Unit 1 Gases

The Nature of Gases Objectives: 1. Use kinetic-molecular theory to explain the behavior of gases. 2. Describe how mass affects the rates of effusion and diffusion 3. Explain how gas pressure is measured and calculate the partial pressure of a gas

Properties of Gases take the shape of their container low density Compressible Mixtures are homogeneous Fluids (flow)

Diffusion and effusion Diffusion: movement of one material through another Effusion: a gas escapes through a tiny opening Graham’s law of effusion Lighter particles effuse faster than heavier particles.

Gas pressure Results from collisions of gas particles with an object. In empty space there are no particles, there is no pressure and (vacuum.) Atmospheric pressure (air pressure): due to atoms and molecules in air. Barometer: used to measure atmospheric pressure.

Units for measuring pressure: Pascal (Pa) Standard atmosphere (atm) Millimeters of mercury (mmHg) 1 atm = 760 mmHg = kPa 1kpa = 1000 pa Standard pressure: 1 atm Factors affecting gas pressure Amount of gas Volume Temperature Standard temperature : 0  C (273K)

Converting between units of pressure 1. A pressure gauge records a pressure of 450 kPa. What is the measurement expressed in atmospheres and millimeters of mercury? For converting to atm: 450 kpa x 1 atm = 4.4 atm 101.3kPa For converting to mmHg: 450kPa x 760 mmHg = 3.4 x 10 3 mmHg kPa

2. What pressure in kilopascals and in atmospheres, does a gas exert at 385 mmHg? 51.3 kPa, atm 3. The pressure on the top of Mount Everest is 33.7 kPa. Is that pressure greater or less than 0.25atm? 33.7 kPa is greater than 0.25 atm

Reaction_to_Air_Pressure_Below_Sea_Level.asf Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture. P total = P 1 + P 2 + P 3 + … P n CW p 405 #1-3 p409 #4-7

Gas Laws Objectives 1. Describe the relationships among the temperature, pressure, and volume of a gas 2. Use the gas laws to solve problems

Boyle’s Law : Pressure and Volume States that for a given mass of gas at constant temperature, the volume of a gas varies inversely with pressure. If pressure increases, volume decreases; if pressure decreases, volume increases. Volume could be in liters (L), mL or cm 3 1L=1000 mL 1 cm 3 = 1 mL

P 1 x V 1 = P 2 x V 2 P: pressure1: initial condition V: volume2: final condition YouTube - Self Inflating a Balloon YouTube - Shaving Cream Under Vacuum

Using Boyle’s Law 1. A balloon with 30.0L of helium at 103kPa rises to an altitude where the pressure is only 25.0kPa. What is the volume of the helium (at constant temperature)? P 1 x V 1 = P 2 x V 2 P 1 = 103 kPaV 1 = 30.0 L P 2 = 25.0 kPaV 2 =? V 2 = P 1 V 1 P 2 = (103 kPa x 30.0L) 25.0 kPa = L

2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure of the container now (at constant temperature)? P 1 x V 1 = P 2 x V 2 P 1 = 205 kPaV 1 = 4.00 L P 2 = ?V 2 =12.0 L P 2 = P 1 V 1 V 2 = (205 kPa x 4.00L) 12.0L = 68.3 kPa

Classwork: pg 443 #1-3

Charles’s Law: Temperature and Volume States that the temperature of an enclosed gas varies directly with the volume at constant pressure. As temperature increases, volume increases. V1 = V2 T1 T2 V1: initial volumeV2: final volume T1: initial temperatureT2: final temperature Temperature has to be in Kelvin scale. K =  C + 273

Volume and Temperature As a gas is heated, it expands. This causes the density of the gas to decrease. YouTube - Balloon in liquid nitrogen

Using Charles’s Law Ex.1 A balloon inflated in a room at 24  C has a volume of 4.00L. The balloon is then heated to a temperature of 58  C. What is the new volume ? T 1 = 24  C T 2 =58  C = 24  C +273 = 58  C +273 = 297K = 331 K V 1 = 4.00 LV 2 = ? Since temperature increases, you expect the volume to increase. V1 = V2 T1 T2 V1T2 = V2 T1 (4.00Lx 331K) = V2 297K 4.56 L = V2

Using Charles’s Law Ex.2 A sample of SO 2 gas has a volume of 1.16L at a temperature of 23  C. At what temperature (in  C) will the gas have a volume of 1.25L? T 1 = 23  C = 296K T 2 =? V 1 = 1.16 LV 2 = 1.25 Since volume increases, you expect the temperature to increase. V1 = V2 T1 T2 T2 = V2 x T1 V1 T2 = 1.25L x 296 K 1.16 L T2 = 319 K -273 = 46.0  C Classwork: p 446 #4-7

Combined Gas Law Describes the relationship among the pressure, temperature and volume, when the amount of gas is constant. P1V1 = P2V2 T1 T2 Standard temperature and pressure (STP): 0  C, 1 atm (101.3 kPa) Useful conversions: 1L =1000 mL ; 1mL =1cm 3 ; 1dm 3 = 1 L

Using the combined gas law: 1. The volume of a gas filled balloon is 30.0L at 313K and 153 kPa. What would the volume be at standard temperature and pressure (STP)? Classwork: p 450 #11,12 p 984 #8,9 T 1 = 313 KT 2 =0  C=273 K (at STP) P 1 = 153 kPaP 2 = kPa (at STP) V 1 = 30.0LV 2 = ? P1V1 = P2V2 T1 T2 P1V1T2 = V2 P2T L = V2 (153kPa x30.0L x 273K) = V2 ( kPa x 313 K)