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Gases Chapter 5
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Measurements on Gases Properties of gases –Gases uniformly fill any container –Gases are easily compressed –Gases mix completely with any other gas –Gases exert pressure on their surroundings Pressure = Force Area
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Measuring pressure –The barometer – measures atmospheric pressure Inventor – Evangelista Torricelli (1643) Measurements on Gases
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–The manometer – measures confined gas pressure Measurements on Gases
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–Units mm Hg (torr) –760 torr = standard pressure Kilopascal (kPa) –101.325 kPa = standard pressure Atmospheres –1 atmosphere (atm) = standard pressure STP = 1 atm = 760 torr = 760 mmHg = 101.325 kPa Measurements on Gases
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Example: Convert 0.985 atm to torr and to kPa. Measurements on Gases 0.985 atm760 torr 1 atm = 749 torr 0.985 atm101.325 kPa 1 atm = 98.3 kPa
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The Gas Laws of Boyle, Charles and Avogadro Boyle’s Law (Robert Boyle, 1627- 1691) –The product of pressure times volume is a constant, provided the temperature remains the same PV = k
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P is inversely related to V The graph of P versus V is hyperbolic Volume increases linearly as the pressure decreases –As you squeeze a zip lock bag filled with air (reducing the volume), the pressure increases making it difficult to keep squeezing The Gas Laws of Boyle, Charles and Avogadro
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–At constant temperature, Boyle’s law can be used to find a new volume or a new pressure P 1 V 1 = k = P 2 V 2 or P 1 V 1 = P 2 V 2 –Boyle’s law works best at low pressures –Gases that obey Boyle’s law are called Ideal gases The Gas Laws of Boyle, Charles and Avogadro
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Example: A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? P 1 = 1.3 atm V 1 = 27 L P 2 = 3.9 atm V 2 = ? The Gas Laws of Boyle, Charles and Avogadro (1.3atm)(27L) = (3.9atm)V 2 V 2 = 9.0 L
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The Gas Laws of Boyle, Charles and Avogadro Charles’ Law (Jacques Charles, 1746 – 1823) –The volume of a gas increase linearly with temperature provided the pressure remains constant V = bT V = b V 1 = b = V 2 T T 1 T 2 or V 1 = V 2 T 1 T 2
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The Gas Laws of Boyle, Charles and Avogadro Temperature must be measured in Kelvin ( K = °C + 273) –0 K is “absolute zero”
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Example: A gas at 30°C and 1.00 atm has a volume of 0.842L. What volume will the gas occupy at 60°C and 1.00 atm? V 1 = 0.842L T 1 = 30°C (+273 = 303K) V 2 = ? T 2 = 60°C (+273 = 333K) The Gas Laws of Boyle, Charles and Avogadro
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0.842L = V 2 303K 333K V 2 = 0.925L
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The Gas Laws of Boyle, Charles and Avogadro Avogadro’s Law (Amedeo Avogadro, 1811) –For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles, n V = an V/n = a V 1 = a = V 2 or V 1 = V 2 n 1 n 2
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The Gas Laws of Boyle, Charles and Avogadro Example: A 5.20L sample at 18°C and 2.00 atm contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? V 1 = 5.20L n 1 = 0.436 mol V 2 = ? n 2 = 1.27 mol + 0.436 mol = 1.706 mol
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The Gas Laws of Boyle, Charles and Avogadro 5.20 L = V 2 0.436 mol 1.706 mol x = 20.3L
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The Ideal Gas Law Derivation from existing laws V=k V=bT V=an P V = kba(Tn) P
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The Ideal Gas Law Constants k, b, a are combined into the universal gas constant (R), V = nRT or PV = nRT P
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The Ideal Gas Law R = PV using standard numbers will give you R nT STP P = 1 atm V = 22.4L n = 1 mol T = 273K Therefore if we solve for R, R = 0.0821 L atm/mol K
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The Ideal Gas Law Example: A sample containing 0.614 moles of a gas at 12°C occupies a volume of 12.9L. What pressure does the gas exert? P = ? V = 12.9L n = 0.614 mol R = 0.0821 Latm/mol K T = 12 + 273 = 285K
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The Ideal Gas Law (P)(12.9L) = (0.614mol)(0.0821 Latm/mol K)(285K) Hint: Rearrange and re-write (watch out for units in numerator and denominator) P =1.11 atm
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The Ideal Gas Law Solving for new volumes, temperature, or pressure (n remaining constant) Combined Law P 1 V 1 = nR = P 2 V 2 or P 1 V 1 = P 2 V 2 T 1 T 2
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The Ideal Gas Law Example: A sample of methane gas at 0.848 atm and 4.0°C occupies a volume of 7.0L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11.0°C? P 1 = 0.848 atmP 2 = 1.52 atm V 1 = 7.0LV 2 = ? T 1 = 4+273 = 277KT 2 = 11+273 = 284K
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The Ideal Gas Law (0.848atm)(7.0L) = (1.52atm)(V 2 ) 277K284K V 2 = 4.0L
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