Gases and the Kinetic Molecular Theory

Transcription

1 Gases and the Kinetic olecular Theory Importance in atmospheric phenomena, gas phase reactions, combustion engines, etc. 5.1 The hysical States of atter The condensed states liquid and solid The gaseous state Gas volume changes greatly with pressure Gas volume changes greatly with temperature Gases have low viscosity (flow easily) Gases have low density (~1000 times lower than liquids and solids) Gases are miscible in all proportions olecular model of the gaseous state olecules are in constant, rapid, random motion (explains the absence of definite shape, miscibility, low viscosity) olecules are widely separated (explains the absence of definite volume, low density, compressibility) 5. ressure Gas molecules collide with each other an the walls of the container molecules exert force on the walls Force ressure Area F A Atmospheric pressure ( atm ) caused by the gravitational pull of the Earth molecules exert force on all objects Barometers -measure the atmospheric pressure (Torricelli) no pressure above the mercury column (vacuum) the weight of the Hg column balances atm the height of the Hg column is proportional to atm g - acceleration of free fall (9.81 m/s ) d - density of Hg (13546 kg/m 3 ) - volume of Hg column h - height of Hg column (0.760 m at sea level) A - area of Hg column base F atm Hg F mhg g A mhg d dha F dhag dhag atm dhg A kg / m s atm anometers measure the pressure of gases in containers Close-end and open-end manometers

2 Units of ressure SI unit pascal (a) 1 a 1 N/m 1 kg/m s atm a (at sea level) Conventional units: atm 1 atm a (exactly) bar 1 bar a (exactly) torr 760 Torr 1 atm (exactly) mm Hg 1 mmhg 1 Torr lb/in 14.7 lb/in 1 atm Example: Convert Torr to atmospheres and kilopascals. 1 atm Torr atm 760 Torr 1 atm a Torr 760 Torr 1 atm a ka 5.3 The Gas Laws Relate the parameters of the gaseous state pressure, volume, temperature, and number of moles Boyle s Law At constant temperature (T) the pressure () of a fixed amount of gas is inversely proportional to its volume () At constant T and n: k k constant (depends on T and n) k constant Assume two states of a gas at constant T state 1 1, 1 state, 1 1 k k 1 1 Example: A.0 L sample of oxygen at 10 atm is transferred to a 15.0 L container at constant temperature. What is the new pressure? 1.0 L 1 10 atm 15.0 L? 10 atm.0 L L 1.3 atm Charles s Law At constant pressure () the volume () of a fixed amount of gas is proportional to its absolute temperature (T) At constant and n: k T k constant (depends on and n) k T T Charles s law helped devise the absolute temperature scale (Lord Kelvin)

3 Assume two states of a gas at constant state 1 T 1, 1 state T, 1 1 k k T T T T 1 1 Example: A balloon is filled with 5.0 L He gas at 15ºC. The temperature is changed to 35ºC. What is the new volume of the balloon? T 1 15ºC 88 K L T 35ºC 308 K? T L 308 K 5.3 L T 88 K 1 ariations of Charles s law Amontons s law At constant volume () the pressure () of a fixed amount of gas is proportional to its absolute temperature (T) At constant and n: k T k constant (depends on and n) T k T Assume two states of a gas at constant state 1 T 1, 1 state T, 1 1 k k T1 T T1 T Example: A cylinder containing N gas at 15ºC and 50 atm is moved to a new location at 35ºC. What is the new pressure in the cylinder? T 1 15ºC 88 K 1 50 atm T 35ºC 308 K? T 1 50 atm 308 K 53 atm T 88 K 1 Avogadro s Law At constant temperature (T) and pressure () the volume () of a gas is proportional to its amount (n) At constant T and : k n k constant (depends on T, ) k n n olar volume ( m ) the volume of one mole of a substance m /n Avogadro s principle At constant T and equal number of moles of different gases occupy equal volumes olar volumes of gases are very similar (/n constant) Assume two states of a gas at constant T and state 1 1, n 1 state, n k k n n n n

4 The Ideal Gas Law 1 k Boyle's Law k T k n Charles's Law Avogadro's Law Combination of the three laws: nt R R proportionality constant nrt ideal gas law R universal gas constant There is no need to memorize the mathematical expression for the individual gas laws since they can all be derived from the ideal gas law Ideal gas obeys the ideal gas law R is determined experimentally R L atm/mol K R J/mol K Assume two states of a gas state 1 1, 1, n 1, T 1 state,, n, T R R nt nt nt nt Note: T must always be in Kelvin Example: A 5.0 L gas sample at 1.0 atm and 10ºC is moved to a.0 L container and heated to 300ºC. What is the new pressure? atm L T 1 10ºC 83 K?.0 L T 300ºC 573 K n 1 n n T nt nt nt T 1.0 atm 5.0 L 573 K atm T 1 83 K.0 L Example: A 3.0 g sample of methane, CH 4, is placed in a.0 L container at ºC. What is the pressure in the container? nrt.0 L T ºC 95 K moles of CH 4 (n): 1 mol CH 4 n 3.0 g CH mol CH g CH4 Latm 0.19 mol K nrt mol K.3 atm.0 L Standard conditions Standard temperature and pressure (ST) 1 atm; T 0ºC K The molar volume of the ideal gas at ST m nrt / RT n n Latm K RT mol K L m.41 1 atm mol

5 5.4 Applications of the Ideal Gas Law The molar mass and density of gasses mass m density d volume mass m molar mass moles n m n nrt n RT m n drt d d RT RT The density of a gas is proportional to its molar mass and pressure and inversely proportional to its temperature Example: Calculate the density of O at ST g/mol 1 atm T 0ºC K (ST) g atm mol g d 1.48 RT Latm L K mol K Finding the molar mass of a volatile liquid Weigh a flask with a known volume Fill the flask with the vapors of the volatile liquid at a known temperature and pressure Cool the flask and let the vapors condense Reweigh the flask to get the mass of the vapors Example: Calculate the molar mass of a liquid if g of its vapors occupy.50 L at 00ºC and 45.0 Torr. d m/ g/.50 L 0.38 g/l T 00ºC 473 K 45.0 Torr [1 atm/760 Torr] atm g L atm K drt L mol K g atm mol ixtures of Gasses ixtures are treated just like pure gases same gas laws apply artial pressure of a gas in a mixture the pressure the gas would exert if it occupied the container alone Dalton s law of partial pressures the total pressure () of a gaseous mixture is the sum of the partial pressures ( i ) of its components A + B + or Σ i ole fraction (χ i ) of a gas in a mixture a fraction of the total number of moles that belongs to that gas ni ni χi ni n χi 1 n n i The sum of all mol fractions is equal to one The ideal gas law can be written for each gas in a mixture in terms of partial pressures i n i RT nrt

6 i nirt i ni χ i nrt n i χ i The partial pressure of a gas is proportional to its mol fraction Example: Calculate the total pressure and the partial pressures of He and Ne in a.0 L mixture containing 1.0 g He and.0 g Ne at 0ºC. grams of He and Ne moles of He and Ne mole fractions of He and Ne total pressure partial pressures 1 mol He 1.0 g He 0.5 mol He 4.00 g He 1 mol Ne.0 g Ne mol Ne 0.18 g Ne χ χ He Ne nhe n + n He Ne nne n + n He Ne n mol nrt nrt nrt Latm 0.35 mol K mol K.0 L 4. atm He Ne χ atm 3.0 atm He χ atm 1. atm Ne Collecting a gas over water total gas + water total atm water (vapor pressure of water) given in tables Example: A.5 L sample of O gas was collected over water at 6 C and 745 torr atmospheric pressure. What is the mass of O in the sample? (The vapor pressure of water at 6 C is 5 torr.) oxygen total - water torr T K 1 atm 70 torr.5 L O 760 torr no mol RT L atm K mol K 3.0 g O mol O 3.1 g O 1 mol O

7 Stoichiometry and the Ideal Gas Law The volume ratios of gases in reactions are the same as their mole ratios (follows from Avogadro s principle) 3H (g) + N (g) NH 3 (g) The ideal gas law can be used to convert between the number of moles of gaseous reactants (or products) and their volumes at certain T and 3 mol H react with 1 mol N 3 L H react with 1 L N Example: How many liters of N are needed to react completely with 5.0 L H? 5.0 L H [1 L N / 3 L H ] 1.7 L N Example: Calculate the volume of CO produced by the decomposition of.0 g CaCO 3 at 5ºC and 1.0 atm. CaCO 3 (s) CaO(s) + CO (g) 1 mol CaCO 1 mol CO.0 g CaCO 0.00 mol CO g CaCO 3 1 mol CaCO 3 L atm 0.00 mol K nrt mol K 1.0 atm 0.49 L Example: Calculate the mass of NaN 3 needed to produce 10 L of N in an air bag at 5ºC and 1.0 atm by the reaction: 6NaN 3 (s) + Fe O 3 (s) 3Na O (s) + 4Fe(s) + 9N (g) T 98 K 1 atm 10 L n? 1 atm 10 L n 0.41 mol RT Latm K mol K 6 mol NaN g NaN mol N 9 mol N 1 mol NaN3 18 g NaN The Kinetic olecular Theory ostulates of the Kinetic olecular Theory The gas particles are negligibly small (their volume can be neglected) The gas particles are in constant, random motion and move in straight lines until they collide The gas particles do not interact except during collisions. The collisions are elastic so there is no loss of energy due to friction The average kinetic energy of gas particles, E k, is proportional to the absolute temperature, T A molecular view of the gas laws Boyle s law ( 1/)

8 A molecular view of the gas laws Charles s law ( T) A molecular view of the gas laws Avogadro s law ( n) Average kinetic energy of the gas particles Example: Calculate the root-mean-square 1 Ek T Ek mu speed of N at 5ºC. T 5ºC 98 K 8.0 g/mol kg/mol u T u T R J/mol K m mass of particles u average square speed J K 3RT J Root-mean-square speed of the gas particles u mol K rms 515 kg kg urm s u urm s T mol 3 RT kg m /s m m u rm s kg s s The axwell Distribution of Speeds Gives the fraction of particles moving at each particle speed Gas molecules travel with a wide range of speeds with a bell-shaped distribution The most probable speed, the average speed and the root-mean-square speed are very close in magnitude The range of speeds widens and u rms increases with increasing the temperature

9 The meaning of temperature RT m3rt Ek mu murms m mn A E k 3RT 3RT and urms N A The range of speeds widens and u rms increases with decreasing the molar mass of the gas Lighter gases have higher molecular speeds u rms depends on T and E k depends only on T T is a measure of the average kinetic energy of the molecular motion Diffusion and Effusion Diffusion gradual dispersal of one substance through another gases diffuse from places with high to places with low concentration Effusion escape of a substance through a small hole into vacuum effusion through porous materials, pin holes, cracks, etc. Graham s Law the effusion rate (ER) of a gas is inversely proportional to the square root of its molar mass (same relation is valid in general for the diffusion rate) 1 ER Can be explained with u rms 3RT/ The time of effusion (t eff ) is inversely proportional to ER 1 t eff teff ER For two gases, A and B: ER A t ER B t B ( ) ( ) ( A) ( ) B eff A A eff B Example: If it takes a certain amount of H 15 s to effuse through a small hole, how long does it take for the same amount of O? t ( O ) ( ) ( O ) O ( ) ( ) eff O teff O teff H eff H H eff 3.00 g/mol 15 s 60 s.0 g/mol t H t 5.6 Real Gases Real gases deviate from ideal behavior Compression factor (Z) Z /nrt For ideal gases: nrt Z /nrt 1 A plot of Z versus gives a straight line for ideal gases, but not for real gases

10 Negative deviations important at moderately high pressures (/nrt < 1) Due to attractive forces between the molecules The molecules attract each other and impact the walls with a weaker force ( and Z decrease) ositive deviations important at very high pressures (/nrt > 1) Due to the actual volume of the molecules The physical volume of the molecules reduces the free volume in the container, but we still use the volume of the entire container,, which is larger than the free volume (Z increases) an der Waals equation: an + ( nb) nrt a, b - van der Waals constants (zero for ideal gases) an / - pressure correction (a depends on the attractive forces between molecules) nb - volume correction (b is a measure for the actual volume of the gas molecules) Real gases approach ideal behavior at low pressures and high temperatures (away from conditions of condensation)