Presentation on theme: "Kinetic Molecular Theory of Gases"— Presentation transcript:
1 Kinetic Molecular Theory of Gases A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.Gas molecules exert neither attractive nor repulsive forces on one another.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
2 Force Pressure = Area Units of Pressure 1 pascal (Pa) = 1 N/m2 BarometerPressure =Units of Pressure1 pascal (Pa) = 1 N/m21 atm = 760 mmHg = 760 torr1 atm = 101,325 Pa
5 Boyle’s Law P a 1/V Constant temperature Constant amount of gas P x V = constantBoyle's Law ( scuba style! ) ACSBR physics mini project – YouTubeP1 x V1 = P2 x V2
6 P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?P1 x V1 = P2 x V2P1 = 726 mmHgP2 = ?V1 = 946 mLV2 = 154 mLP1 x V1V2726 mmHg x 946 mL154 mL=P2 == 4460 mmHg
8 Variation of gas volume with temperature at constant pressure. Charles’ & Gay-Lussac’s LawV a TTemperature must bein KelvinV = constant x TV1/T1 = V2/T2T (K) = t (0C)
9 V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? V2 x T1 V1 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?V1/T1 = V2/T2V1 = 3.20 LV2 = 1.54 LT1 = KT2 = ?V2 x T1V11.54 L x K3.20 L=T2 == 192 Kcharles law demonstration - Google VideosGiant Koosh Ball in Liquid Nitrogen! - YouTube
10 Avogadro’s Law V a number of moles (n) V = constant x n V1/n1 = V2/n2 Constant temperatureConstant pressureV = constant x nV1/n1 = V2/n2
11 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?4NH3 + 5O NO + 6H2O1 mole NH mole NOAt constant T and P1 volume NH volume NO
12 Ideal Gas Law 1 Boyle’s law: V a (at constant n and T) P Charles’ law: V a T (at constant n and P)Avogadro’s law: V a n (at constant P and T)V anTPV = constant x = RnTPR is the gas constantPV = nRT
13 PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K) The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).Experiments show that at STP, 1 mole of an ideal gas occupies L.PV = nRTR =PVnT=(1 atm)(22.414L)(1 mol)( K)R = L • atm / (mol • K)
14 PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.6 L What is the volume (in liters) occupied by 49.8 g of HCl at STP?T = 0 0C = KP = 1 atmPV = nRTn = 49.8 g x1 mol HCl36.45 g HCl= 1.37 molV =nRTPV =1 atm1.37 mol x x KL•atmmol•KV = 30.6 L
15 PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 = Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?PV = nRTn, V and R are constantnRV=PT= constantP1 = 1.20 atmT1 = 291 KP2 = ?T2 = 358 KP1T1P2T2=P2 = P1 xT2T1= 1.20 atm x358 K291 K= 1.48 atmEgg in a Bottle.wmv - YouTube
16 d is the density of the gas in g/L Density (d) Calculationsm is the mass of the gas in gmV=PMRTd =M is the molar mass of the gasMolar Mass (M ) of a Gaseous SubstancedRTPM =d is the density of the gas in g/L
17 Deviations from Ideal Behavior of Gases Deviation from ideal behavior is large at high pressure and low temperatureAt lower pressures and high temperatures, the deviation from ideal behavior is typically small, and the ideal gas law can be used to predict behavior with little error.
18 Deviation from ideal behavior as a function of temperature As temperature is decreased below a critical value, the deviation from ideal gas behavior becomes severe, because the gas CONDENSES to become a LIQUID.
19 J. D. van der Waals corrected the ideal gas equation in a simple, but useful, way (a) In an ideal gas, molecules would travel in straight lines. (b) In a real gas, the paths would curve due to the attractions between molecules.
20 Mixtures of Gases Dalton's law of partial pressure states: the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
21 Dalton's Law of Partial Pressure PT = P1 + P2 + P3 + …….
22 Partial Pressure in terms of mole fraction ORXA Ptotal = PA(XA = mole fraction of A and PA = partial pressure of gas A)
23 Mole fraction What is the mole Fraction of Gas A in mixture I? What is the mole fraction of Gas B in mixture II?
24 XA = 3 = 0.2 3+4+5 PA = 2.5 atm PT = PA/ XA = 2.5/0.2 = 12.5 atm Example: If there are 3 moles of gas A, 4 moles of gas B and 5 moles of gas C in a mixture of gases and the pressure of A is found to be 2.5 atm, what is the total pressure of the sample of gases?XA = = 0.23+4+5PA = 2.5 atmPT = PA/ XA = 2.5/0.2 = 12.5 atm
25 PT = PO + PH O 2KClO3 (s) 2KCl (s) + 3O2 (g) Bottle full of oxygen gas and water vapor2KClO3 (s) KCl (s) + 3O2 (g)PT = PO + PH O2
26 Graham’s Law of Effusion Graham’s law states that the rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure:Graham’s Law of Effusiondiffusion of a gas - YouTube
27 Graham’s Law of Effusion The velocity of effusion is also inversely proportional to the molar masses:
28 Graham’s Law of Effusion But the time required for effusion to take is directly proportional to the molar masses:The density of the gas is also directly proportional to the molar masses:
29 Graham’s Law of Effusion Compare the rate of effusion for hydrogen and oxygen gases.