 or NH 3? Why? NH 3, has less mass. Compare the rates of effusion for F 2 and Cl 2. F Cl 2 7 g/mol 38g/mol 2.37
At a certain temperature and pressure, Cl 2 has a velocity of.038m/s. What is the velocity of SO 2 at the same condition? x.038m/s 7 g/mol 64g/mol x.05.038m/s ( ) ( ) x.05.038m/s.040m/s
Determining the Molar Mass An unknown gas was placed into a container with N 2 gas. The nitrogen was found to travel.2 times faster than the unknown gas. What is the molar mass of this unknown gas? X.2 > 28g/mol (.2 ) X.44 > 28g/mol 28g/mol 2 X x (.44) (28g/mol) 40.3 g/mol 2