What is Pressure? Force of gas particles running into a surface. It’s measured by a barometer

Pressure is measured by a Barometer Pressure changes with altitude Barometric pressure at sea level is (1 atm) or 760 mm Hg

Pressure and Moles (# of Molecules) If pressure is molecular collisions with the container… As number of molecules increase, there will be more molecules to collide with the wall THUS, collisions between molecules and the wall increase Pressure increases As # of moles increase, pressure increases Think about blowing up a balloon!

# of Gas Particles vs. Pressure

Pressure & Volume If pressure is molecular collisions with the container… As volume increases, molecules can travel farther before hitting the wall THUS collision frequency between molecules & the wall decreases Pressure decreases As volume increases, pressure decreases. Think about how your lungs work! http://www.youtube.com/watch?v=q6-oyxnkZC0

What is “Temperature”? Temperature – measure of the average kinetic energy of the molecules Energy due to motion (Related to how fast the molecules are moving) As KELVIN temperature increases, Average Kinetic Energy Increases and Molecular motion increases

Pressure and Temperature If temperature is related to molecular motion… and pressure is molecular collisions with the container… As temperature increases, molecular motion increases Thus collision frequency increases Pressure increases As temperature increases, pressure increases

Volume and Temperature and volume is the amount of space the gas occupies… If temperature is related to molecular motion… As temperature increases, molecular motion increases THUS molecules will move farther away from each other Volume increases As temperature increases, volume increases Think of liquid nitrogen and the balloon. http://www.youtube.com/watch?v=QEpxrGWep4E

Volume and Moles (# of Molecules) As number of molecules increases, there will be more molecules to collide with the wall Thus, collision frequency increases Volume increases As # of moles increase, volume increases

Pressure In Versus Out A container will expand or contract until the pressure inside equals the atmospheric pressure outside A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain or in an airplane. Example: The internal pressure of the bag at low altitude is high At high altitude there is lower pressure Lower pressure Higher pressure Lower pressure The internal pressure is higher than the external pressure. The bag will expand in order to reduce the internal pressure.

When Expansion Isn’t Possible Rigid containers cannot expand Example: An aerosol can is left in a car trunk in the summer. What happens? The temperature inside the can begins to rise. As temperature increases, pressure increases. Lower pressure Higher pressure Can Explodes! The internal pressure is higher than the external pressure. The can is rigid—it cannot expand, it explodes!

Air Pressure Crushing Cans http://www.csun.edu/scied/4-discrpeant-event/the_can_crush/index.htm

Air Pressure Crushing “Cans” http://www.youtube.com/watch?v=Zz95_VvTxZM Another cool video http://www.youtube.com/watch?v=JsoE4F2Pb20

Section 2: Kinetic Molecular Theory(KMT): explains gas behavior based upon the motion of molecules based on an ideal gas IDEAL gases are IMAGINARY gases that follow the assumptions of the KMT

Assumptions of the KMT All gases are made of atoms or molecules that are in constant, rapid, random motion 1 The KELVIN temperature of a gas is proportional to the average kinetic energy of the particles 2 3 Gas particles are not attracted nor repelled from one another *** All gas particle collisions are elastic (no kinetic energy is lost to other forms) 4 5 The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant***Gases have no Volume!

So what is a “REAL” gas? Real gases, (like nitrogen), will eventually condense into a liquid when the temperature gets too low or the pressure gets too high BECAUSE: Assumption #3 Gas particles do have attractions and repulsions towards one another Assumption #5 Gas particles do take up space

Real Gases Deviate from Ideal Gas Behavior when at HIGH pressure The gas molecules are compressed making the volume they take up more significant than if they were spread out

Real Gases Deviate from Ideal Gas Behavior when at LOW temperature. The lower kinetic energy causes the molecules to move slower and ATTRACTIVE FORCES that really exist start to take effect

A Molecule’s POLARITY can cause a deviation from Ideal Gas Law behavior Polar gases (HCl) deviate more than nonpolar gases (He or H2)- due to innate strong attractive forces of polar molecules

Gas Movement: Effusion vs Diffusion Effusion –gas escapes from a tiny hole in the container Effusion is why balloons deflate over time!

Diffusion –gas moves across a space from high to low concentration Diffusion is the reason we can smell perfume across the room

Effusion, Diffusion & Particle Mass How are particle size (mass) and these concepts related? As particle size (mass) increases, the particles move slower, and the rate of effusion/diffusion is lower it takes them more time to find the hole or to go across the room As mass of the particles increases, rate of effusion and diffusion is lowered.

Rate of Diffusion & Particle Mass H2 Watch as larger particles take longer to get to your nose CO2

Section 3—Gas Laws How can we calculate Pressure, Volume and Temperature of a gas?

Pressure Units 1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi Several units are used when describing pressure Unit Symbol atmospheres atm Pascals, kiloPascals Pa, kPa millimeters of mercury mm Hg pounds per square inch psi 1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi

Conversions Between Different Pressure Units 1 atm = 760 mmHg = 101.3 kPa Examples Convert 654 mm Hg to atm 2.Convert 879 mm Hg to kPa 3.Convert 15.6 atm to kPa 654 mmHg x 1atm = 760 mmHg .861 atm 879 mmHg x 101.3 Kpa = 760 mmHg 117 Kpa 15.6 atm x 101.3 Kpa = 1atm 1580 Kpa

Temperature Unit used in Gas Laws Kelvin (K)– temperature scale with an absolute zero Temperatures cannot fall below an absolute zero Examples Convert 15.6 °C into K 2. Convert 234 K into °C 15.6 + 273 = K 288.6  289 K °C + 273 = 234 -39 °C

Standard Temperature & Pressure (STP) the conditions of: 1 atm (or the equivalent in another unit) 0°C (273 K) Problems often use “STP” to indicate quantities…don’t forget this “hidden” information when making your list!

GAS LAWS: “Before” and “After” This section has 5 gas laws which have “before” and “after” conditions. For example: P= Pressure V= Volume T=Temperature n= moles(molecules) 1= initial amount 2= final amount

Boyle’s Law Pressure Increases as Volume Decreases

Boyles’ Law Volume & Pressure have an INVERSELY proportional when temperature and moles are held constant P = pressure V = volume The two pressure units must match and the two volume units must match! Example: A gas sample is 1.05 atm when at 2.5 L. What volume is it if the pressure is changed to 0.980 atm?

Boyles’ Law V2 = 2.7 L P1 = 1.05 atm V1 = 2.5 L P2 = 0.980 atm ***The two pressure units must match & the two volume units must match! Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 0.980 atm? P1 = 1.05 atm V1 = 2.5 L P2 = 0.980 atm V2 = ? L V2 = 2.7 L

Boyles Law: Graph

Charles’ Law Volume Increases as Temperature Increases

Charles’ Law Volume & Temperature are DIRECTLY proportional when pressure and moles are held constant. The two volume units must match & temperature must be in Kelvin! V = Volume T = Temperature What is the final volume if a 10.5 L sample of gas is changed from 25C to 50C? Example: Temperature needs to be in Kelvin! 25C + 273 = 298 K V1 = 10.5 L T1 = 25C V2 = ? L T2 = 50C 50C + 273 = 323 K

Charles’ Law V1 = 10.5 L T1 = 25.0C = 298 K V2 = ? L T2 = 50.0C ***The two volume units must match & temperature must be in Kelvin! Example: What is the final volume if a 10.5 L sample of gas is changed from 25C to 50C? V1 = 10.5 L T1 = 25.0C V2 = ? L T2 = 50.0C = 298 K = 323 K V2 = 11.4 L

Charles Law: Graph

Gay-Lussac’s Law Temperature decreases as Pressure decreases

Gay-Lussac’s Law Pressure & temperature are DIRECTLY proportional when moles and volume are held constant P = Pressure T = Temperature The two pressure units must match and temperature must be in Kelvin! Example: A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be? Temperature needs to be in Kelvin! P1 = .329 atm T1 = 47C P2 = ? atm T2 = 77C 47C + 273 = 320 K 77C + 273 = 350 K

Gay-Lussac’ Law Example: A sample of hydrogen gas at 47C exerts a pressure of .329 atm. The gas is heated to 77C at constant volume and moles. What will the new pressure be? P1 = .329 atm T1 = 47.0C P2 = ? atm T2 = 77.0C = 320 K = 350 K P2 = .360 atm

Gay Lussac Law: Graph

Avogadro’s Law Moles and Volume are directly proportional when temp. & pressure are held constant V = Volume n = # of moles of gas The two volume units must match! A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles? Example:

Avogadro’s Law The two volume units must match! A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles? Example: n1 = 0.15 moles V1 = 2.5 L n2 = 0.55 moles V2 = ? L V2 = 9.2 L

Combined Gas Law Example: P = Pressure V = Volume n = # of moles T = Temperature Each “pair” of units must match and temperature must be in Kelvin! Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298 K is changed to STP?

Combined Gas Law P = Pressure V = Volume T = Temperature Moles is not mentioned so remove it from equation! What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298K is changed to STP? Example: STP is standard temperature (273 K) and pressure (1 atm) P1 = 755 mmHg V1 = 15.5 L T1 = 298 K P2 = 760mmHg V2 = ? L T2 = 273 K V2 = 14.1 L

Why you really only need 1 of these The combined gas law can be used for all “before” and “after” gas law problems! For example, if volume is held constant, then and the combined gas law becomes:

Transforming the Combined Law Watch as variables are held constant and the combined gas law “becomes” the other 3 laws Hold pressure and temperature constant Avogadro’s Law Hold moles and temperature constant Boyles’ Law Hold pressure and moles constant Charles’ Law

Section 4—Other Gas Laws

Dalton’s Law Each gas in a mixture exerts its own pressure called a partial pressure Total Pressure = PT Partial Pressure= P1, P2….

Dalton’s Law Example: A gas mixture is made up of oxygen(2.3 atm) and nitrogen(1.7 atm) gases. What is the total pressure? PT = 2.3 atm + 1.7 atm = 4.0 atm

Modified Daltons Law When a gas is collected over water, the total pressure of the mixture collected is a combination of water vapor and the gas you are collecting! 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO + PH O 2

Modified Dalton’s Law Example: What is the pressure of the water vapor if the total pressure of the flask is 17.5 atm and the pressure of the oxygen gas is 16.1 atm? 17.5 = 16.1 atm + PH2O =1.4 atm

The Ideal Gas Law (an“AT NOW”equation) The volume of a gas varies directly with the number of moles and its Kelvin temperature P = Pressure V = Volume n = moles R = Gas Law Constant T = Temperature Choose the one with units that match your pressure units! There are three possibilities for “R”! Volume must be in Liters when using “R” to allow the unit to cancel!

The Ideal Gas Law Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy?

The Ideal Gas Law The Ideal Gas Law does not compare situations—it describes a gas in one situation. A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy? Example: Chosen to match the kPa in the “P” above n = 0.55 moles P = 105.7 kPa T = 27°C + 273 = 300 K V = ? R = 8.31 L kPa / mole K V2 = 13 L

The Ideal Gas Law Example 2: What mass of hydrogen gas in grams is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure? n = ? P = 3.50 atm T = 27°C + 273 = 300 K V = 10.0 L R = .0821 L atm /mole K n = 1.42 mol NOW CHANGE TO GRAMS Chosen to match the atm in the “P” above 1.42 mol x 2.02 g = 2.87 g 1 mol