Presentation on theme: "6 Gases1 A gas expand to occupy the entire volume it is placed in. Molecules in a gas translate freely between collisions, and they all behave alike regardless."— Presentation transcript:
6 Gases1 A gas expand to occupy the entire volume it is placed in. Molecules in a gas translate freely between collisions, and they all behave alike regardless of their type. What are some of the properties of gases? Pressure, temperature, heat capacity, volume, density, molar volume, color, average speed of molecules, solubility (in water or other liquid), absorption, compressibility, gas-liquid equilibrium, composition, identity (compound or element), chemical properties, combustibility, stability Which of these properties are intensive properties, and which are extensive properties?
6 Gases2 Announcement Appointments for Winter '04 enrolment are now posted on your QUEST account – Each student has only a three day appointment time. If missed, you will have to wait until all students have completed their appointments. Open enrolment begins November 3rd, but courses could be full by then CHECK YOUR QUEST ACCOUNT FOR YOUR APPOINTMENT as soon as possible.
4 Pressure Pressure: force per area (1 Pascal = 1 N m –2 ) Liquid pressure (explain these formulation in terms of physics) F W g * m g * V * d g * h * A * d P = --- = ---- = = = = g * h * d A A A A A These equivalences are useful for unit conversions: 1 atm = kPa = 76 cm Hg = 760 mm Hg (torr in honor of Torricelli) = b = mb (bar & m bar) = pounds / sq. inch
6 Gases5 Torricellis Barometer Barometric pressure Explain Torricellis work ( ) P = g h d kg 1 m 3 Hg 1 atm = 0.76 m Hg N 1 kg = N m –2 (Pascal) = k Pa
6 Gases6 Torricelli Mercury Barometer Evangelista Torricelli invented the Torricelli Mercury Barometer in He used a long glass tube, closed at the upper end, open at the lower and filled with mercury.
6 Gases7 Pump Water from a Well The specific gravity of mercury is If water is used for a barometer, what is the height of water corresponding to 1.00 atm? Solution: 76 cm Hg g 1 cm 3 Hg 1 cm 3 H 2 O 1 g = cm H 2 O = m H 2 O Explain water pump and depth of well 10.33mWater10.33mWater No water!! Water water everywhere! What about a diver under water? Be sure to get that during lecture.
6 Gases8 Pressure of Skater A skater weighing 70 kg stands on one foot and the contact between the blade and ice is 1 cm 2. What is the pressure in torr sustained by the ice? Solution: 70 kg 1 cm cm 2 1 m mm 1 m 1 m 3 Hg kg = mm Hg = torr = 67.8 atm
6 Gases9 Avogadros Law The ABCD of gas laws are Avogadros, Boyles, Charles & Daltons laws of gases. Avogadros hypothesis: proposed in 1811 at the same temperature T and pressure P, equal volumes contain equal amounts of gases in moles n. at the same temperature T and pressure P, the volume V of a gas is proportional to the number of molecules or number of moles n. V P,T = k n(k, a constant, 22.4 L at STP) Explain Avogadros hypothesis and implications Avogadros scientific contributions to science will be given. Explain Avogadros law in your language
6 Gases10 Boyles Law For a certain amount (constant n ) of gas at constant temperature T, the volume V times the pressure P is a constant. P V n, T = constant = P 1 V 1 = P 2 V 2 How do you graph the Boyles law? State the law in another way. What curves are P-V plots? P V Robert Boyle, ( ) Mathematical aspects of PV product will be discussed T2 n1T2 n1 T1 n1T1 n1 P 1 V 1 P 2 V 2
6 Gases11 Charles Law (law of Charles-Gay-Lussac) For a certain amount of gas at constant pressure, its volume, V, is directly proportional to its temperature T in Kelvin. V n, P = b T ( b is a constant) o r P n, V = b T ( b is a constant) State the Charles law in another way V 1 V = = b, V 1 T 2 = V 2 T 1 T 1 T 2 T V n, P or P n, V Charles, Jacques-Alexandre-César ( , top) first to ascend in a H 2 -balloon, developed this law in Later Joseph Louis Gay-Lussac (lower) published a paper citing Charless law.
6 Gases12 General Gas Equation Combining ABC gas laws, we have P 1 V 1 T 1 P 2 V 2 T 2 = Subscripts 1 and 2 refer to different conditions for the same quantities of gases ( n ). Experiments show that one mole of gas at STP occupies 22.4 L. = n R
6 Gases13 The Ideal Gas Equation The ABC laws of gases can be combined into one and the result is an ideal gas equation. A+B+C ideal P V = n R T 1 atm * L R = = L atm mol –1 K –1 1 mol K kPa * L R = = L kPa mol –1 K –1 1 mol K Please confirm that 1 kPa L = 1 J(1 L = 1e-3 m 3 )
6 Gases14 Daltons law The Partial pressure P i is the pressure of a component in a mixture as if others dont exist in that system – due to the fact all gases behave as if they are independent of each other. P i = n i Daltons law: The total pressure P total, of a mixture of gases is the sum of the partial pressures of the components. P total = P 1 + P 2 + … + P n = ( n 1 + n 2 + … + n n ) ( V is common to all) R T V correction
6 Gases15 Application of the Ideal gas Law Parameters of the ideal gas law: P, V, T, n and a constant R ideal P V = n R T Ideal gas law A+B+C At constant n and T, P 1 V 1 = P 2 V 2 Boyles law At constant n and V P = ( n R / V ) TP 1 / T 1 = P 2 / T 2 Charles law At constant n and P V = ( n R / P ) TV 1 / T 1 = V 2 / T 2 ditto At constant P and T V = ( R T / P ) nV = k nAvogardros law
6 Gases16 Gas Densities Evaluate the density of O 2 (molar mass M = 32.0) at 300 K and 2.34 atm. Hint : Density d of a gas with mass m (= n M) and volume V m n M n d d = = = VV VM Thus n R Td R T P = = V M d = P M / R T Find relationship between density d and M. Manipulate symbols to get a useful formula before you calculate the quantities. Include and work out the units plse d = 2.34 *32.0 / ( *300) = ___
6 Gases17 Reactions Involving Gases How much NaN 3 is required to produce 12.0 L N 2 gas at 302 K and 1.23 atm for the air bag in your designed Autotie? Solution: Equations: 2 NaN 3 = 2 Na + 3 N 2 ; n = V ( P / R T ) 12.0 L = ___ N atm mol K L atm * 302 K (23+3*14) g NaN 3 1 mol NaN 3 2 mol NaN 3 3 mol N 2 Work out N 2 volume for 51 g NaN 3 used under the same condition.
6 Gases18 - reaction involving gases NO is made from oxidizing NH 3 at 1123 K with platinum as a catalyst. How many liter of O 2 at 300 K and 1 atm is required for each liter of NO measured also at 300 K and 1 atm? Solution: The reaction is: 4 NH O 2 4 NO + 6 H 2 O Since n = ( P/RT ) V molar relationships are the same as volumetric relationship, providing T and P are the same. 1 L NO 5 mol O 2 4 mol NO = 1.25 L O 2 Complicated problem may have a simple solution, Further oxidation of NO leads to NO 2, which is used to make HNO 3, a valuable commodity. How much H2O is produced?
6 Gases19 A Mixture of Gases What is the pressure exerted by the gas when 1.0 g of H 2, 2.0 g of O 2, and 0.1 g of CO 2 are all confined in a 10.0 L cylinder at 321 K? Solution: Daltons law : n total = i n i (count molecules non-discriminately) P = n R T / V; n = 1.0 g H g O g CO J * 321 K 10 L mol K 1 mol 2 g H 2 1 mol 32 g O 2 1 mol 44 g CO 2 P = mol = 126 kPa (note 1 J = 1 kPa L) = mol Calculate partial pressures of each gas plse
6 Gases20 Vapor pressure The (saturated) vapor pressure is the partial pressure that is at equilibrium with another phase. Vapor pressure of ice Vapor pressure of water Explain Structure of water molecule Hydrogen bonding Structure of water Structure of ice Vapor pressure of ice and water Relative and absolute humidity
6 Gases21 Collecting Gas Over Water When g of a sample containing Ag 2 O is heated, 40.6 mL of O 2 is collected over water at 296 K, and the atmosphere is 751 mmHg. Vapor pressure of water at 296 K is 21.1 mmHg. What is the percentage of Ag 2 O in the sample? Solution: Ag 2 O = 2 Ag O 2 n = P V / R T;R = L atm mol -1 K -1 P = (751 – 21.1) mmHg / 760 mmHg = atm ( P O2 = P total – P water ) V = 40.6 mL = L Mass of Ag 2 O = atm* L O L atm mol -1 K -1 *296 K 1 mol Ag 2 O 0.5 mol O g Ag 2 O 1 mol Ag 2 O = g Ag 2 O Percentage of Ag 2 O = g Ag 2 O / g = Ag 2 O = 60.3 % Ag 2 O
6 Gases22 Assumptions of Kinetic-molecular Theory 1. Gas is composed of tiny, discrete particles (molecules or atoms). 2. Particles are small and far apart in comparison to their own size. 3. Ideal gas particles are dimensionless points occupying zero volume. 4. Particles are in rapid, random, constant straight line motion. 5. There is no attractive force between gas molecules and between molecules and the sides of the container. 6. Molecules collide with one another and the sides of the container. 7. Energy is conserved but transferred in these collisions. 8. Energy is distributed among the molecules in a particular fashion known as the Maxwell-Boltzmann Distribution.
6 Gases23 Kinetic-molecular Theory of Gases For N gas molecules, molecule mass = m, molecular mass = M speed = u, average speed = u, Avogadros number N volume = V, temperature = T, Pressure = P, Kinetic energy = ½ m u 2 Collision frequency u N / V Pressure ( m v ) ( u ) ( N / V ) ( N / V ) m u 2 = 1 / 3 ( N / V ) m u 2 (1/3 due to 3-Dimensional space) P V = 1 / 3 N m u 2 = n R T (Meaning of T) Thus, 3 R T = N A m u 2 correction Furthermore, u 2 = 3 R T / M (Temperature and speed) Explain the significances of and apply these formulas for sciences
6 Gases24 Molecular Speeds Distributions of speed of various gases will be demonstrated using a simulation program, and for each gas, three speeds are indicated. In the following: m = mass of a molecule, M = molar mass, R = gas constant, and k = R / N avogadro = Boltzmann constant. The most probable speed u mp = (2 k T / m ) 1/2 = (2 R T / M ) 1/2 The mean or average speed u ave = ( 8 k T /π m ) 1/2 = (8 RT / π M ) 1/2 The root-mean-square speed u rms = ( 3 k T / m ) 1/2 = (3 RT / M ) 1/2
6 Gases25 Diffusion of Gases All gases move together because they are subjected to the same pressure head. Different gases diffuse at different rates Diffusion contributes to net movement of O 2 and CO 2 across the alveolar-capillary membrane (breathe). Constant molecular motion. Diffusion from higher to lower concentration regions. Since u ave = ( 8 k T /π m ) 1/2 = (8 RT / π M ) 1/2, (slide 24) 1 Diffusion rate Grahams law M Discuss diffusion rates of H 2, He, CH 4, N 2, O 2, CO 2, 235 UF 6, 238 UF 6 at lecture
6 Gases26 Diffusion problems Problems are usually to compare diffusion or effusion rates in the following terms: u rms = ( 3 k T / m ) 1/2 = u ave = ( 8 k T /π m ) 1/2 = diffusion rate effusion time for same amount distance traveled by molecules in certain period amount of gas effused 3 RT M 1 M 8 RT π M M
6 Gases27 Comparing Effusion Rates If 1e20 N 2 molecules effuse from an orifice in 1.0 min, how many H 2 molecules will effuse the same orifice at the same condition ( T P )? How many minutes will be required for the same number of H2 molecules to effuse? H 2 effusion rate = M N2 / M H2 (N 2 effusion rate ) = ( 28 / 2 )1e20 molecules/min) = __________ figure out value and units Time for 1e20 H 2 molecules to effuse = ( 2 / 28 ) 1.0 min = __________ What is the effusion rate ratio of N 2 and H 2 or any two gases?
6 Gases28 Effusion and Life O2O2 CO 2 T P1P1 P2P2 A Breathing chemistry is complicated, and we can only scratches the surface!
6 Gases29 The van der Waals equation for real gases Gases tend to behave ideally at high T and low P. Required T and P for ideality depends on gas properties and molar mass, and van der Waals proposed correction terms for the ideal gas equation for real gases. ( P + ) ( V – n b ) = n R T n 2 a V 2 Correction for intermolecular forces Correction for volume of molecules Explain the meaning of vdW eqn where a and b are gas-dependant constants. Gas a L 2 atm mol -2 b L mol -1 He Ne N O CO Cl Note units for a and b
6 Gases30 Application of van der Waals Equation What is the pressure of Cl 2 at 300 K occupying 20.0 L according to vdW and ideal gas laws? Solution : Look up data for Cl 2, a = 6.49 L 2 atm mol -2, b = L mol -1, n = 1 mol, R = L atm K -1 mol -1 P =- n R T V – n b n 2 a V 2 1 mol * L atm mol -1 K -1 *300 K 20.0 L – 1 mol mol -1 L 1 2 mol 2 * 6.49 L 2 atm mol L 2 =-=- Please calculate the results, and P from the ideal gas law. = _____________________
6 Gases31 Problems related to van der Waals Equation What is the molar volume of Cl 2 at 300 K and 1 atm according to vdW? Solution: Look up data for Cl 2, a = 6.49 L 2 atm mol -2, b = L mol -1, n = 1 mol, R = L atm K -1 mol -1 n R T P + n 2 a / V 2 V = + n b = / ( ) = L Calculate P for a definite volume is easier, and using the successive method for V is interesting, but its a challenge. try V = 22 L = 24 L try V = L and calculate V = / ( ) = L Find out how engineers deal with real gases. = ?
6 Gases32 Volume of vdW eqn What is the volume of occupied by 132 g CO 2 gas at 12.5 atm and 300 K? Solution : Solve volume of van der Waals equation for V R T + b P P n 2 a P n 2 a b P V 3 – n () V 2 + () V – () = 0Derive please a = 3.59 L 2 atm mol -2, b = L mol -1, n = 1 mol, R = L atm K -1 mol -1 This is similar to problem 6 –106, a question for practicing the successive approximation method.
6 Gases33 Successive Method again V = + n b n R T P + n 2 a / V 2 What is the volume of occupied by 132 g CO 2 gas at 12.5 atm and 300 K? Solution : a = 3.59 L 2 atm mol -2, b = L mol -1, T = 300 K n = 132/44 = 3 mol, R = L atm K -1 mol -1 = 3* *300 / ( * 3.59 / 1 2 ) + 3*.0427 = 1.78 L = / ( *3.59 / 3 2 ) = 4.7 L = / ( *3.59 / 4 2 ) = 5.2 L = / ( *3.59 / 5 2 ) = 5.5 L = / ( *3.59 / ) = 5.58 L
6 Gases34 Molecular Formula of Gas Combustion of g hydrocarbon produces CO 2 and g H 2 O. A g sample of the same has a volume of 131 mL at 298 K and 753 mmHg. Find the molecular formula. Solution : C : H = : = : = 1 : 1.5 = 2 : *2 18 Empirical formula is C 2 H 3 M == = 54.3_______ work out units C 4 H 6 has a molar mass of Confirm and conclusion please! (0.288 g / L –1 ) * L atm mol –1 K –1 * 298 K (753 / 760) atm d R T P A 2-step problem, similar to one in Advanced Exercises
6 Gases35 ROOMS FOR TEST #1 (CHEM 120) – Wed., Oct. 8 th Write the test during your regular lecture time (10:30 am) on Wed., Oct. 8 th. Go to DC 1350 and ESC 146/149 according to your Surnames SurnamesRoom(s) For 8:30 class A – LDC 1350 M – ZESC 146 & 149 For 9:30 class A – MoDC 1350 Mu – ZESC 146 & :30 am A – MaDC 1350 (For you)Mc – ZESC 146 & 149 (First Year Chem Lab) For 11:30 class A – MaDC 1350 Mc – ZESC 146 & 149
6 Gases36 Regarding Test 1 Do not enter the room until directed to do so by the proctors. We need space and time to set out the test booklets and computer answer cards. Bring a calculator and a pencil for filling out the computer answer card. Do NOT bring your own scrap paper or periodic table. All work must be done on the test booklet. A periodic table will be supplied.
6 Gases37 Some concepts to review Convert between mass and mole and vice versa Find empirical and molecular formulas Figure out limiting and excess reagent, calculate theoretical and percent yields Calculate concentrations in molarity, mass percentage, etc even when solutions are combined (dilution) Analyze binary mixture: extra problems B2 and B3 (handout page 8) Figure out the net ionic reaction equations Balance redox reaction equations (figure out oxidation states, balance half-reaction equations and balance equations)
6 Gases38 Concepts to review – cont. Apply ideal gas low to various problems Calculate stoichiometric quantities using on gas law and reaction equations. Apply Daltons partial pressure equation Compare effusion or diffusion rates of gases