Presentation on theme: "Avogadros Law. What is Avogadros Law Avogadros Principle – equal volumes of gases at the same temperature and pressure contain equal numbers of particles."— Presentation transcript:
What is Avogadros Law Avogadros Principle – equal volumes of gases at the same temperature and pressure contain equal numbers of particles
Avogadros Formula Formula: V 1 = V 2 n 1 n 2 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 __________. increase decrease * n represents the amount of gas
Graph for Avogadros Law What Laws have we learned were also direct relationships, in which their graphs were similar to Avogadros?
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 0 o C 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!)
Avogadros Law: Example 1 Calculate the volume that 0.881 moles of oxygen gas at STP will occupy. (This can be solved using the Avogadros law formula or using the dimensional analysis method)
Avogadros Law: Example 1 Answer Formula: V 1 = V 2 n 1 n 2 What do we know? V 1 = ? L n 1 = 0.881 mol O 2 V 2 = 22.4 L n 2 = 1 mol O 2 Remember, at STP: 22.4 L/mol First way to solve, using the formula: V 1 = 22.4 L.881 mol O 2 1 mol O 2 After cross multiplying you end up with V 1 (1 mol O 2 ) = (22.4 L)(.881 mol O 2 ) 1 mol O 2 1 mol O 2 V 1 = ___________ ---------- ----------- ---------- 19.73 L Second way to solve, using the dimensional analysis:.881 mol O 2 | 22.4 L = | 1 mol O 2 --------- 19.73 L ---------
Avogadros Law: Example 2 How many grams of N 2 will be contained in a 2.0 L flask at STP?
Avogadros Law: Example 2 Answer Formula: V 1 = V 2 n 1 n 2 What do we know? V 1 = 2.0 L n 1 = ? g N 2 V 2 = 22.4 L n 2 = 1 mol N 2 Remember, at STP: 22.4 L/mol First, solve for the number of moles of N 2 : 2 L N 2 | 1 mol N 2 =.089 mol N 2 | 22.4 L N 2 Then, use dimensional analysis to convert from moles of N 2 to grams of N 2 :.089 mol N 2 | 28.014 g N 2 = 2.50 g N 2 | 1 mol N 2
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 pressures 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 30 o C 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 = 30 o C + 273 = 303K 0.0821 (L*atm)/(mol*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 (3.52)(.05) = n(.0821)(303) (.0821)(303) n = ______________ -------- ------- -------- -----.0071 mol
Ideal Gas Law Example 2 (finding grams) Avogadros 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 N 2 gas present in a 0.600 L sample kept at 1.00 atm and a temperature of 22.0 o C.
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 N 2 R = T = 22.0 o C + 273 = 295 K 0.0821 (L*atm)/(mol*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) n = ______________ ------- ------- -------- -----.025 mol N 2 But, they want the answer in grams, so we need to do dimensional analysis using molar mass:.025 mol N 2 | 28.014 g N 2 = __________ | 1 mol N 2 ---------.694 g N 2 ----------
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 doesnt 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.
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 gass liquefication point), the molecules do not have enough energy to continue on their path to avoid the attraction.
Daltons Law Analogy In other words, Daltons Law of Partial Pressure would look like this:
Daltons Law Formula Formula: P t = P 1 + P 2 + P 3 +... P t = the total pressure P 1 = Partial Pressure of gas 1, P 2 = Partial Pressure of gas 2, etc. **ALL Units, MUST be the same for each pressure!!!**
What is Daltons Law? Daltons 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) Daltons Law of Partial Pressure assumes each gas in the mixture is behaving like an ideal gas
Demo Videos 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=Q MQXe2SzEucYVMXVK3i8VQ