Example Problems: 1.) What is the partial pressure of: Total moles = 13.2 moles 5.0 mol A 7.0 mol B 1.2 mol C Total Pressure = 3.

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1 5.6 Dalton s Law of Partial Pressures Dalton s Law of Partial Pressure; The total pressure of a gas is the sum of all its parts. P total = P 1 + P + P 3 + P n Pressures are directly related to moles: n Total = n 1 + n + n 3 + n n + Will not work with gases that react with each other. Example Problems: 1.) What is the partial pressure of: Total moles = 13. moles 5.0 mol A 7.0 mol B 1. mol C Total Pressure = 3.0 atm P A= (5/13.)* 3.0atm = 1.1 atm P B= 1.6 atm P C= 0.7 atm.) 75.0 ml of O gas is collected over water at 3.0ºC and 753 torr. Vapor Pressure of water is 1.05 mm Hg at 3.0ºC. Calculate pressure of dry gas. (P tot= 743 torr = P O + P HO ; P O dry = = 7 torr 3.) How many molecules are in your lungs? Assume that the volume is 3 liters and the air pressure of 1 atm at 37ºC. (3L)(5x10 7x1atm) See P. 4, Table 5.4, X H = 5x10-7 : n = = 9x10-8 mol R(310K) 9x10-8 mol (6.0x103 molecules) = 5x10 16 molecules H 4.) (P. 33 #59), n= mol, P CO = nrt/v = 1.09 atm, b.) 740/760 =0.974 atm atm =.06 atm 5.) N (g) + 3H (g) NH 3(g) 1 atm N + 4 atm H. If reaction goes to completion, what is total pressure in tank? XS H by (4 atm H total need - 3 atm needed = excess 1atm H ) + atm NH 3 = 3 atm 6.) Tanks below are connected and allowed to mix. What is the partial pressure of each gas and the total pressure of the system when the valves are open? A.0 L 3. atm Boyles Law (P 1V 1 = P V ), Total volume = 7.0 L B 1.0 L.0 atm C 8.0 L 1.8 atm P A = (.0L)(3.atm) ( ) = 0.91atm P B= 0.9 atm P C=.1 atm

2 5.7 The Kinetic Molecular Theory of Gases (KMT) Gas laws predict properties of gases but do not explain why they behave as they do. Discussion in 5.7 (read!) explains how the effects on a gas are understood in terms of the Kinetic Molecular Model. Most important to this discussion is temperature. Temperature: A measure of the average kinetic energy of the particles, or an index of the random motion of the particles; i.e. higher temperature, greater motion. Kinetic Energy is derived as: KE (avg) = 1 mv = 1 m = 3 RT (R= J/mol K) Boyles Law: Decreased volume increases gas density. Hence the collision rate, more frequent collisions. Pressure increases. Charles Law: An increase in temperature increases the speed of molecules, more frequent and greater collision force. Therefore, the volume of a gas will expand until balanced by the external pressure. Avogadro s Law: More moles, greater collision frequency and/or volume until equal to external pressure. Dalton s Law of Partial Pressures: If molecules don t repel or attract, then the pressure is a sum of all of the gases present. Distribution of Molecular Speeds: (Maxwell (Boltzmann) Speed Distribution Curves) Peak of curve represents most probable speed. Lighter molecules move faster than heavier. At greater temperatures the distribution of speeds is also greater. Are all gases moving at this speed? Yes, but only for a very short distance until they collide with another gas molecule. The mean free path is the average distance a molecule travels between collisions. Very small distance. A range of velocities is observed due to collisions amongst particles which slows the mixing process. At higher temperatures, the curve s peak moves toward the higher value and the range of the velocities becomes greater. See figure below What is the average speed of a molecule at a given temperature? Root Mean Square Velocity: The average of the squares of particle velocities, (from ½ mv ) u rms =, root mean square velocity, average speed of particle u rms = = 3RT = M m/s units M= molar mass in kg/mol R = J/K. mol Ex. Calculate rms of He at 5ºC. (1300 m/s) Predict the speed of a molecule that is times heavier than He. (slower, but not ½ the speed. It would be slower by, or 960 m/s.

3 Effusion and Diffusion of Gases Diffusion: refers to the mixing of gases, a complicated treatment. Effusion: Passage of a gas through a tiny orifice into an evacuated chamber. Rate (speed) in ml/min is measured here. Thomas Graham s Law of Effusion: (at same temperature and pressure) The rate of effusion is inversely proportional to the square roots of the molecular mass. υ υ rms rms 1 = M M 1 Compare s the rate of effusion. Ex. Under a set of conditions, SO gas (64 g/mol) effuses at 4 ml/min. What rate would O gas (3 g/mol) effuse? Ans: O gas is lighter and is predicted to travel faster, but only x faster. Compare correctly 34 ml/min, not 48 ml/min. 5.8 Real Gases Real gases deviate from ideal behavior. Real gases approach ideal behavior at conditions of high temperature and low pressure. Ideal Gas: a.) Point mass, no volume because the distances between particles is so great that it is assumed that their volumes are negligible. In reality, the volumes of the gas particles do occupy space of the container. b.) The average kinetic energy of the particles is directly proportional to the absolute temperature. c.) Gas particles are independent of each other. Identities of particles do not matter since volume is unimportant. d.) Particles are in constant motion. Collisions create pressure. Collisions are eleastic which is to say that no energy is lost. e.) No intermolecular attractions or repulsions. H O vapor is a terrible ideal gas. PV f.) = 1 nrt

4 van der Waals Equation: Modified ideal gas equation for real gases. 1.) First, account for the finite volumes of gas. Gases do occupy volume of the container. V-nb, where n is the number of moles and b is an empirical constant. See Table 5.3 nrt P = note: as the value of b increases, so does the size of particle. (V nb).) Second, account for the intermolecular attractions. The real effect of this is to make the pressure lower. This is affected by the concentration of molecules, (n/v), and is less significant at higher temperatures in which the particles are moving faster and will not apply an attractive or repulsive force on each other. nrt n Pobs = a ( V nb) v note: the value of a increases with amount of intermolecular attractions. van der Waals equation: n + a x V nb V ( ) nrt Pobs =

5 VAN DER WAALS EQUATION OF STATE The Ideal Gas Law, PV = nrt, can be derived by assuming that the molecules that make up the gas have negligible sizes, that their collision with themselves and the wall are perfectly elastic, and that the molecules have no interactions with each other. The van der Waal's equation is a second order approximation of the equation of state of a gas that will work even when the density of the gas is not low. Here a and b are constants particular to a given gas. Substance Some van der Waals Constants a (J. m 3 /mole ) b (m 3 /mole) Pc (MPa) Air x K Tc (K) Carbon Dioxide (CO) x K Nitrogen (N) x K Hydrogen (H) x K Water (HO) x K Ammonia (NH3) x K Helium (He) x K Freon (CClF) x K The parameter b is related to the size of each molecule. The volume that the molecules have to move around in is not just the volume of the container V, but is reduced to ( V - nb ). The parameter a is related to intermolecular attractive force between the molecules, and n/v is the density of molecules. The net effect of the intermolecular attractive force is to reduce the pressure for a given volume and temperature. When the density of the gas is low (i.e., when n/v is small and nb is small compared to V) the van der Waals equation reduces to that of the ideal gas law. One region where the van der Waals equation works well is for temperatures that are slightly above the critical temperature Tc of a substance

6 Observe that inert gases like Helium have a low value of a as one would expect since such gases do not interact very strongly, and that large molecules like Freon have large values of b. There are many more equations of state that are even better approximation of real gases than the van der Wall equation. Selected Practice Problems: P. 19# 70, 71, 74, 76, 81, 83, 87, 93, 96, 99, 103, 110, 111 (wrong equation), 113, 117, 15, 16, 130, 134, 136, 138, 140, 147, 150, 157, 159, 16, 163, 164, 166, 168, 170