27 Avogadro Law ProblemA 3.0-liter gas sample contains 7.0 mol. How much gas will there be for the sample to be 2.3 litres? P&T does not change Given the unknown equation replace and solve P1V1 = P2V2 n1T n2T2 V1 = 3.0 L n1 = 7.0 mol V2 = 2.3 L n2 = ? mol 3.0 L = 2.3 L = 7.0 mol N2 mol 5.4 mol 19 Charles`s Law Learning CheckA 25 L balloon is released into the air on a hot afternoon (42ºC). The next morning, the balloon is recovered from the ground. It is a very cold morning and the balloon has decreased to 22 L. What is the temperature in º C? Given the unknown equation, replace and solve V1 = 25 L T1 = 42 oC +273= 315 K V2 = 22 L P1V1 = P2V2 T V1T2 T2 = ? ºC 25 L = 22 L = 315 K T2 277.2 K = 4.2 ºC. 17 Karls` Law The egg in the bottleHot air rises and the gases expand when heated. Charles conducted experiments to quantify the relationship between temperature and volume of a gas and showed that a diagram of the volume of a given gas sample relative to temperature (in ºC) at constant pressure is a straight line. Gay-Lussac showed that a V-vs. T diagram is a straight line that can be extrapolated at -273.15°C to zero volume, a theoretical state. The slope of the diagram from V to T varies for the same gas at different pressures, but the intersection remains constant at -273.15 °C. The representations of V vs T for different amounts of gas are straight lines with different slopes but the same intersection on the T-axis. The importance of the invariant intersection T in V-T diagrams was recognized by Thomson (Lord Kelvin), who postulated that -273.15°C was the lowest possible temperature that could theoretically be reached, and he called it absolute zero (0 K).
The results of Charles and Gay-Lussac can be formulated as follows: at constant pressure, the volume of a fixed quantity of a gas is directly proportional to its absolute temperature (in K). This relation is called Charlemagne`s law and is mathematically given by V = (constant) [T (in K)] or V T (in K, at constant P). Image courtesy of Christy Johannesson 18 Charles`s Law Problem Substitute and Solve T1 = 27.0oC +273= 300 KMrs. Rodriguez inflates a balloon for a party. It is located in an air-conditioned room at 27.0 ° C and the balloon has a volume of 4.0 l. A curious and intrepid chemistry teacher, she heats the balloon to a temperature of 57.0°C. What is the new volume of the ball if the pressure remains constant? Given the unknown equation, replace and solve T1 = 27.0 oC + 273 = 300 K V1 = 4.0 L T2 = 57.0 oC + 273 = 330 K P1V1 = P2V2 T V1T2 V2 = ? L 4.0 L = V2 = 300 K K 4.4 L 43 Gas stoichiometry perfect gas lawPV=nRT Are you looking for grams or moles of gas? Step 1: Start with the law of ideal gases to find moles of gas Step 2: 1 Switch to gram of gas gram/mol? 1) Use the law of perfect gases 2) Perform stoichiometric calculations courtesy of Christy Johannesson. 38 Welcome to Mole Island 1 mol = 6.02 x 1023 particles 42 Gas stoichiometry problemWhat volume of methane (CH4) is required in the following combustion reaction to produce 26 L of water vapour? CH4 (g) + 2O2(g) ͢ CO2(g) + 2H2O(g) x L ͢ L 1 mol ͢ mol 1 L ͢ L x L = 26 L 1L L x= 13 L Image courtesy of Christy Johannesson 28 Gay-Lussac law P1 = P2 T1 T2 Direct relation PT PT PV & n constant Direct relation PT PT 20 Boyle`s law P↓ V ↑ & P↑ V ↓ P 1/V (pressure is inversely proportional to volume) P1V1 = P2V2 T and n = constant P V 16 RESPIRATORY MECHANICS KHARLESS` LAW BREATHING MECHANICS The gas changes from high pressure to low pressure.
This is also responsible for all weather conditions. Timberlake, Chemistry 7th Edition, page 254 31 COMBINED LAW OF PERFECT GASES P1V1 = P2V2 n1T1 n2T2If P, V, n or T are constant, they cancel each other out of the equation. n normally constant (unless you add or remove gas), so T T2 12 Volume = how much space a gas occupiesUnits L, ml, cm3 1000 mL = 1 L 1 mL = 1 cm3 35 Law of perfect gases Replacement and solution of the problemA rigid steel cylinder with a volume of 20.0 L is filled with nitrogen gas to an ultimate pressure of atm at 27.0 oC. How many moles of gas does the cylinder contain? If the equation is unknown, V = 20.0 l p = atm T = 27.0oC + 273 = 300 K moles of nitrogen? PV=nRT R= atm L/K Mol n 0821 atm L/K Mol x 300 K = atm x 20.0L= 162 Mol 26 AVOGADRO LAW Vn Vn V n (direct) V1 = V2 n1 n2 T & P Constant VV n 36 Law of perfect gases Learning controlOne balloon contains 2.00 mol of nitrogen at a pressure of atm and a temperature of 37°C. What is the volume of the balloon? Given the unknown equation, replace and solve n = 2.00 mol P = atm T = 37.0oC + 273 = 310 K V in L? PV=nRT R= atm L/K Mol 0.980 atm x V= 2.00 mol x atm L/K Mol x 310 K = 51.9 L 14 Charlemagne`s law V1 = V2 T1 T2 T is always in K V TT V (temperature is directly proportional to volume) T ↑ V↑ & T↓ V↓ V1 = V2 T T2 T is always in K K = °C + 273 P and n = constant V T 21 Boyle`s law P1V1 = P2V2 (temperature is constant)When the volume of a gas decreases, its pressure increases as long as the temperature or amount of gas does not change.