Chapter 13 Calculating Gases 1 Section 12.1 Pressure + Temperature conversions, Dalton’s + Graham’s Laws Section 13.1 The Gas Laws Section 13.2 The Ideal Gas Law Section 13.3 Gas Stoichiometry

2 Example problem: 17.6 psi How much is 17.6 psi expressed in kPa?Conversion factor kPa 14.7 psi known pressure Start with known pressure and cancel out old unit 17.6 psi101.3 kPa121 kPa 17.6 psi x kPa = 121 kPa 14.7 psi Converting Pressure Units kPa = 760 mm Hg (torr) = 1 atm = bar = 14.7 psi kPa = 1 1 atm = kPa = 1 1 atm = 1 etc. any combination 760 mm Hg 14.7 psi 760 mm Hg 14.7 psi because each refers to the same pressure

33 Amount of Gas and Gas Pressure Twice the number of gas particles doubles the pressure. When a gas is pumped into a closed rigid con- tainer, the pressure increases proportional to the number of particles added. Twice the number of gas particles doubles the pressure.

44 The ball that contains more air has higher pressure and greater mass.

5 Dalton’s Law of Partial Pressure At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases. P total = P 1 + P 2 + P 3 + … + +=

6 Graham's Law of Effusion The rate of effusion of a gas is inversely proportional to the square root of the gas's molar mass. Effusion Process of a gas escaping through a tiny hole Remember that KE = ½ mv 2 the smaller the mass, the faster the average particle travels at a given temperature

7 HCl and NH 3 vapors react, forming a white “smoke” above the bottles. a) Above which bottle is the reaction mostly visible? Explain. b) What is the relative effusion rate between HCl and NH 3 ? Graham's Law of Effusion ____________ Rate A =  molar mass B Rate B =  molar mass A ______ Rate NH 3 =  = 1.46 Rate HCl =  17.04

88 Boyle’s Law For a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. P 1 x V 1 = P 2 x V 2 P 1 = 100 kPa P 2 = 50 kPa P 3 = 200 kPa V 1 = 1.0L V 2 = 2.0L V 3 = 0.5 L P 1 x V 1 = P 2 x V 2 = P 3 x V 3 = 100 (Lx kPa) 1/2 the volume doubles the pressure!!!

9 Jacques Charles ( ) observed that at constant pressure the graph of gas volume versus temperature yields a straight line. The lines extended (extrapolated) to zero volume all intersect the temperature axis at the same point, C. Absolute Zero Lord Kelvin identified C, as absolute zero, the lowest temperature, at which the average kinetic energy of gas particles would be zero and stop moving (cannot be reached). Zero Kelvin = 0K = C

10 Kelvin Temperature (SI unit) of a substance is directly related to the average kinetic energy of the substance A substance has at 200K twice the average kinetic energy as it does at 100K All calculations must use Kelvin for temperature!!

11

12 Charles’ Law The volume of a gas is directly proportional to its Kelvin temperature if pressure and mass are kept constant. V 1 = 1L T 1 = 300 K V 2 = 2 L T 2 = 600 K 100 kPa V 1 V 2 T 1 T 2 = Twice the Kelvin Temperature will double the volume

13 The pressure of a fixed amount of gas varies directly with the Kelvin temperature when the volume remains constant. Gay-Lussac’s Law Twice the kinetic energy of gas particles (Kelvin temperature scale) doubles the pressure.

14 The Combined Gas Law is a combination of Boyle’s, Charles’ and Gay-Lussac laws, when pressure, temperature and volume all change at the same time (but amount of gas unchanged). P 1 x V 1 = P 2 x V 2 T 1 T 2 Boyle’s Charles’ Gay-Lussac

15 How does the pressure change when only one variable changes?

16 Avogadro’s Principle Equal volumes of gases at the same temperature and pressure contain equal number of particles. Molar Volume At STP one mole of any gas occupies 22.4 L (STP 0  C, 1 atm) because particles are tiny compared to the amount of empty space around them without intermolecular forces … but different masses

17 The Ideal Gas Law P x V = n x R x T with R = 8.31 L x kPa K x mol or R = L x atm mol x K R = ideal gas constant allows you to solve also for the number of moles n of gas of fixed V, T and P. Example: A helium balloon is filled with 565 mol He at 23.0  C to a pressure of 152 kPa. What is the volume of the balloon under those conditions? (all units must match units of R!!!)

18 The Ideal Gas Law derived from P 1 xV 1 = P 2 x V 2 T 1 x n 1 T 2 x n 2 if 1 mole at STP P 1 xV 1 = 101 kPa x 22.4 L = 8.31 kPa L T 1 x n K x 1 mol K mol P x V = R P x V = n x R x T T x n

Memorize only one formula… 19 Combined, Boyle’s, Charles’ and Gas-Lussac’s Laws Can be derived from the Ideal GasLaw : PxV 1 = nRT 1 and PxV 2 = nRT 2 If conditions in red remain unchanged: V 1 = nR and V 2 = nR T 1 P T 2 P  V 1 = V 2 Charles’ Law T 1 T 2 Apply: Derive Boyle’s, Gay-Lussac’s and the Combined Law the same way!

20 The Ideal Gas Law, Molar Mass (M) and Density (D) Molar mass and the ideal gas law Density and the ideal gas law

21 Ideal Gases versus Real Gases No gas is truly ideal, but most behave as ideal gases at a wide range of temperatures and pressures. Real gases deviate most from ideal gases at high pressures and low temperatures. Polar molecules have larger attractive forces between particles. Polar gases do not behave as ideal gases. Large nonpolar gas particles occupy more space and deviate more from ideal gases. Ideal gases follow the assumptions of the kinetic theory:  There are no intermolecular attractive or repulsive forces between particles or with their containers.  The particles are in constant random motion.  Collisions are perfectly elastic.

mol of O 2 at STP occupies how much volume? A.30.0 L B.22.4 L C.25.4 L D.67.2 L

23 The law of combining gas volumes states that in chemical reactions involving gases, the ratio of the gas volumes is a small whole number, which reflect the coefficients in the balanced equation. (use coefficients of balanced equation for volumes) 22 2 H 2 (g) + O 2 (g)  2 H 2 O (g) 2 liters Hydrogen gas + 1 liter oxygen gas produce only 2 liters of water vapor

24 Gas Stoichiometry (Volume-Volume Problems) Coefficients in a balanced equation represent volume ratios for gases. Stoichiometry and Volume-Mass Problems A balanced chemical equation allows you to find ratios for only moles and gas volumes, not for masses. All masses given must be converted to moles or volumes before being used as part of a ratio.

25 How many mol of hydrogen gas are required to react with 1.50 mol oxygen gas in the following reaction? 2H 2 (g) + O 2 (g) → 2H 2 O(g) A.1.00 B.2.00 C.3.00 D.4.00 How many liters of hydrogen gas are required to react with 3.25 liters of oxygen gas in the above reaction? A.2.00 B.3.25 C.4.00 D.6.50