Chapter Elements That Exist as Gases at 25 C, 1 atm. 5.2 Pressure basic physics. Gas Properties

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1 5.1 Elements That Exist as Gases at 25 C, 1 atm Chapter 5 The Gaseous State YOU READ AND BE RESPONSIBLE FOR THIS SECTION! Gaseous compounds include CH 4, NO, NO 2, H 2 S, NH 3, HCl, etc. Gas Properties 1. Weak attractive forces molecules move freely 2. Fill their container no definite volume or shape 3. Very compressible 4. Gases mix rapidly and completely 4. Low density O 2 (g) is 99.96% empty space. 0.04% molecules at 273 K,1 atm 5.2 Pressure basic physics Sea of gas molecules exerts pressure on us Downward force or air on an area = mass of column of air / area = 14.7 psi (lb/in 2 ). Barometer Measures atmospheric pressure P air = P Hg in tube Downward P of Hg column balanced by atmospheric pressure on Hg in dish One atmosphere (atm) = P that supports 760 mm Hg column (0 o C, sea level) = 1.01 x 10 2 kpa Units of Pressure mmhg = height of supported Hg column (mm) Torr = mmhg Atmospheres (atm) 1 atm is pressure at 0 o C, sea level 1 atm = 760 mmhg Pascals (Pa) - SI pressure unit 1 atm = kpa

2 Why Hg in Barometer? Liquid over wide range, very dense, low reactivity Pressure on a barometer supports a 560-mm Hg column. What is the height in a barometer w/ H 2 O instead of Hg? Mass of supported liquid = (height)(cross-section area)(density) (h Hg )(area Hg )(d Hg ) = (h H2O )(area H2O )(d H2O ) h H2O = (h Hg )(area Hg )(d Hg ) / (area H2O )(d H2O ) h H2O = (h Hg )(d Hg ) / (d H2O ) = (560 mm)(13.6 g cm -3 ) / (1.00 g cm -3 ) = 7616 mm = 7.62 m 5.3 Gas Laws 4 macroscopic properties of a gas sample V = volume T = temperature P = pressure n = amount (number of moles, measured indirectly) Each gas law holds 2 variables constant observes effect of other 2 on one another Boyle s Law: P vs. V Relate P and V w/ T and n constant [Demonstration] Volume is inversely proportional to pressure V 1/P V = k 1 /P PV = k 1 PV = k 1 P 1 V 1 = k 1, P 2 V 2 =k 1 P 1 V 1 = P 2 V 2 Boyle s law: P vs. V Independent of identity of gas Constant T and n Excellent approximation at low P & high T P = k 1 /V or V = k 1 /P The volume of a sample of neon is 2.0 L at 0.50 atm of pressure. What is the final pressure if the volume increases to 5.0 L (at constant T)? 0.20 atm Gay-Lussac s Law: P vs. T Gay-Lussac related P and T [Demonstration] As T increases, P increases P T P = k 2 T P/T = k 2 P 1 /T 1 = P 2 /T 2

3 Charles Law: T & V Charles, hot-air balloonist (1800 s) related V and T w/ P and n constant. [Demonstration] As T decreases, V decreases T V T = k 3 V T/V = k 3 Charles Law T = k 2 V T/V = k 2 k 2 = proportionality constant independent of identity of gas requires constant P and n T 1 /V 1 = k 3 and T 2 /V 2 = k 3 so T 1 /V 1 = T 2 /V 2 excellent approximation at low P & high T Kelvin Temperature At 100 o C a sample of nitrogen gas occupies 420 ml. What volume does it occupy at 400 o C? 758 ml T(K) = t( o C) o C Avogadro s Law: V vs. n Relate V and n w/ T and P constant [Blow up balloon.] Number of moles of gas increases, V increases. V n V = k 4 n V/n = k 4 V 1 /n 1 = V 2 /n 2 Upshot of Avogadro s Law Coefficients in a balanced equation function correspond to volumes: 3H 2 (g) + N 2 (g) --> 2NH 3 (g) 3 moles + 1 mol --> 2 moles 3 L + 1 L --> 2 L How does the pressure vary as this reaction proceeds to the right? H 2 + Cl 2 > 2 HCl (g)

4 How many L of NH 3 are produced when 2.9 L of H 2 reacts with an excess of N 2 (at constant T and P)? 5.4 The Ideal Gas Equation Boyle: V 1/P Charles: V T Avogadro: V n V nt(1/p) PV nt or PV = nrt R = ideal gas constant = L. atm/(mol. K) combines k 1, k 2, k 3 and k 4 into one constant Molar volume PV = nrt Standard Temperature and Pressure (STP) = 0 o C, 1 atm V = nrt/p V = (1 mol)( L. atm/mol. K)(273 K)/1 atm = 22.4 L Same V regardless of identity of gas! 22.4 L of H 2 weighs 2.0 g 22.4 L of SF 6 weighs g Problem Solving w/ Ideal Gas Equation Given 3 of P, V, n, T, calculate 4th Before and after problems Derive the sub-laws Calculate gas density Calculate molar mass Gas Stoichiometry Calculate the density of the gas in g/l if a 2.10 L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 o C. What is its density at STP? What is its molar mass? Calculate the volume of HCl produced when 5.6 L of H 2 (measured at STP) react with an excess of Cl 2 gas. What is the mass of HCl produced?

5 5.6 Dalton s Law of Partial Pressures For gas mixtures Partial Pressure pressure of an individual gas component in a mixture Dalton s Law of Partial Pressure total pressure of a mixture of gases is the sum of the pressures that each gas would exert if it were present alone P T = P 1 + P 2 + P Mole fraction -a dimensionless quantity n A X A = n A + n B + n C +... = n A n T n A = P A V/ RT for any component n T = P T V/ RT for all gases X A = P A P T or P A = X A P T Saturated with H 2 O (g) A mixture of helium and neon gases is collected over water at 28.0 o C and 745 mmhg. If the partial pressure of helium is 368 mmhg, what is the partial pressure of neon? (Vapor pressure of water at 28.0 o C = 28.3 mmhg.) P Ne = P total - P He - P H2O = 745 mmhg mmhg mmhg P Ne = 349 mmhg A mixture of three gases has a total pressure at 398 K of 1380 mmhg. Analysis shows 1.27 mol CO 2, 3.04 mol CO, and 1.50 mol Ar. What is the partial pressure of Ar? P Ar = X Ar P total = [1.50 mol/ ( )mol] 1380 mmhg P Ar = 356 mmhg 5.7 Kinetic Molecular Theory Physical properties of gases are explained by motion of individual molecules. 1. Volume occupied by gas molecules themselves is negligible compared to the total volume of gas. Gas molecules approach point masses w/ 0 volume. At 273K and 1 atm pressure, O 2 molecules take up about 0.04% of a container 99.96% is empty space.

6 Kinetic Molecular Theory 2 2. The molecules of a gas are in constant, rapid, random straight-line Brownian motion. collisions are perfectly elastic (no energy lost in collisions of molecules, walls) total E of gas molecules is constant 3. No attractive nor repulsive forces between molecules Kinetic Molecular Theory 3 4. Average kinetic energy of the molecules is proportional to the Kelvin T of gas KE = 1/2 mu 2, average kinetic E of molecule u 2 = S (u N2 /N), mean square speed KE T, T = absolute temperature 1/2 mu 2 µ T 1/2 mu 2 = CT, C = constant Maxwell speed distribution Peak of curve is most probable speed Fraction of molecules with speed above a threshold increases rapidly with T For a particular gas, molecular speed increases with T Strong T dependence of fraction of molecules above a threshold speed Maxwell speed distribution 2 Root-Mean-Square Speed Measure of average speed of molecules Total kinetic E of a gas: N A (1/2 mu 2 ) = 3/2 RT but N A m = M = molar mass u 2 = 3RT / M For different gases, molecular speed decreases with (molecular mass) 1/2 At a particular T, kinetic energy of all molecules is equal u rms = 3RT M R = J/mol. K

7 Significance higher T, faster molecular speed heavier gas (M), slower molecular speed Units u rms = 1 J = 1 kg m 2 /s 2 M in kg/ mol u in m/ s 3RT M Pressure and Temperature Gas pressure results from collisions between molecules and walls of container pressure depends on frequency of collisions per unit area and how hard molecules strike wall Temperature is a measure of average kinetic energy of molecules Kinetic Molecular Theory explains Compressibility of gases large spaces between molecules Boyle s law P 1/V because decreasing V increases number density, increasing collision rate Charles law V T because increasing T increases frequency and force of collision; gas expands until P re-balances external P Kinetic Molecular Theory explains Avogadro s law V n because under same conditions of P and t, molecules have same spaces between them regardless of identity Dalton s law of partial pressure molecules don t attract or repel one another, so each component exerts P independent of others The temperature in the stratosphere is -23 o C. Calculate the root-mean square speed of an O 3 molecule at that temperature. 360 m/s Diffusion Gradual mixing of molecules of one gas with another From high concentration to low Slow because multiple collisions with other molecules slowing a molecule s progress Light gas diffuses faster than heavy gas

8 5.8 Deviation From Ideality Two ideal assumptions molecules do not exert any force on one another volume of the molecules is negligibly small compared to the volume of the container Real gases deviate from assumptions at high P (molecular volume is significant) at low T (molecules begin to interact) Introduce correction terms Van der Waals Equation P + an 2 V 2 Corrects for encounters between molecules, which depends on number per unit volume. a is a constant (V - nb) = nrt Corrects for volume of molecules. b is a constant P + an 2 V 2 (V - nb) = nrt Table 5.4 Van der Waals Constants Size of a depends on attractions between molecules (atoms). Larger attractions decrease the force molecules exert on wall (P) Size of b depends mainly on size of molecules (atoms). Larger molecules occupy more V. Gas a (atm L 2 /mol 2 ) b (L/mol) He Xe CO CCl H 2 O Calculate P of 2.50 mol of CO 2 in a 5.00 L container at 450 K. Compare ideal and van der Waals behavior. P ideal = atm P vdw = atm