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Chapter 5 - Gases
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1)fluidity –ability to flow –mainly empty space –random arrangement 2)low density –part. very spread out 1000x further apart than solid or liquid –example – oxygen density –D–O 2(gas) = 1.43 g/L or 0.00143 g/mL –D–O 2(liquid) = 1140 g/L or 1.14 g/mL –D-O 2(solid) = 1429 g/L or 1.429 g/mL
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3)highly compressible –no definite volume –mainly empty space 4)fill any container(no def. shape) –no attractive forces, independent part. –expand to occupy all spaces of container
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5)exert pressure –gas pressure – force created when gas particles collide with a surface press = force/area –force – measured in newtons »1 N = 1 kg▪m/s 2 pascal(Pa) = newton/m 2
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–standard measures of pressure 101,325 Pa or 101.325 kPa 760 mm Hg = 29.92 in. Hg 760 torr 14.7 psi 1 atm –measured w/ barometer discovered by Torricelli air pressure, on an avg. day at sea level, will support 760 mm(29.92 in.) of Hg
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aneroid barometer
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kinetic molecular theory(KMT) –molecular model that describes behavior of gases –postulates of KMT 1)gas part. in constant random motion –no intermolecular forces 2)volume of a gas particle is negligible 3)collisions are elastic, no attractive or repulsive forces 4)avg. KE is directly related to Kelvin temp.
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measurable properties of gases 1)pressure(P) kPa, mm Hg, atm 2)temp.(T) K, o C 3)volume(V) mL, L, cm 3 4)moles(n)
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Boyle’s law V P or V P –volume of a gas is inversely related to the pressure
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k = PVor V 1 P 1 = V 2 P 2
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sample problem: A balloon has a volume of 5.75 L and a pressure of 0.975 atm. If Aunt Edith sits on the balloon and decreases the volume to 1.75 L what is the pressure of the gas inside the balloon? k = PV k= 0.975 atm x 5.75 L k = 5.61 L x atm P = k V P = 5.61 L x atm 1.75 L P = 3.21 atm P 1 V 1 = P 2 V 2 V 2 V 2 P 2 = P 1 V 1 V2V2 P 2 = 0.975 atm x 5.75 L 1.75 L P 2 = 3.20 atm
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Gay-Lussac’s Law T P or T P –the pressure of a gas is directly related to the temperature of the gas
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k = Por P 1 P 2 TT 1 T 2 =
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Sample problem – A balloon has a pressure of 135.5 kPa at 25 o C, what is the pressure if the balloons temperature was at -15 o C? k = P T k = 135.5 kPa 25 o C k = 5.4 kPa/ o C k x T = P x T T P = k x T P = 5.4 kPa x -15 o C o C P = -81 kPa P = 0.455 kPa x 258 K K P = 117 kPa 298 K k = 0.455 kPa/K
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Sample problem – A spray paint can has a bursting pressure of 5.65 atm. If the can has a pressure of 1.30 atm at 25 o C, what temperature would the can burst? k = P T k = 1.30 atm 298 K k = 0.00436 atm/K k = P T T x k = P k k T = P k T = 5.65 atm 0.00436 atm/K T = 1.30 x 10 3 K = 1020 o C x T T x
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When a gas is heated, the volume of the gas __________ ??INCREASES
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Charles’ Law T V or T V – the volume of a gas is directly related to the temperature of the gas
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k = VorV 1 V 2 TT 1 T 2 Charles also proved the change in volume of a gas was 1/273 the original volume for every o C change –what would happen to 1.0 L of gas if the temperature decreased by 273 o C???? =
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Sample problem: If a weather balloon contains 45 L of hydrogen gas at 15 o C, what is the volume when it gets up in the atmosphere and the temperature is -45 o C? k = V T k = 45 L 288 K k = 0.16 L K k x T = V V = 0.16 L x 228 K K V = 36 L
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Boyle’s Law - k = PV Gay-Lussac’s Law - k = P/T Charles’ Law - k = V/T Combined Gas Law - k = _____ or P 1 V 1 P 2 V 2 T 1 T 2 = PV T T
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Sample problem – An aerosol can has propane gas as a propellant. What would the volume of propane be at STP if the volume of propane in the can is 135 mL at a temperature of 22 o C and pressure of 1.73 atm?
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Sample problem – When I was 6, I accidentally let go of my helium balloon from the circus. The volume of the balloon when I let go was 2.49 liters at 18 o C and 1.22 atm of pressure. When the balloon was struck by a jet passing over at higher altitude, what was its volume if the temperature was 24 o C less and the pressure was 674 torr?
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Avogadro’s law –equal volumes of gases, at the same T and P, contain the same # of particles (Avogadro’s principle) –V of a gas is directly related to the particles of gas(moles) n V or n V k = Vor V 1 V 2 nn 1 n 2 –basis for Avogadro’s #(6.023 x 10 23 ) –molar volume – the volume of 1 mol of any gas @ STP 22.41 L of any gas @ STP = 1 mol any gas =
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Ideal gas law -describes the behavior of a gas that always and perfectly follows the gas laws -ideal gas properties -no IM forces -particles have no volume -no condensation -real gases follow gas laws most of the time -except at high P -except at low T -except any time intermolecular forces affect gas part.
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Ideal Gas Law PV = nRT P = pressure V = volume n = moles T = temperature R = 0.08206 L atm/mol K R = 8.314 L kPa/mol K R = 62.36 L torr/mol K R = PV nT
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Sample problem – How many moles of carbon dioxide gas are there if 35.6 liters has T = 35 o C and 96.3 kPa?
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Sample problem – How many liters of helium gas are there if 15.5 grams has T = 15 o C and 105.3 kPa?
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Dalton’s law of partial pressure – the total P of a mixture of gases is the sum of the partial P of each gas P total = P 1 + P 2 + P 3 + P n Sample problem
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diffusion –the even spreading out of a substance from an area of high concentration to areas of low concentration effusion – passage of a gas thru a small opening or hole
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Graham’s law of diffusion –the rate of diffusion of a gas is inversely related to the square root of the gases molar mass –compare rates of diffusion of different gases ט A M B ט B M A =
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Sample problem – If helium gas and carbon dioxide gas effuse out of the same hole, how do their rates compare? rate He 44.01 g/mol 3.32 rate CO 2 4.00 g/mol 1.00 He effuses 3.32 times faster than CO 2 = =
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Gas Stoichiometry gas volumes correspond to mole ratios 3 H 2 (g) + N 2 (g) 2 NH 3 (g) coefficients represent –3 moles hydrogen react with 1 mole nitrogen to form 2 moles ammonia –3 liters hydrogen react with 1 liter nitrogen to form 2 liters ammonia(for gases only) –67.23 liters hydrogen @ STP react with 22.41 liters nitrogen @ STP to form 44.82 liters ammonia @ STP
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sample problem: How many liters of hydrogen gas will be produced at 280.0 K and 96.0 kPa if 1.74 moles of sodium react with water? answer: 21.1 L H 2
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sample problem: What volume of oxygen, collected at 25 o C and 101 kPa, can be prepared by decomposing 37.9 grams of potassium chlorate? answer: 11.4 L
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