chapter5 - Chapter 5 Gases 5.1 Pressure 1 atm= 760 Torr...

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Chapter 5: Gases 5.1 Pressure 1 atm= 760 Torr =760 mmHg = 101.325 kPa Pressure= Force/Area= N/m 2 = Pascal 5.2 The Gas Laws of Boyle, Charles, and Avagadro Boyles Law 1 PV = k (k = constant) @ constant T& -there±s an inverse relationship between V²P -an ³ideal gas´ strictly obeys Boyle±s law P 1 V 1 = P 2 V 2 Boyle±s law II -PV is equal to 22.4 L atm at zero pressure for 1 mol of a gas Charle±s Law -volume of gas is directly proportional to temperature and is zero when temperature is 273.15&C (0& K) K = C&+ 273.15 V α T V=bT (b is proportionality constant) (F.Y.I>).000001 K has been reached V/T = b so V 1 /T 1 = V 2 /T 2 (T must be in Kelvin) Avogadro±s Law - equal volumes of gases contain the same number of particles at same T&²P V = an (n= # of moles, a= proportionality constant) V α n @ (constant T& ² P) V/n = a, so V 1 /n 1 = V 2 /n 2 5.3 The Ideal Gas Law Boyle±s law: V = k/P Charle±s law: V = bT Avogadro±s law: V = an V = R(Tn/P) R = universal gas constant R = 0.08206 L atm/K mol -this gas equation works the best for ideal gases at low temperatures -(not to yourself) explain how the Ideal Gas Law can be used to derive Boyle±s, Charle±s, and the combined gas laws. 5.4 Gas Stoichiometry V = nRT/P = (1.000 mol)(0.08206 L atm/K mol)(273.2K)/1.0000 atm = 22.42 L Molar volume of an ideal gas at STP = 22.42 L (0&C ² 1 atm) Molar Mass of a Gas -N = grams of gas/molar mass = mass/ molar mass = m/ molar mass -P = nRT/V = (m/molar mass)TR/V = m(RT)/V (molar mass) -m/V = d, so P = dRT / molar mass
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5.5 Dalton&s law of Partial Pressure
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This note was uploaded on 09/27/2010 for the course CHEM 210 taught by Professor Mcomber during the Spring '10 term at Skyline College.

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chapter5 - Chapter 5 Gases 5.1 Pressure 1 atm= 760 Torr...

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