Avogadro’s Law

What is Avogadro’s Law Avogadro’s Principle – equal volumes of gases at the same temperature and pressure contain equal numbers of particles

Avogadro’s Formula Formula: V1 = V2 n1 n2 This is a direct relationship! So if the amount of gas increases, then the volume will ___________. If the amount of gas decreases, then the volume will __________. * n represents the amount of gas increase decrease

Graph for Avogadro’s Law What Laws have we learned were also direct relationships, in which their graphs were similar to Avogadro’s?

Avogadro’s Law

Avogadro’s Law Molar Volume – for a gas is the volume that one mole of that gas occupies at STP (STP is standard temperature and pressure which is 0oC and 1 atm) Avogadro showed experimentally that 1 mole of any gas will occupy a volume of 22.4L at STP **Conversion Factor: 1 mol (any gas) = 22.4 L at STP ** (This is on your formula chart under constants and conversions!)

Avogadro’s Law: Example 1 Calculate the volume that 0.881 moles of oxygen gas at STP will occupy. (This can be solved using the Avogadro’s law formula or using the dimensional analysis method)

Avogadro’s Law: Example 1 Answer Formula: V1 = V2 n1 n2 What do we know? V1 = ? L n1 = 0.881 mol O2 V2 = 22.4 L n2 = 1 mol O2 Remember, at STP: 22.4 L/mol Second way to solve, using the dimensional analysis: .881 mol O2 | 22.4 L = | 1 mol O2 First way to solve, using the formula: V1 = 22.4 L .881 mol O2 1 mol O2 After cross multiplying you end up with V1(1 mol O2) = (22.4 L)(.881 mol O2) 1 mol O2 1 mol O2 V1 = ___________ --------- 19.73 L --------- ---------- ---------- ----------- ---------- 19.73 L

Avogadro’s Law: Example 2 How many grams of N2 will be contained in a 2.0 L flask at STP?

Avogadro’s Law: Example 2 Answer Formula: V1 = V2 n1 n2 What do we know? V1 = 2.0 L n1 = ? g N2 V2 = 22.4 L n2 = 1 mol N2 Remember, at STP: 22.4 L/mol First , solve for the number of moles of N2: 2 L N2 | 1 mol N2 = .089 mol N2 | 22.4 L N2 Then, use dimensional analysis to convert from moles of N2 to grams of N2: .089 mol N2 | 28.014 g N2 = 2.50 g N2 | 1 mol N2

Ideal Gas Law

What is the Ideal Gas Law? It is a good approximation to the behavior of many gases under many conditions

Ideal Gas Law Formula

Example: R = 0.0821 (Liters)(atm) (moles)(Kelvin) Ideal Gas Law Formula Formula: PV = nRT (called Piv-Nert formula) R is called the Ideal Gas Constant R is dependent on the units of the variables for P, V, and T Temperature is always in Kelvin Volume is always in Liters Pressure is either in atm, mmHg, or kPa. Because of the different pressure units, there are 3 possibilities of the ideal gas constant (refer to the EOC Chart under constants and conversions) Example: R = 0.0821 (Liters)(atm) (moles)(Kelvin) *We would use this value for R if the given pressure’s units are in atm*

Ideal Gas Law Simulation http://phet.colorado.edu/en/simulation/gas-properties

Ideal Gas Law Example 1 (using moles) If the pressure by a gas at 30oC in a volume of .05 L is 3.52 atm, how many moles of the gas is present? **To know which R value to use, look at what units pressure is in**

Ideal Gas Law Example 1 Ans (finding moles) Formula: PV = nRT What do we know? P = 3.52 atm V = .05 L n = ? moles R = T = 30oC + 273 = 303K What R value will we use? (Hint: Look at the units for pressure) 0.0821 (L*atm)/(mol*K) Now lets plug in the information and solve: P * V = n * R * T (3.52)(.05) = n(.0821)(303) (.0821)(303) (.0821)(303) n = ______________ -------- ------- -------- ----- .0071 mol

Ideal Gas Law Example 2 (finding grams) Avogadro’s Law allows us to write a gas law that is valid not only for any P, V, or T but also for any mass of any gas! Example: Calculate the grams of N2 gas present in a 0.600 L sample kept at 1.00 atm and a temperature of 22.0oC.

Ideal Gas Law Example 2 Answer (using moles) Formula: PV = nRT What do we know? P = 1.00 atm V = 0.600 L n = ? g N2 R = T = 22.0oC + 273 = 295 K What R value will we use? (Hint: Look at the units for pressure) Now lets plug in the information and solve: P * V = n * R * T (1.00)(.600) = n(.0821)(295) (.0821)(295) (.0821)(295) n = ______________ 0.0821 (L*atm)/(mol*K) ------- ------- -------- ----- .025 mol N2 But, they want the answer in grams, so we need to do dimensional analysis using molar mass: .025 mol N2 | 28.014 g N2 = __________ | 1 mol N2 --------- .694 g N2 ----------

Ideal vs. Real Gas

Ideal vs. Real Gas An ideal gas obeys all the assumptions of the kinetic theory. (Atoms or molecules are non-interacting particles, etc.) NO gas is IDEAL (ideal gas doesn’t exist). They all take up space and interact with other molecules (attraction, repultion) but most gases will behave like ideal gases at the right conditions of temp. and pressure. Real gases do NOT behave like ideal gases at extremely high pressures and extremely low temperatures.

Ideal vs. Real Gas

Explanation Ideal gases do not have molecular volume and show no attraction between molecules at any distance Real gas molecules have volume and show attraction at short distances. High P will bring the molecules very close together. This causes more collisions and also allows the weak attractive forces to come into play. With low temperatures (close it gas’s liquefication point), the molecules do not have enough energy to continue on their path to avoid the attraction.

Dalton’s Law of Partial Pressure

Dalton’s Law Analogy In other words, Dalton’s Law of Partial Pressure would look like this:

Dalton’s Law Formula Formula: Pt = P1 + P2 + P3 + ... Pt = the total pressure P1 = Partial Pressure of gas ‘1”, P2 = Partial Pressure of gas “2,” etc. **ALL Units, MUST be the same for each pressure!!!**

What is Dalton’s Law? Dalton’s Law of Partial Pressures – the total pressure exerted on a container by several different gases is equal to the sum of the pressures exerted on the container by each gas (Partial pressure of a gas in a mixture of gases is the pressure which that gas would exert if it were the only gas present in the container) Dalton’s Law of Partial Pressure assumes each gas in the mixture is behaving like an ideal gas

Dalton’s Law Example What is the atmospheric pressure (outside pressure) shown in the picture? (739.6mmHg) What is the pressure of the gas in the tube? (722.3 mmHg) Is this gas the only gas that has been collected? (No) According to the picture, what other gases can be detected in the tube? (water vapor) What is the pressure of the water vapor in the tube? (17.3mmHg) What is the temperature at which the gases have been collected? (20oC) What is the total pressure inside the tube if the atmospheric pressure is 739.6 mmHg? (same as outside pressure)

Demo - Gas collection over water Demo Video’s Simulation (or demo - or show both): Collecting a Gas Over Water http://www.kentchemistry.com/links/GasLaws/dalton.htm Demo - Gas collection over water (or use Butane gas from a cigarette lighter) http://www.youtube.com/watch?v=UQRCUxrLU1c&edufilter=QMQXe2SzEucYVMXVK3i8VQ A more exciting Demo  – Oxygen Collection Over water http://www.youtube.com/watch?v=2Tj3Ir0brco&edufilter=QMQXe2SzEucYVMXVK3i8VQ

Gas Stoichiometry