Sit down and get ready for the test, I will hand them out when the bell rings. You have until 8:10 DMA 3/21/11

You can have a few minutes to finish your test, if you are already done, work on something quietly DMA 3/21/11

When you are done with your test, begin the following: In your book, read (it is important that you actually read this) Ch 13-1 (pg. 417-423) and take notes which include the following info: The 6 physical properties that are common to all gases What it means when the terms “atom” “diatomic molecule” or “polyatomic molecule” are used when taking about gases The 6 postulates of the Kinetic-molecular theory of gases Answer question #5 on pg. 423 This is homework if you don’t get it done today

Name 4 physical properties of gases DMA 3/22/11 Name 4 physical properties of gases

The Nature of Gases-physical properties Gases expand to fill their containers Gases are fluid – they flow Gases have low density 1/1000 the density of the equivalent liquid or solid Gases are compressible Gases effuse and diffuse

Kinetic Molecular Theory Gases are made of particles, which have mass Particles of matter are ALWAYS in motion Volume of individual particles is  zero. Collisions of particles with container walls cause pressure exerted by gas. Particles exert no forces on each other. Average kinetic energy proportional to temperature in Kelvin of a gas.

Measuring Gases When you are measuring a gas, you want to know four things about it: 1. n --Amount of the gas-this means in moles represented by the letter n 2. V—volume 3. T—temperature 4. P—pressure

Temperature When using the gas laws, temperature is measured in Kelvin To find Kelvin: Add 273 to Celcius. T(K)=T(oC) + 273

Converting Celsius to Kelvin Gas law problems involving temperature require that the temperature be in KELVINS! Kelvins = C + 273 °C = Kelvins - 273

Celcius & Kelvin Temperature Scales

Converting Temperature Convert the following temperatures to the Kelvin scale. 20 C 85 C –15 C –190 C

An Early Barometer Mercury Barometer Atmospheric pressure– the pressure exerted by the air in the atmosphere. Barometer-measures air pressure The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

Which shoes create the most pressure?

Standard Temperature and Pressure “STP” P = 1 atmosphere (760 torr) T = 0°C (273 Kelvins) The molar volume of an ideal gas is 22.42 liters at STP

Pressure Is caused by the collisions of molecules with the walls of a container is equal to force/unit area SI units = Newton/meter2 = 1 Pascal (Pa) 1 standard atmosphere = 101,325 Pa= 101.3 kPa 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr

Converting Pressure Units Change 5 atmospheres (atm) into millimeters mercury (mm Hg). 1 atm = 760 mm Hg 5 atm x 760 mm Hg = 1 atm 3800 mm Hg

Converting Pressure Units Change 1900 mm mercury (mm Hg) into atmospheres (atm). 1 atm = 760 mm Hg 1900 mm Hg x 1 atm = 760 mm Hg 2.5 atm

Converting Pressure Units Change 560 kilopascals into millimeters mercury (mm Hg). 101.3 kPa = 760 mm Hg 560 kPa x 760 mm Hg = 101.3 kPa 4201 mm Hg

Converting Pressure Units Change 1013 kilopascals into millimeters mercury (mm Hg). 1 atm = 101.3 kPa 1013 atm x 1 atm = 101.3 kPa 10 atm

Boyle’s Law* Pressure is inversely proportional to volume when temperature is held constant. Pressure ´ Volume = Constant P1V1 = P2V2 (T = constant)

If pressure increases then volume decreases. A Graph of Boyle’s Law If pressure increases then volume decreases. If pressure decreases, then volume increases. Inversely proportional.

Charles’s Law The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. (P = constant)

Practice Problem P1 = 710 mm Hg, V1 = 2.40 L, P2 = 75 mm Hg V2 = ? A gas at constant temperature occupies a volume of 2.40 L and exerts a pressure of 710 mm Hg What volume will the gas occupy at a pressure of 75 mm Hg? P1 = 710 mm Hg, V1 = 2.40 L, P2 = 75 mm Hg V2 = ? P1V1 = P2V2 (Boyle’s Law) 710 mm Hg x 2.40 L = 75 mm Hg x V2 V2 = 22.7 L The new volume will be 22.7 liters.

Practice Problem P1 = 3.4 atm, V1 = 14 L, P2 = 5 atm V2 = ? Sulfur dioxide gas at a pressure of 3.4 atm and a volume of 14 liters increases pressure to 5 atm. What is the resulting volume? P1 = 3.4 atm, V1 = 14 L, P2 = 5 atm V2 = ? P1V1 = P2V2 (Boyle’s Law) 3.4 atm x 14 L = 5 atm x V2 V2 = 9.5 L The new volume will be 9.5 liters.

Practice Problem P1 = 11.8 kPa, V1 = 420 ml, V2 = 630 ml P2 = ? Carbon monoxide at a pressure of 11.8 kPa in a vessel of 420 ml expands to a new volume of 630 ml. What is the new pressure in kPa? P1 = 11.8 kPa, V1 = 420 ml, V2 = 630 ml P2 = ? P1V1 = P2V2 (Boyle’s Law) 11.8 kPa x 420 ml = P2 x 630 ml P2 = 7.9 kPa The new pressure is 7.9 kilopascals.

Practice Problem Given 90 ml of H2 gas collected when the temperature is 27 C, how many ml will H2 occupy at 42 C? V1 = 90 ml, T1 = 27 C, T2 = 42 C, V2 = ? T1 = 27 C + 273 = 300 K T2 = 42 C + 273 = 315 K V1 = V2 (Charles’ Law) T1 T2 90 ml = V2 V2 = 90 ml x 315 K 300 K 315 K 300 K V2 = 94.5 ml

Why Don’t I Get a Constant Value for PV = k? Air is not made of ideal gases 2. Real gases deviate from ideal behavior at high pressure

What you need to do

DMA 3.24.11 Using the conversions you have in your notes, convert the following pressure units: 20 mm Hg = ? atm 3 atm = ? kPa 3 atm = ? mm Hg 500 mm Hg = ? kPa 200 kPa = ? atm

DMA 3.25.11 What is the kinetic molecular theory?

Gay Lussac’s Law The pressure and temperature of a gas are directly related, provided that the volume remains constant.

The Combined Gas Law The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas. Why would you need this law when you already have all the others? What is different here?

Combined Gas Law If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = P1 V1 P2 Boyle’s Law Charles’ Law Gay-Lussac’s Law V2 T1 T2

Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20atm? What do you need to do first? Set up Data Table P1 = 0.800 atm V1 = 180 mL T1 = 302 K P2 = 3.20 atm V2= 90 mL T2 = ??

Calculation = 604 K P1 = 0.800 atm V1 = 180 mL T1 = 302 K P1 V1 P2 V2 = P1 V1 T2 = P2 V2 T1 T1 T2 T2 = P2 V2 T1 P1 V1 T2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL T2 = 604 K - 273 = 331 °C = 604 K

Learning Check A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

One More Practice Problem A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

To Do Today Finish Boyle’s Law and Charles’ Law homework-due today! Combined gas law practice problems-due Monday In book-review Ch 13 to help your understanding 

DMA 3/28/11 A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

Dalton’s Law John Dalton 1766-1844

Dalton’s Law of Partial Pressures For a mixture of gases in a container, PTotal = P1 + P2 + P3 + . . . This is particularly useful in calculating the pressure of gases collected over water.

Dalton’s Law of Partial Pressures 2 H2O2 (l) ---> 2 H2O (g) + O2 (g) 0.32 atm 0.16 atm What is the total pressure in the flask? Ptotal in gas mixture = PA + PB + ... Therefore, Ptotal = PH2O + PO2 = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures.

Avogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related. twice as many molecules

Avogadro’s Hypothesis and Kinetic Molecular Theory The gases in this experiment are all measured at the same T and V. P proportional to n

P V = n R T IDEAL GAS LAW Brings together gas properties. Can be derived from experiment and theory. BE SURE YOU KNOW THIS EQUATION!

Using PV = nRT L • atm Mol • K P = Pressure V = Volume T = Temperature R = 0.0821 n = number of moles R is a constant, called the Ideal Gas Constant Instead of learning a different value for R for all the possible unit combinations, we can just memorize one value and convert the units to match R. L • atm Mol • K

Using PV = nRT How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 oC? Solution 1. Get all data into proper units V = 27,000 L T = 25 oC + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm And we always know R, 0.0821 L atm / mol K

Using PV = nRT 2. Now plug in those values and solve for the unknown. How much N2 is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC? Solution 2. Now plug in those values and solve for the unknown. PV = nRT RT RT n = 1.1 x 103 mol (or about 30 kg of gas)

Learning Check Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?

Try This One A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

DMA 3/29/11 A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C? The first thing you need to do is decide which law to use!

At STP For example What is the volume at STP of 4.00 g of CH4? At STP determining the amount of gas required or produced is easy. 22.4 L = 1 mole For example What is the volume at STP of 4.00 g of CH4? How many grams of He are present in 8.0 L of gas at STP?

Gas Stoichiometry: Practice! How many liters of O2 at STP are required to produce 20.3 g of H2O?

Gases and Stoichiometry 2 H2O2 (l) ---> 2 H2O (g) + O2 (g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

Gases and Stoichiometry 2 H2O2 (l) ---> 2 H2O (g) + O2 (g) Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP? Solution 1.1 g H2O2 1 mol H2O2 1 mol O2 22.4 L O2 34 g H2O2 2 mol H2O2 1 mol O2 = 0.36 L O2 at STP

Not At STP Chemical reactions happen in MOLES. If you know how much gas - change it to moles Use the Ideal Gas Law n = PV/RT If you want to find how much gas - use moles to figure out volume V = nRT/P

Example #1 HCl(g) can be formed by the following reaction 2NaCl(aq) + H2SO4 (aq)2HCl(g) + Na2SO4(aq) What mass of NaCl is needed to produce 340 mL of HCl at 1.51 atm at 20ºC?

Example #2 2NaCl(aq) + H2SO4 (aq) 2HCl(g) + Na2SO4 (aq) What volume of HCl gas at 25ºC and 715 mm Hg will be generated if 10.2 g of NaCl react?

GAS DIFFUSION AND EFFUSION diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container.

Diffusion Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

Effusion Effusion: describes the passage of gas into an evacuated chamber.