P T = P A + P B + P C..P i Boyle's Law The volume of a given quantity of gas varies inversely with the pressure of the gas, at a constant temperature.

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1 CHEM/TOX 336 Winter 2004 Lecture 2 Review Atmospheric Chemistry Gas Chemistry Review The Gaseous State: our atmosphere consists of gases Confined only by gravity force of gas on a unit area is due to the mass alone that force will as altitude, since gas density Atmospheric Pressure force / area = pressure exerted by the atmosphere = ATMOSPHERIC PRESSURE Barometer: P a = F a / A = F g / A = ρ g h Dalton's Law of Partial Pressures Each gas of a mixture exerts its own pressure independent of the other gases in the mixture. This is known as its PARTIAL PRESSURE. Total pressure (mixture) = Σ partial pressures

2 P T = P A + P B + P C..P i Boyle's Law The volume of a given quantity of gas varies inversely with the pressure of the gas, at a constant temperature. P 1 V 1 = P 2 V 2 = a constant Charles' Law The volume of a given quantity of gas at constant pressure varies directly with the absolute temperature. - gas volume increases as temperature increases but not proportionally to C - introducing K = C = absolute temperature Temperature expressed in K = T T 1 /T 2 = V 1 /V 2 Avogadro's Principle One mole of all gases at standard temperature and pressure occupy the same volume. = L V α n (P, T constant)

3 Combined Gas Law By combining each of these laws we get: V α nt / P PV / nt = R (a constant) PV = nrt (you've seen this before!) Combined Gas Law For one mole of gas under standard conditions: R = P o V o / T o R = 1 atm x L/mole = cm 3 atm K mole K = L atm / mole K = x 10 7 ergs/ mole K = kcal/ mole K Using PV = nrt for mixtures of gases P T = n T RT / V P T = RT/V Σ n i Mole fraction = mole fraction of one gaseous species in a mixture = n i / n T sum of mole fractions = 1

4 Problem #1 If 10 g H 2 and 64 g O 2 are contained in a 10L flask at 200 C, calculate the total pressure of the mixture. m=10 g and MW=2 g/mol n=10g / 2g/mol =5 mol of H 2 gas m= 64 g O 2 and MW = 32 g/mol, n= 2 mol O 2 gas n H 2 =5 mol, n O 2 =2 mol, V=10 L T=200 C = K P T = RT/V Σ n i = L atm/mole K x K x (5+2) mol/ 10 L = 27.2 atm Upon sparking, what is the final pressure in the container? 2H 2 + O 2 2H 2 O 5 mol 2 mol 4 mol H 2 O + 1 mol H 2 excess limiting reagent 1 T is K >> K which is the boiling pt of water, so product is water vapour not liquid So Σn = = 5 P T = L atm/mole K x K x 5 mol/ 10 L = 19.4 atm 1 Determining the limiting reagent: If hydrogen is the limiting reagent, then all of the hydrogen (5 mol) will be consumed, and 5/2x1=2.5 mol of oxygen will be required. But there are only 2 mol of oxygen, so oxygen is the limiting reagent. Performing the same comparison, starting with oxygen, 2 mol of oxygen will require 2/1x2 = 4 mol of hydrogen to react with, and there is excess hydrogen (5-4=1 mol). Therefore oxygen is the limiting reagent.

5 Problem #2 Seven grams of nitrogen, 16 g oxygen and 3.03 g hydrogen are introduced into an evacuated container of 80 L capacity at 50 C. What is the partial pressure of each gas in the mixture? 7g N 2, MW = 28 g/mol n = 7 g / 28 g/mol = 0.25 mol N 2 16g O 2, MW = 32 g/mol n = 16 g / 32 g/mol = 0.50 mol O g H 2, MW = 2 g/mol n = 3.03 g/ 2 g/mol = 1.5 mol H 2 n T = = 2.25 mol P T = n T RT/V = 2.25 x x ( )/80 = 0.64 atm P N = 0.25/2.25 x 0.64 = P O = 0.5/2.25 x 0.64 = 0.14 P H = 1.5/2.25 x 0.64 = = 0.64 atm total!

6 General Thermodynamics the relationship between the energy changes that occur in chemical or physical processes. how much "work" is done, energy added or released. First Law of Thermodynamics Energy is neither created nor destroyed. H = E + nrt (for constant PV) (page 9 of the textbook) E = change in internal energy n = #moles gaseous product - #moles gaseous reactants Example: vaporization of a liquid vapour has to do work against a constant P is a reversible reaction, P vapour = P atmosphere w = P V = P(V v - V l ) but V l is negligible compared to V v so (if vapour behaves as an ideal gas): V v = nrt/p and w = PV v = P (nrt/p) = nrt therefore work of vaporization depends ONLY on temperature

7 Atmospheric Chemistry Atmospheric composition Major components N 2 78% O 2 21% (Unique property of earth's atmosphere; biological processes) Ar ~1% Atmospheric composition "Medium" components - CO ppmv - CH ppbv - H 2 O variable (range ~ 0 to 0.4% (0.004 atm) Definitions: ppmv and ppbv ppmv = 10-6 P T - varies with altitude because P T varies Atmospheric composition Trace components H ppbv N 2 O 310 ppbv CO ~100 ppbv O 3 <30 ppbv (clean troposphere) 80 ppbv (Canadian guideline) some hydrocarbons: low ppbv NO x low pptv - 10 ppbv SO 2 <1 ppbv

8 Regions of the atmosphere and environmental problems Troposphere (lower atmosphere, to ~ 15 km) Stratosphere (~ km) Mesosphere (~ km) Thermosphere (>90 km) Most of the mass of the atmosphere (~ tonnes) is present in the troposphere (~ 99% below 30 km) beyond 150 km is essentially a vacuum, P T ~ 10-9 atm Tropospheric problems Greenhouse warming Photochemical smog Particulates Acid precipitation Stratospheric problems Ozone depletion Residence Times J = amount of substance in a "reservoir rate of inflow to, or outflow from a reservoir J = 1/ E k' where k' is the rate of change or the rate of the substance from the reservoir constant for loss Definitions: reservoir, sources, sinks

9 Residence Times Residence times of atmospheric gases - mostly biologically regulated O 2 ( t) / ( t / yr) = yr O 2 sinks (respiration, decay, combustion, weathering) balance the source (photosynthesis) N 2 ( t) / ( t / yr) = yr sinks: N fixation, lightening and combustion, industrial) sources: denitrification and decay CO 2 ( t) / ( t / yr) = 1.7 yr sinks: photosynthesis, ocean (not well understood) sources: respiration, combustion, decay pronounced 1 year cycle: CO 2 is removed faster than added in summer; reverse trend in winter rate of increase 1.5 ppmv / yr H 2 O ( t) / ( t/ yr) = yr = 10 days

10 Question #1 Consider all the sulfur compounds in the troposphere, under steady state conditions. If the mixing ratio is 1 ppb by mass what is the mass of sulfur compounds in the troposphere. (mass of troposphere = g) Answer: 1 ppb = 1x10-9 4x10 21 g x 1x10-9 = g If the total natural and anthropogenic production is equal to g per year, what would the residence time be assuming steady-state conditions? Answer: τ (tau) = g / g per year = 0.02 year, or about 1 week

11 Definition: Relative humidity P(H 2 O) expressed as a percent of the Saturation or Equilibrium Vapour Pressure for a given temperature eg. 80% relative humidity at 0 C = atm 80% relative humidity at 25 C = atm