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H 3 the Gas Laws

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H 3 the Gas Laws The Gas Laws Apparatus for Studying the Relationship Between Pressure and Volume of a Gas As P (h) increases V decreases Boyle’s Law P ∝ 1/V P x V = constant Constant temperature Constant amount of gas P1 x V1 = P2 x V2 Example 1 A helium balloon was compressed from 4.0L to 2.5L at a constant temperature. If the pressure of the gas in the 4.0L balloon is 210 kPA, what will the pressure be at 2.5L? Given: V1 = 4.0L V2 = 2.5L P1= 210 kPa P2 = ? P1V1 = P2V2 (210 kPa) × (4.0L) = P2 × (2.5L) P2 = 336 kPa ≈ 340 kPa Example 2 A sample of neon gas occupies 0.200L at 0.860 atm. What will be its volume at 29.2 kPa pressure? Given: P1V1 = P2V2 V1 = 0.200L V2 = ? (87.1 kPa) × (0.200L) = (29.2 kPa) × V2 P1= 0.860 atm P2 = 29.2 kPa V2 = 0.597L **Units must match for each variable (doesn’t matter which one is converted) 0.860 atm 101.3 kPa = 87.1 kPa 1 atm Day 2 • Comparing volume and temperature • Keep pressure constant Variation in Gas Volume with Temperature at Constant Pressure As T increases V increases Charles’s Law **Temperature must be in Kelvin K = 0C + 273 푉1 = 푉2 푇1 푇2 푉 푉 ∝ 푇 = 푐표푛푠푡푎푛푡 Constant pressure 푇 Constant amount of gas Example 1 A gas sample at 40.0°C occupies a volume of 2.32L. If the temperature is raised to 75.0°C, what will the volume be, assuming the pressure remains constant? Given: T1 = 40.0°C = 313K T2 = 75.0°C = 348K V1= 2.32L V2 = ? V1 V2 = T1 T2 2.32L V2 313 = 348 V2 = 2.58L Example 2 A gas sample at 55.0°C occupies a volume of 3.50L. At what new temperature in Celcius will the volume increase to 8.00L? Given: T1 = 55.0°C = 328.0K T2 = ? V1= 3.50L V2 = 8.00L V1 V2 = T1 T2 3.50L 8.00 = 328.0 T2 T2 = 750K-273=477°C Day 3 Gay-Lussac’s Law 푃 ∝ 푇 푃1 푃2 푃 = = 푐표푛푠푡푎푛푡 푇1 푇2 푇 Constant volume Constant amount of gas **Temperature must be in Kelvin K = 0C + 273 Example 1 The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank? Given: P1 = 3.20 atm P2 = ? T1= 22.0°C = 295.0K T2 = 60.0°C =333.0K P1 P2 = T1 T2 3.20 atm P2 295.0K = 333.0K P2 = 3.61 atm Example 2 A rigid container has a gas at constant volume at 665 torr pressure when the temperature is 22.0C. What will the pressure be if the temperature is raised to 44.6C? Given: P1 = 665 torr P2 = ? T1= 22.0°C = 295K T2 = 44.6°C =317.6K P1 P2 = T1 T2 665 torr P2 295K = 317.6K P2 = 716 torr STOP Avogadro’s Law and Combine Gas Law What is the relationship between number of moles and volume? What is the same about these gases? What is different about these gases? Avogadro’s Law V ∝ n (number of moles) 푉 Constant temperature = constant 푉1 푉2 Constant pressure 푛 = 푛1 푛2 1 mol of gas at STP = 22.4L Gas Stoichiometry When gases are involved, the coefficients in a balanced chemical equation represent not only molar amounts (mole ratios) but also relative volumes (volume ratios). Example 1 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what will be the new volume? Assume constant temperature and pressure. Given: V1 = 5.00 L V2 = ? n1= 0.965 mol n2 = 1.80 mol V1 V2 = n1 n2 5.00 L V = 2 0.965 mol 1.80 mol V2 = 9.33 L Example 2 Calculate the volume that 0.881 mol of gas at STP will occupy. 0.881 mol 22.4 L = 19.7 L 1 mol Example 3 How many grams of carbon dioxide gas are in a 0.75 L balloon at STP? 0.75 L CO2 1 mol CO2 44.01 g CO2 = 1.5 g CO2 22.4 L CO2 1 mol CO2 Example 4 What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C3H8)? Assume constant temperature and pressure. 퐶3퐻8 + 5 푂2 → 3 퐶푂2 + 4 퐻2푂 4.00 L C H 5 L O 3 8 2 = 20.0 L 1 L C3H8 Gas Law Summary Charles’s Law Gay-Lussac’s Law Combined Gas Law 퐵표푦푙푒′푠 푙푎푤: 푃 × 푉 = 푐표푛푠푡푎푛푡 푎푡 푐표푛푠푡푎푛푡 푛 푎푛푑 푇 푉 퐶ℎ푎푟푙푒′푠 푙푎푤: = 푐표푛푠푡푎푛푡 푎푡 푐표푛푠푡푎푛푡 푛 푎푛푑 푃 푇 푃 퐺푎푦 − 퐿푢푠푠푎푐′푠 푙푎푤: = 푐표푛푠푡푎푛푡 푎푡 푐표푛푠푡푎푛푡 푛 푎푛푑 푉 푇 푉 퐴푣표푎푑푟표′푠 푙푎푤: = 푐표푛푠푡푎푛푡 (푎푡 푐표푛푠푡푎푛푡 푃 푎푛푑 푇) 푛 Combine all 4 to make one master equation: 푃푉 푃1푉1 푃2푉2 푃 푉 푃 푉 = 푐표푛푠푡푎푛푡 = 1 1 = 2 2 푛1푇1 푛2푇2 푛푇 푇1 푇2 Example 1 A gas at 110 kPa and 30.0˚C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0˚C and the pressure increased to 440 kPa, what is the new volume? Given: 110 (2.00) 440 (V2) P1 = 110 kPa P2 = 440 kPa = T1 = 30.0˚C =303K T2 = 80.0˚C = 353K 303 353 V1 = 2.00L V2 = ? V2 = 0.583L ≈ 0.58L 푃 푉 푃 푉 1 1 = 2 2 푇1 푇2 Example 2 An unopened bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm and temperature of 5.00˚C. If the bottle is dropped into a lake and sinks to a depth at which the pressure and temperature changes to 1.52 atm and 2.90˚C, what will be the volume of gas in the bottle? 푃 푉 푃 푉 1.30 (46.0) 1.52 (V ) 1 1 = 2 2 = 2 Given: 푇1 푇2 278 275.9 V1 = 46.0 mL V2 = ? P1 = 1.30 atm P2 = 1.52 atm V2 = 39.0mL T1 = 5.00˚C = 278K T2 = 2.90˚C = 275.9K STOP Warm Up A sample of gas of unknown pressure occupies 3.2 L at a temperature of 302 K. The sample of gas is then tested under known condition and has a pressure of 22.6 kPa and occupies 1.2 L at 310 K. What was the original pressure of the gas? The Ideal Gas Law Ideal Gas Real Gas Follows some gas laws under Follows all gas laws under some conditions of temperature all conditions of temperature and pressure. and pressure. Does not conform to the Kinetic Molecular Theory. Real gases Follows all conditions of the have a volume and attractive and Kinetic Molecular Theory repulsive forces. (KMT) • A real gas differs from an ideal gas the most at low temperature, An ideal gas does not exist high pressure, and high in real life concentration. o Low T gas is moving less High P gas is more attractive High Conc has more collisions Ideal Gas law 푃푉 = 푐표푛푠푡푎푛푡 (R is the universal gas constant) 푛푇 PV = nRT PV (1 atm)(22.4L) R = = nT (1 mol)(273K) 0.0821퐿 푎푡푚 8.314퐿 푘푃푎 62.4퐿 푚푚퐻 62.4퐿 푡표푟푟 푅 = = = = 푚표푙 퐾 푚표푙 퐾 푚표푙 퐾 푚표푙 퐾 Example 1 Calculate the number of moles of gas contained in a 3.00L vessel at 298K with a pressure of 1.50 atm. PV=nRT V = 3.00L T = 298K (1.50)(3.00) = n (0.0821) (298) P = 1.50atm n= 0.184 mol n = ? R = 0.0821 L atm mol K Example 2 What will the pressure (in kPa) be when there are 0.400 mol of gas in a 5.00L container at 17.0˚C? V = 5.00L T = 17.0˚C = 290K PV=nRT P = ? kPa P(5.00) = (0.400)(8.314)(290) n = 0.400 mol R = 8.314 L kPa P= 193 kPa mol K STOP Stoichiometry Example 1 If 5.00L of nitrogen reacts completely with excess hydrogen at a constant pressure and temperature of 3.00 atm and 298K, how many grams of ammonia are produced. 푁2 + 3퐻2 → 2푁퐻3 PV=nRT 퐿 푎푡푚 (3.00 atm)(5.00 L) = (n)(0.0821 )(298 K) 푚표푙 푘 n= 0.613 mol N2 0.613 mol N2 2 mol NH3 17.04 g NH3 = 20.9 g NH3 1 mol N2 1 mol NH3 Example 2 How many grams of calcium carbonate will be needed to form 6.75L of carbon dioxide at a pressure of 2.00 atm and 298K? CaCO3(s) CO2(g) + CaO(s) PV=nRT 퐿 푎푡푚 (2.00 atm)(6.75 L) = (n)(0.0821 )(298 K) 푚표푙 푘 n= 0.552 mol CO2 0.552 mol CO2 1 mol CaCO3 100.09 g CaCO3 = 55.2 g CaCO3 1 mol CO2 1 mol CaCO3 Example 3 How many liters of chlorine will be needed to make 95.0 grams of C2H2Cl4 at 3.50 atm and 225K? 2Cl2(g) + C2H2(g) C2H2Cl4(l) 95.0 g C2H2Cl4 1 mol C2H2Cl4 2 mol Cl2 = 1.13mol Cl2 167.84 g C2H2Cl4 1 mol C2H2Cl4 PV=nRT 퐿 푎푡푚 (3.50 atm)(V) = (1.13 mol)(0.0821 )(225 K) 푚표푙 푘 V= 5.97 L Cl2 Graphic Organizer for PV=nRT and Stoichiometry Density (d) Calculations P d = M P m P RT n = = V RT so M V RT m is the mass of the gas in g n = m d = m M V M is the molar mass of the gas d is the density of the gas in g/L Molar Mass (M ) of a Gaseous Substance dRT M = P 47 Example 1 What is the molar mass of a pure gas that has a density of 1.40gL at STP? P d = M RT Example 2 Calculate the density a gas will have at STP if its molar mass is 39.9g/mol.
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