Ch. 5 Gases
Loyola Marymount University
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This 3 page Class Notes was uploaded by Alyssa Weisblatt on Friday October 7, 2016. The Class Notes belongs to Chem 110 at Loyola Marymount University taught by Dr. Jennifer Casey in Fall 2016. Since its upload, it has received 9 views. For similar materials see General Chemistry I in Science at Loyola Marymount University.
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Date Created: 10/07/16
Ch. 5 Gases Gases have low density Common gases: noble gases, colored red on the periodic table, CO, CO₂, H₂O, NO, NO₂ Covalent/molecular compounds Pressure: P=F/A F= force A=area atm= atmosphere 760mmHg = 1 atm 1 atm= 101,325 Pa Pa= Pascal (mmHg= mm of Mercury) Ex: P=6.82 x 10³ mmHg, Pa? 6.82 x 10³ (1 atm/760 mmHg) (101,325 Pa/1 atm) = 9.09 x 10⁵ Pa The Simple Gas Laws: useful for undergoing a change 1. Boyle’s Law: P α (1/V) α= proportional to V increases as P decreases P= x/V x=PV P₁V₁= P₂V₂ 2. Charle’s Law: V α T T increases as V increases V=yT y= V/T (V₁ / T₁) = (V₂ / T₂) Ex: V₁ = 2.80 L , P₁ = 1.02 atm , V₂ = 3.25 L , P₂ = ? 1.02(2.75) = P(3.25) P₂ = 0.863 atm Ex: V₁ = 2.80 L , T₁ = ? , V₂ = 2.57 L , T₂ = 0.00° C (2.80/T) = (2.57/0) *can’t have 0 as denominator, so change to Kelvin by adding 273.15 to °C (2.80/T) = (2.57/273.15) T₁ = 298 K 3. Avogadro’s Law: V α n n = moles V=zn z=V/n (V₁ / n₁) = (V₂ / n₂) Ex: V₁ = 6.15 L , n₁ = .254 mol air , V₂ = 2.55 L , n₂ = ? (6.15/.254) = (2.55/n) n₂ = .105 mol air in that volume Mol that’s left? .254 - .105 = .149 mol air exhaled Ideal Gas Law: PV= RnT R= ideal gas constant = .08206 L atm/ mol K Ex: .845 mol, P= 1.37 atm, T= 315 K, V= ? V= [(.0845 mol)(.08206 L atm/ mol K)(315 K)] / 1.37 atm V= 15.9 L Ex: T= 125° C , P= 755 mmHg , density of N₂ = ? density = m/v = mol/V = n/V 755 mmHg (1 atm/ 760 mmHg) = .993 atm T= 125 + 273.15 = 398 K (n/V) = .993 atm/(.08206 L atm/ mol K)(398 K) = .0304 mol/L .0304 mol N₂/ L (28.014 g N₂/ 1 mol N₂) = .852 g/L Ex: 0.311 g, V= 0.225 L, T= 55°C (55 + 273= 328 K), P= 886 mmHg , molar mass= ? 886mmHg (1 atm/760 mmHg) = 1.17 atm (1.17 atm) (.225 L) / (.08206 L atm/ mol K)(328 K) = .00978 mol .311 g / .00978 mol = 31.8 g/mol Ex: CO (g) 2H₂ (g) → CH₃OH (g) V of H₂ when T= 355 K and P= 738 mmHg for H₂; 35.7 g CH₃OH 738 mmHg (1 atm/760 mmHg) = .971 atm 35.7 g CH₃OH (1 mol/32.042 g)( 2 mol H₂/ 1 mol CH₃OH) = 2.23 mol H₂ (2.23 mol)(.08206 L atm/ mol K)(355 K) / .971 atm = 66.9 L Ex: 2H₂ (g) + O₂ (g) → H₂O (g) wants g H₂O 1.24 L H₂ at STP STP= standard temperature and pressure: P= 1 atm and T= O°C (1 atm)(1.24 L) / (.08206 L atm/ mol K)(273 K) = .0554 mol H₂ .0554 mol H₂ (2 mol H₂O/ 2 mol H₂)(18.015 g H₂O/ 1 mol H₂O) = .997 g H₂O Gas Mixtures: Pₐ = nₐRT/V Pₑ = nₑRT/V P₊ = Pₐ + Pₑ P₊ = (nₐRT/V) + (nₑRT/V) = n(total) (RT/V) Dalton’s Law: Pₐ/P₊ = (nₐRT/V) / [(n total)R/V] = nₐ/ n total = χₐ χₐ= mol fraction Pₐ = χₐP₊ Ex: 1.00 L of He, Ar, and Ne P₊ = 662 mmHg T= 298 K P(He) = 341 mmHg P(Ne) = 112 mmHg g Ar=? P(Ar) = 662 mmHg – 341 mmHg + 112 mmHg = 209 mmHg 209 mmHg Ar (1 atm/760 mmHg) = .275 atm (.275 atm)(1.00 L) / (.08206 L atm/ mol K)(298 K) = .0112 mol Ar .0112 mol Ar (39.95 g Ar/1 mol Ar) = .449 g Ar Kinetic Molecular Thoery of Gases: assumptions 1. Size of particles is negligible (particles as points) 2. Kinetic energy is proportional to temperature (KE= (3/2)RT) 3. Collisions are elastic (no energy loss) 4. No interactions between particles (chemically) Boyle’s Law: P α (1/V) lower volume, pressure increase because collisions increase Charle’s Law: V α T increase T, KE and V increase (or pressure if V can’t expand) Avogadro’s Law: V α n increase n, V increase (or pressure if V can’t expand) Root mean square speed (rms speed) = U rms U rms = √3RT/MM MM = molar mass in kg Ex: O₂ at 25° C? R = 8.314 J/mol K J= Jules= kg*m²/s² 32 g O₂ (1 kg/ 1,000 g) = .03200 kg/mol U rms= √[3(8.314 J/mol K)(25 + 273 K)]/ .03200 kg/mol = 482 m/s Diffusion (a.k.a. drunken walk): process by which gas molecules spread out Effusion: process by which a gas escapes from a container through a small hole Rate A/ Rate B = √ MM(B)/ MM(A) Ex: unknown gas effuses at a rate that is 0.462 times that of N₂ (at same temp); calculate MM of unkown gas .462 (unknown) / 1 (N₂) = √28.02 g/mol / MM(A) = 131 g/mol Real Gases 1. At high pressure, size does matter Ex: in an airplane, tall people take up more space 2. At high pressure or low temperature, interactions matter [P + a(n/v)²][V – nb] = nRT Van der Waals Equation of State P V = nRT P = Accounts for attraction V= accounts for size Ex: 27° C, 10.0 mol, 1.50 L, 130 atm; is this gas ideal? Have all parts of PV = nRT, so nothing to solve for; need “a” and “b” for Waals equation, so we can pretend we don’t know P and see if we get the same value P= 10(.08206)(27 + 273) / 1.50 = 164 atm not ideal because atm is different
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