Chapter 3: Gases PDF

Chapter (3): Gases Properties that distinguish gases from liquids and solids: 1. Gas volume changes greatly with pressure.  Chapter (3): Gases Properties that distinguish gases from liquids and solids: 2. Gas volume changes greatly with temperature. 3. Gases have relatively low viscosity a...

Chapter (3): Gases Properties that distinguish gases from liquids and solids: 1. Gas volume changes greatly with pressure.  Chapter (3): Gases Properties that distinguish gases from liquids and solids: 2. Gas volume changes greatly with temperature. 3. Gases have relatively low viscosity and flow rapidly. ( due to weak interaction between gas molecules)  Chapter (3): Gases Properties that distinguish gases from liquids and solids: 4. Most gases have relatively low densities under normal conditions. Small amount of gas can occupy a large volume  Chapter (3): Gases Properties that distinguish gases from liquids and solids: 5. Gases are miscible.( mix in any proportion and no phase boundaries) Chapter (3): Gases Gas Laws Gaseous samples can be described by 4 variables: Volume V (L) Amount n (number of moles ) Temperature T (K) Pressure P (atm) Chapter (3): Gases Gas Laws Boyle’s law Chapter (3): Gases Gas Laws Chapter (3): Gases Gas Laws Charles’s Law  Chapter (3): Gases Gas Laws Charles’s Law N.B. :The absolute or Kelvin temperature, TK = t°C + 273.15  Chapter (3): Gases Gas Laws Avogadro’s Law  The Mole:  Mass of one mole is the atomic mass or molecular mass expressed in grams. “one mole of any substance contains Avogadro’s number of particles” This number (6.022x1023) is called Avogadro’s number.  “ One mole of a substance consists of 6.022x1023 units of that substance.”  The same as” one dozen of something contains 12 units of that thing.”  Chapter (3): Gases Gas Laws Avogadro’s Law “One mole of every gas under the standard conditions of temperature and pressure (STP) occupies the same volume (22.414 litres)” Standard molar volume is 22.414L STP: “ It is a standard conditions at which T= 0 oC or (273.15 oK), and P=1 atm.”  Chapter (3): Gases Gas Laws Avogadro’s Law  Another form for Avogadro’s law: “Equal volumes of gases at the same temperature and pressure contain the same number of particles”. These balloons each hold 1.0 L of gas at 25C and 1 atm. Each balloon contains 0.041 mol of gas, or 2.5x 1022 molecules. Chapter (3): Gases Gas Laws Avogadro’s Law  Chapter (3): Gases Ideal Gas Law Boyl’s Law V α 1/P Charles Law V α T Avogadro’s Law V α n Combining the three Laws VαnT P V=R nT P Chapter (3): Gases Ideal Gas Law PV= nRT Ideal gas equation atm Liter mol Kilven R is general gas constant  Chapter (3): Gases Ideal Gas Law *Values of general ( universal ) gas constant (R): for 1 mole of ideal gas at STP; ( T = 0 oC = 273 oK, and P= 1 atm.): = 8.314 (N.m/mol.K) (J/mol.K) ⸪Pa=N/m2 ⸫Pa. m3= (N/m2). m3 = N.m = j Chapter (3): Gases Chapter (3): Gases Other laws derived from ideal Gas Laws Gas density Gas molar mass  Chapter (3): Gases Dalton’s law of partial pressure Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases: Chapter (3): Gases - Problems Chapter (3): Gases - Problems Chapter (3): Gases - Problems Chapter (3): Gases - Problems  Chapter (3): Gases - Problems Argon is an inert gas used in light bulbs to retard the vaporization of the filament. A certain light bulb containing argon at 1.20 atm and 18 C is heated to 85 C at constant volume. What is the final pressure of argon in the light bulb (in atm)? PV = nRT; (n, V and R are constant) (P1V1/n1T1) = (P2V2/n2T2) (P1/T1)=(P2/T2)  [P2 = (P1xT2/T1) = (1.2 atm x 358K/291K)] = 1.48 atm  Chapter (3): Gases - Problems A 250-mL flask, open to the atmosphere, contains 0.0110 mol of air at 0°C. On heating, part of the air escapes; how much remains in the flask at 100°C? (P1V1/n1T1) = (P2V2/n2T2) at 0°C (P, V and R are constant) (n1T1)=(n2T2)  [n2 = (n1xT1/T2) = (0.011mol x 273K/373K)] = 0.008 mol  Chapter (3): Gases - Problems Example #3: A mixture of gases contains 2.14 g of N2, 5.85 g of H2, and 4.18 g of NH3. If the total pressure of the mixture is 4.58 atm, what is the partial pressure of each component? Example #4: A mixture of 14.0 grams of hydrogen, 84.0 grams of nitrogen, and 2.00 moles of oxygen are placed in a flask. When the partial pressure of the oxygen is 78.00 mm of mercury, what is the total pressure in the flask? N=14, H=1, O=16 Chapter (3): Gases Deviation of real gases from ideal behavior n= 1 in all cases  Chapter (3): Gases Ideal gas behavior and Real gases At ordinary conditions: relatively high temperatures and low pressures—most gases exhibit nearly ideal behavior. While at extreme conditions: low temperature and high pressure—gases deviate from ideal behavior. Corrections to Ideal Gas Law !!  Chapter (3): Gases Ideal gas behavior and Real gases Corrections to Ideal Gas Law Intermolecular Interaction (Attractive forces between molecules are significant) Molar Volume decrease their pressure (Gas molecules has a significant value)  Chapter (3): Gases Ideal gas behavior and Real gases Corrections to Ideal Gas Law Van der Waals Equation for real Gases Intermolecular Interaction (Attractive forces between molecules are significant) Molar Volume (Gas molecules has a significant value) Preal = Pideal – P decreased Videal = Vreal - V Molecules Pideal = Preal + P decreased Videal = Vreal - nb Pideal = Preal + a (n2/V2)  Chapter (3): Gases Ideal gas behavior and Real gases – problem Determine the pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C. Carry out two calculations: in the first calculation, assume that methane behaves as an ideal gas; in the second calculation, assume that methane behaves as a real gas and obeys the van der Waals equation. [a= 2.303 liter2.atm/mol2 and b = 0.0431 L/mol] 1- PV=nRT P*0.2446=1*0.0821*298.15 P=100.1 atm [4sf] 2- (P+a*n^2/V^2)(V-nb) = nRT (P+2.303/0.2446^2)(0.2446-0.0431)=0.0821*298.15 P=82.9865=82.99 [4SF]