Gas particles move in straight line paths. As they collide, they create a force, pressure.


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1 #28 notes Unit 4: Gases Ch. Gases I. Pressure and Manometers Gas particles move in straight line paths. As they collide, they create a force, pressure. Pressure = Force / Area Standard Atmospheric Pressure = 14.7 psi (1b./ in 2 = force / area) (at sea level) = Pa (pascals) { millibars} = 1 atm (atmospheres) = 760 mm Hg* = 760 torr* *equivalent A. Closed Manometers: B. Open Manometers: Ex. 1) Find the pressure in Pa and atm, if the height of the mercury is 45 mm. Height of Mercury = the Pressure 45 mm Hg Pa = 6.0 X10 3 Pa 760 mm Hg 45 mm Hg 1 atm = 5.9 X102 atm 760 mm Hg Ex. 2) The mercury is 45 mm higher in the atmospheric arm. Find the pressure in Pa and atm, if the atmospheric pressure is Pa. 45 mm Hg Pa = 6.0 X10 3 Pa 760 mm Hg P atm + P Hg = P gas The gas pushes the hardest (strongest), so it is alone! Pa X10 3 Pa = P gas Pa = P gas 1.1 X 10 5 Pa = P gas 1.1 X10 5 Pa 1 atm = 1.1 atm Pa
2 Ex. 3) Same as Ex. 2, except the mercury is higher in the gas arm. (height of Hg = 45mm, P atm = Pa) 45 mm Hg Pa = 6.0 X10 3 Pa 760 mm Hg P atm = P Hg + P gas The air pushes the hardest (strongest), so it is alone! Pa = 6.0 X10 3 Pa + P gas Pa = P gas 9.4 X10 4 Pa 9.4 X10 4 Pa 1 atm = 0.93 atm Pa II. Ideal Gas Law: In this ideal gas law, the gas molecules are spread out. There will not be any attractions, so P observed will not need any adjustments. The molecules will be much smaller than the distances separating them. They are treated as point masses. They have mass, but negligible volume. (This works at low pressures and high temperatures.) Because of this the volume of the container can be used without subtracting the volume of the gas particles. PV=nRT where R= L atm/(mol K) o C = K Ex. 1) What pressure is exerted by mol of CO 2 contained in a 922 ml vessel at 16 o C? P =?, n = mol, V = 922 ml, T = 16 o C 922 ml 1 X103 L = L 16 o C = 289 K 1 ml PV = nrt P (0.922 L) = (0.622 mol) ( L atm/(mol K)) (289K) P = 16.0 atm
3 #29 Notes III. More Gas Laws (Derived from the Ideal) Ex. 1) A gas in a container at 25 o C, having a volume of 300 cm 3 exerts a pressure of 520 torr. What is its volume at standard atmospheric pressure? T 1 = 25 o C V 1 = 300 cm 3 V 2 =? P 1 = 520 torr P 2 = 760 torr = 1atm etc. PV = nrt n, R, T are constant PV = constant so, P 1 V 1 = constant and P 2 V 2 =constant P 1 V 1 = P 2 V 2 Boyle s Law (520 torr) (300 cm 3 ) = (760 torr) V cm 3 = V 2 Ex. 2) Find the volume at standard temperature for a sample of H 2 that at 9.00 o C has a volume of 613 cm 3. **Standard Temperature = 273 K V 1 =? V 2 = 613 cm 3 T 1 = 273 K T 2 = 9.00 o C 9.00 o C = 282 K PV = nrt P, n, R are constant *Temperatures must be in Kelvin! (Adding conversions will not cancel.) V = nrt P V = nr V = constant so, V 1 = V 2 Charles Law T P T T 1 T 2 V 1 = (613 cm 3 ) (273K) (282 K) **Cross Multiply!! V 1 = 593 cm 3
4 Ex. 3) 45.3 mol of CH 4 gas occupies 916 L of space. What volume would mol occupy? n 1 = 45.3 mol n 2 = mol V 1 = 916 L V 2 =? PV = nrt P, R, T are constant V = nrt P V = RT V = constant n P n V 1 = V 2 n 1 n 2 Avogadro s Law (916 L) = V 2 (45.3 mol) (0.214 mol) 4.33 L = V 2 Ex. 4) CO 2 in a 45 ml vessel at 0.98 atm and 25 o C is transferred to a L container at 35 o C. What is its new pressure? V 1 = 45 ml V 2 = L 45 ml 1 X103 L = L P 1 = 0.98 atm P 2 =? 1 ml T 1 = 25 o C T 2 = 35 o C 25 o C = 298 K 35 o C = 308 K PV = nrt PV = nr T n, R are constant PV = constant T P 1 V 1 = P 2 V 2 T 1 T 2 (0.98 atm) (0.045 L) = (P 2 ) (0.067 L) (298 K) (308 K) 0.68 atm = P 2
5 #30 Notes IV. Gas Density PV = nrt mols = mass molar mass PV = (mass) RT (molar mass) (molar mass) = (mass) RT PV (molar mass) = (mass) RT V P Molar mass = drt but density is in g/l!!! P Ex. 1) A compound has the empirical formula CH 2. A 200 ml flask at 298 K and 755 torr contains 0.57 g of the gaseous compound. Give the molecular formula. d = m/v 200 ml 1 X103 L = L 1 ml d = 0.57 g L 755 torr 1 atm = atm 760 torr d = 2.85 g/l mm = drt/p mm = (2.85 g/l) ( L atm/(mol K)) (298 K) (0.993 atm) mm = 70. g/mol Empirical Formula = CH 2 = g/mol 70. g/mol = g/mol X5 Molecular Formula = C 5 H 10
6 #31 Notes V. Dalton s Law of Partial Pressures In a mixture of gases, each particular gas will exert the same pressure it would, if it were alone. In this room P O2 = 0.20 atm, P N2 = 0.79 atm, What is P tot? P tot = P 1 + P 2 + P 3 + The total pressure is the sum of the partial pressures. Ex. 1) CO 2 is collected over H 2 O at 10 o C in a 20.0 cm 3 vessel. The manometer indicates a pressure of 60.0 kpa. What is the mass of gas collected? ( P H2O at 10 o C = 1.2 kpa) P tot = P gas + P H2O 60 kpa = P gas kpa 58.8 kpa = P gas P = 58.8 kpa 1000 Pa 1 atm = atm 1kPa Pa T = 10 o C = 283 K V = 20.0 cm 3 1 ml 1 X103 L = L 1 cm 3 1 ml n =? PV = nrt (0.580 atm) ( L) = n ( L atm/(mol K)) (283 K) X104 mol = n X104 mol CO g = 2.2 X102 g CO 2 1 mol ** Remember: All gases diatomic (N 2, H 2, O 2 ) except noble gases (He, Ne, Ar).
7 Ex. 2) 6.00 dm 3 of H 2 and 2.00 dm 3 of O 2 at 25 o C and 760. torr are pumped into a L container. What is the pressure in the container? H 2 O 2 new container V = 6.00 dm 3 V = 2.00 dm 3 V = L T = 25 o C T = 25 o C P = 760. torr P = 760. torr P =? Constant T, so ignore PV = nrt n, R, T are constant P 1 V 1 = P 2 V 2 H 2 O 2 P 1 V 1 = P 2 V 2 **1L =1dm 3 P 1 V 1 = P 2 V 2 (760 torr) ( 6.00 dm 3 ) = P 2 (0.250 L) (760 torr) ( 2.00 dm 3 ) = P 2 (0.250 L) torr = P 2 of H torr = P 2 of O 2 P tot = P H2 + P O2 = torr torr = = 2.43 X10 4 torr
8 #32 Notes VI. Gas Stoichiometry Derive a conversion factor between mols and volume at STP for 1 mol of any gas. STP = standard Temperature (273 K) and standard Pressure (1atm) P = 1 atm V =? n = 1 mol T = 273 K PV = nrt (1 atm) V = (1 mol) ( L atm/(mol K)) (273 K) V = 22.4 L ** 1 mol of any gas = 22.4 L at STP Ex. 1) An excess of nitrogen gas reacts with 5.0 g of hydrogen gas. What volume of ammonia is produced at STP? Write the reaction: N 2(g) + 3 H 2(g) 2 NH 3(g) 5.0 g H 2 1 mol H 2 2 mol NH L NH 3 = 37 L NH g H 2 3 mol H 2 1 mol NH 3 How many grams of H 2 are produced from 39 L NH 3? 39 L NH 3 1 mol NH 3 3 mol H g H 2 _ = 5.3 g H L NH 3 2 mol NH 3 1 mol H 2 If it is not at STP, do stoichiometry, but stop at mols (after the mol ratio) and then use PV = nrt. OR Change given amounts to STP by using gas equations (T = 273 K, P = 1atm), then do stoichiometry.
9 #33 Notes VII. Kinetic Molecular Theory of Gases 1) The gas particles are so small compared to the distances separating them that we can assume their volume is negligible (zero). Ideal Gases 2) The gas particles are in constant motion and their collisions cause the pressure exerted by the gas. 3) The particles are assumed to exert no attractive/ repulsive forces on each other. Ideal Gases 4) The average Kinetic Energy (KE) is directly proportional to temperature. KE = ½ m v 2 mass velocity *high temperature, high velocity of particles, high KE KE = 3/2 RT where R = 8.31 J/(mol K) 1 joule = 1 Newton meter 1 Newton = 1 (kg m) /s 2 VIII. Root Mean Square Velocity KE = 1/2 mv 2 = 3/2 RT mv 2 = 3RT v 2 = 3RT m v rms = (3RT/m) ** molar mass must be in kg/mol, velocity in m/s IX. Effusion/ Diffusion Effusion: filling of a gas into a chamber by a hole. Diffusion: mixing of 2 gases.
10 Graham s Law of Effusion (works for diffusion too.) Rate of effusion for gas #1 = v rms #1 = (3RT/m 1 ) = (1/m 1 ) = m 2 Rate of effusion for gas #2 = v rms #2 = (3RT/m 2 ) = (1/m 2 ) = m 1 Rate gas #1 = m 2 Rate gas #2 m 1 **Rate must be in volume/time, but units don t matter if they agree. Ex. 1) Find KE and v rms for F 2 at 25 0 C. KE = 3/2 RT = 3/2 (8.31 J/(mol K)) (298 K) = 3.71 X10 3 J/mol V rms = 3RT/m = (3) (8.31 J/(mol K)) (298 K) = 442 m/s ( kg/mol) F 2 = g 1 kg = kg/mol F 2 mol 1 X10 3 g Ex. 2) A gas has an effusion rate of 15 ml/min. If CH 4 has a rate of 45 ml/min, what is the molar mass of the other gas? R 1 = m 2 CH 4 = g 1 kg = kg/mol R 2 m 1 mol 1 X10 3 g Rate? gas = m CH4 Rate CH 4 = m?gas 15 ml/min = kg/mol CH 4 45 ml/min CH 4 m?gas = m?gas m?gas = m?gas = m?gas = kg/mol
11 #34 Notes IX. Real Gases PV = nrt is for Ideal Gases The particle volume was assumed to be zero, but molecules really do take up space. V volume around the gas = V volume of the container nb volume of the gas particles Where n = mols and b = an experimentally found empirical constant The attractive forces were also assumed to be zero, but attractive forces between molecules will keep them from hitting the container as hard as predicted. P observed pressure = P the real pressure a (n/v) 2 decrease in pressure caused by the attractive forces Where a = an experimentally found proportionality constant and n = mols and V is volume so P real = P observed + a (n/v) 2 P real V around the gas = nrt [P observed + a (n/v) 2 ] (V nb) = nrt a,b: see Table in textbook Van der Waals Equation for Real Gases Ex.1) Calculate the pressure exerted by mol of Krypton in a L container at 25.0 o C. P =? n = mol Look up for Kr: a = 2.32 atm L 2 /mol 2 V = L b = L/mol T = 25 o C = 298 K [P observed + a (n/v) 2 ] (V nb) = nrt [P + (2.32 atm L 2 /mol 2 ) ( mol/ L) 2 ] [1.000 L ( mol)( L/mol)] = ( mol)( L atm/(mol K))(298 K) [P atm] [ L] = ( L atm) [P atm] = atm P = 6.03 atm *End of Notes* (Assignments #3536 are Review Assignments. There are no notes for these assignments.)
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