Substances that are Gases under Normal Conditions

Chapter 5: Gases 5.1 Early Experiments 5.2 The gas laws of Boyle, Charles, and Avogadro 5.3 The Ideal Gas Law 5.4 Gas Stiochiometry 5.5 Dalton s Law of Partial Pressures 5.6 The Kinetic molecular Theory of Gases 5.7 Effusion and Diffusion 5.8 Collisions of Gas Particles with the Container Walls 5.9 Intermolecular Collisions 5.10 Real Gases 5.11 Chemistry in the Atmosphere

Substances that are Gases under Normal Conditions Substance Formula M (g/mol) Helium He 4.0 Neon Ne 20.2 Argon Ar 39.9 Hydrogen H 2 2.0 Nitrogen N 2 28.0 Nitrogen Monoxide NO 30.0 Oxygen O 2 32.0 Hydrogen Chloride HCl 36.5 Ozone O 3 48.0 Ammonia NH 3 17.0 Methane CH 4 16.0

Important Characteristics of Gases 1) Gases are highly compressible An external force compresses the gas sample and decreases its volume, removing the external force allows the gas volume to increase. 2) Gases are thermally expandable When a gas sample is heated, its volume increases, and when it is cooled its volume decreases. 3) Gases have low viscosity Gases flow much easier than liquids or solids. 4) Most Gases have low densities Gas densities are on the order of grams per liter whereas liquids and solids are grams per cubic cm, 1000 times greater. 5) Gases are infinitely miscible Gases mix in any proportion such as in air, a mixture of many gases.

Pressure = force per unit area (Definition!) Pressure = P = Force Area Gas samples exert a force pushing against any surface they contact, due to the impact of molecules striking the surface. Per unit area, it s the same for all surfaces in or walls containing the same gas sample, unless the gas is flowing somewhere collectively. That pressure increases with molecular velocity and mass, and with the number of molecules per unit volume (number density). A sheet suspended within a gas sample gets same force on both sides, so it does not move. Average molecular velocity increases with temperature but decreases with molecular mass, in such a way that the mass effect cancels. This leaves pressure nearly proportional to temperature and to number density, but independent of type gas (or molecular mass): P = R(T)(n/V), where R is the gas constant, n = # moles = N/N A

Pressure of the Atmosphere Called Atmospheric pressure, or the force exerted on earth s surface by the gases in air The force per unit area of these gases: Pressure = Force Area

The Magdeburg sphere experiment

Pressure of the Atmosphere Called Atmospheric pressure, or the force exerted on earth s surface by the gases in air The force per unit area of these gases. Pressure = P = Force Area Can be measured using a barometer:

Figure 5.1: A Torricellian barometer. Density of Mercury = 13.6 g/cm 3 760 mm column of 1 cm 2 area (76 cm) weight = volume x density = (76 cm x 1 cm 2 ) x 13.6 g/cm 3 = 1030 g = 1.03 kg = 2.28 lbs P = force / area = weight / area = 2.28 lbs / cm 2 = 14.7 lbs / in 2 = 1.00 atm.

Construct a Barometer using Water! Density of water = 1.00 g/cm 3 Density of Mercury = 13.6 g/cm 3 Height of water column = H w Height Water Height Mercury = Density Mercury Density Water H w = Height of Hg x Density of Mercury Density of Water H w = 760 mm Hg x (13.6 g/cm 3 ) /1.00 g/cm 3 ) = 1.03 x 10 4 mm = 10.3 m = 33.8 ft

Figure 5.2: A simple manometer, a device for measuring the pressure of a gas in a container. Units of pressure here are mm of Hg or mm of water, depending on which liquid is used in the tube.

Converting Units of Pressure Problem: A chemist collects a sample of Carbon dioxide from the decomposition of Lime stone (CaCO 3 ) in a closed-end manometer (i.e, vacuum is the reference pressure), the height of the mercury is 341.6 mm Hg. Calculate the CO 2 pressure in torr, atmospheres, and kilopascals. Plan: The pressure is in mmhg, so we use the conversion factors from Table above to find the pressure in the other units. Solution: converting from mmhg to torr: P CO2 (torr) = 341.6 mm Hg x 1 torr = 341.6 torr 1 mm Hg converting from torr to atm: P CO2 ( atm) = 341.6 torr x 1 atm = 0.4495 atm 760 torr converting from atm to kpa: P CO2 (kpa) = 0.4495 atm x 101.325 kpa = 45.54 kpa 1 atm

Common Units of Pressure Unit Atmospheric Pressure Scientific Field Used Pascal (Pa) = N/m 2 ; 1.01325 x 10 5 Pa SI unit; physics, kilopascal (kpa) 101.325 kpa Chemistry bar 1.01325 bar Meteorology, Chemistry atmosphere (atm) 1 atm Chemistry millimeters of mercury 760 mmhg Chemistry, medicine, (mmhg), also called 760 torr biology torr pounds per square inch 14.7 lb/in 2 (psi or lb/in 2 ) Engineering

Boyle s Law : P - V inversely proportional when n and T are constant in a gas sample Pressure is inversely proportional to Volume k k P = or V = or PV=k V P Change of Conditions Problems if n and T are constant! P 1 V 1 = k P 2 V 2 = k Then : P 1 V 1 = P 2 V 2

Figure 5.3: A J-tube similar to the one used by Boyle. P = P atm + h

Figure 5.4: Plotting Boyle s data from Table 5.1. Syringe Demo

Boyle s Law : Balloon A balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to an altitude of 6.5 km, where the pressure is 0.40 atm. Assume that the temperature remains constant (which obviously is not true), what is the final volume of the balloon? P 1 = 1.0 atm P 2 = 0.40 atm V 1 = 0.55 L V 2 =? V 2 =

Applying Boyles Law to Gas Problems Problem: A gas sample at a pressure of 1.23 atm has a volume of 15.8 cm 3, what will be the volume in L if the pressure is increased to 3.16 atm? Plan: We begin by converting the volume that is in cm 3 to ml and then to liters, then we use the pressure ratio to obtain the final volume! Solution: V 1 (cm 3 ) P 1 = 1.23 atm P 2 = 3.16 atm V 1cm 3 = 1 ml 1 = 15.8 cm 3 V 2 = unknown T and n remain constant V 1 (ml) V 1 = 15.8 cm 3 x x = L 1000mL = 1L V 1 (L) V 2 = V 1 x = P 1 P 2 V 2 (L) x P 1 /P 2

Applying Boyles Law to Gas Problems Problem: A gas sample at a pressure of 1.23 atm has a volume of 15.8 cm 3, what will be the volume in L if the pressure is increased to 3.16 atm? Plan: We begin by converting the volume that is in cm 3 to ml and then to liters, then we do the pressure change to obtain the final volume! Solution: V 1 (cm 3 ) P 1 = 1.23 atm P 2 = 3.16 atm V 1cm 3 = 1 ml 1 = 15.8 cm 3 V 2 = unknown T and n remain constant V 1 (ml) V 1 = 15.8 cm 3 x 1 ml x 1 L = 0.0158 L 1 cm 3 1000mL 1000mL = 1L V 1 (L) P 1 P 2 V 2 = V 1 x = 0.0158 L x 1.23 atm = 0.00615 L 3.16 atm V 2 (L) x P 1 /P 2

Boyle s Law - A gas bubble in the ocean! A bubble of gas is released by the submarine Alvin at a depth of 6000 ft in the ocean, as part of a research expedition to study under water volcanism. If the gas bubble has an initial volume of 1.00 cm 3, what size will it be at the surface at a pressure of 1.00 atm? Assume that the ocean is isothermal (i.e., the same temperature throughout), and that the density of sea water is 1.026 g/cm 3. Step 1. Determine the pressure at the submarine from the depth of sea water, using the density of Hg in a barometer for comparison! (Remember to add 1 atmosphere!) Step 2. Use Boyle s law. Initial Conditions Final Conditions V 1 = 1.00 cm 3 P 1 =? V 2 =? P 2 = 1.00 atm

Calculation Continued P 1 =1 atm + 6 x 10 3 ft x x x P 1 = 1 atm + mm Hg But 760 mm Hg = 1.00 atm, so P 1 = 1 atm + V V 2 = 1 x P 1 P = = 2

Calculation Continued P 1 =1 atm + 6 x 10 3 ft x 0.3048 m x 1000 mm x 1 ft 1 m 1.026 g H 2 O/cm 3 13.6 g Hg/cm 3 P 1 = 1 atm + 138,000 mm Hg But 760 mm Hg = 1.00 atm, so P 1 = 1 atm + (138,000 mm Hg) x (1.00 atm / 760 mm Hg) P 1 = 1 atm + 182 atm = 183 atm! V V 2 = 1 x P 1 = 1.00 cm 3 x 183 atm P = 183 cm 3 = 0.183 liters 2 1.00 atm

Figure 5.7: Plots of V versus T ( C) for several gases.

Figure 5.8: Plots of V versus T as in Fig. 5.7 except that here the Kelvin scale is used for temperature.

Charles Law - V - T- proportional At constant pressure and for a fixed amount (# moles) of gas: Volume is proportional to Temperature : V= constant x Absolute Temperature V = k T Change of conditions problems: Since V = k T or V 1 / T 1 = V 2 / T or: 2 T 1 V 1 = T or 2 T 1 = V 1 x V 2 Temperatures must be expressed in Kelvin! T 2 V 2

Charles Law Problem A sample of carbon monoxide, a poisonous gas, occupies 3.20 L at 125 o C. Calculate the temperature ( o C) at which the gas will occupy 1.54 L if the pressure remains constant. V 1 = 3.20 L T 1 = 125 o C = 398 K V 2 = 1.54 L T 2 =? V 1 / T 1 = V 2 / T 2 So T 2 =

Charles Law Problem - I A balloon in Antarctica in a building is at room temperature ( 75 o F ) and has a volume of 20.0 L. What will be its volume outside where the temperature is -70 o F? V 1 = 20.0 L V 2 =? T 1 = 75 o F T 2 = -70 o F o C = ( o F - 32 ) 5/9 T 1 = ( 75-32 )5/9 = 23.9 o C K = 23.9 o C + 273.15 = 297.0 K T 2 = ( -70-32 ) 5/9 = - 56.7 o C K = - 56.7 o C + 273.15 = 216.4 K

Antarctic Balloon Problem - II V 1 / T 1 = V 2 / T 2 so V 2 = V 1 x ( T 2 / T 1 ) V 2 = x V 2 =

Antarctic Balloon Problem - II V 1 / T 1 = V 2 / T 2 V 2 = V 1 x ( T 2 / T 1 ) V 2 = 20.0 L x 216.4 K 297.0 K V 2 = 14.6 L The Balloon shrinks from 20 L to 15 L!!!!!!! Just by going outside!!!!!

Avogadro s Law - Moles and Volume The volume of gas is directly proportional to the amount of gas (in moles), when measured at the same P and T, and independent of the type of gas. This math symbol means is proportional to. V α n or V = some constant x n n 1 = initial moles of gas V 1 = initial volume of gas n 2 = final moles of gas V 2 = final volume of gas n 1 = n 2 or: n V 1 V 1 = n 2 x 2 V 1 V 2

Avogadro s Law: Volume and Amount of Gas Problem: Sulfur hexafluoride is a gas used to trace pollutant plumes in the atmosphere, if the volume of 2.67 g of SF 6 at 1.143 atm and 28.5 o C is 2.93 m 3, what will be the mass of SF 6 in a container whose volume is 543.9 m 3 at 1.143 atm and 28.5 o C? Plan: Since the temperature and pressure are the same it is a V - n problem: Use Avogadro s Law to calculate the moles of the gas, then use the molecular mass to calculate the mass of the gas. Solution: Calculate the mole ratio from the volume ratio n = n x V 2 2 1 V = n x 543.9 m 3 = 186 x n 1 1 1 2.93 m 3 For the same type gas, mass is proportional to moles, so mass 2 = 186 x mass 1 = 186 x 2.67 g = 496 g SF 6

Volume - Amount of Gas Relationship Problem: A balloon contains 1.14 moles (2.298 g H 2 ) of Hydrogen and has a volume of 28.75 L. What mass of Hydrogen must be added to the balloon to increase it s volume to 112.46 Liters. Assume T and P are constant. Plan: Volume and amount of gas are changing at constant T and P, so we will use Avogadro s law, and the ratio of volumes. Solution: n 1 = 1.14 moles H 2 n 2 = 1.14 moles +? moles V 1 = 28.75 L V 2 = 112.46 L T = constant P = constant n n 1 2 V V = 1 V n 2 2 = n 1 x 2 = V 1

Volume - Amount of Gas Relationship Problem: A balloon contains 1.14 moles(2.298g H 2 ) of Hydrogen and has a volume of 28.75 L. What mass of Hydrogen must be added to the balloon to increase it s volume to 112.46 Liters. Assume T and P are constant. Plan: Volume and amount of gas are changing at constant T and P, so we will use Avogadro s law, and the ratio of volumes. Solution: n 1 = 1.14 moles H 2 n 2 = 1.14 moles +? moles V 1 = 28.75 L V 2 = 112.46 L n n 1 2 V V = 1 V n 2 2 = n 1 x 2 = 1.14 moles H 2 x V 1 n 2 = 4.4593 moles = 4.46 moles added mass = 8.99g - 2.30g = 6.69g T = constant P = constant 112.46 L 28.75 L mass = moles x molecular mass mass = 4.46 moles x 2.016 g / mol mass = 8.99 g H 2 gas

IDEAL GAS LAW The ideal gas equation combines both Boyles Law and Charles Law into one easy-to-remember law: PV = nrt n = number of moles of gas in volume V R = Ideal gas constant R = 0.08206 L atm / (mol K) = 0.08206 L atm mol -1 K -1 Later R = 8.314 J / (mol K) = 8.314 J mol -1 K -1 An ideal gas is one for which both the volume of molecules and forces between the molecules are so small that they have insignificant effect on its P-V-T behavior. Independent of substance, in the limit when n/v 0, all gases behave ideally. Pretty accurate below 10 atm.

Variations on Ideal Gas Equation During chemical and physical processes, any of the four variables in the ideal gas equation may be fixed. Thus, PV=nRT can be rearranged for the fixed variables: for a fixed amount at constant temperature PV = nrt = constant Boyle s Law for a fixed amount at constant pressure V/T = nr/p = constant Charles Law for a fixed pressure and temperature V = n (RT/P) or V n Avogadro s Law for constant volume P/T = nr/v = constant when amount constant P = nrt/v so P n when T constant JUST REMEMBER: PV = nrt and rearrange as needed

Standard Temperature and Pressure (STP) A set of Standard conditions have been chosen to make it easier to quantify gas amounts (i,.e., liters at STP ). Standard Temperature = 0 0 C = 273.15 K Standard Pressure = 1 atmosphere = 760 mm Mercury At these standard conditions, if you have 1.0 mole of any ideal gas, it will occupy a standard molar volume : V = nrt/p = Standard Molar Volume = 22.414 Liters = 22.4 L or L/mol

Table 5.2 (P 150) Molar Volumes for Various Gases at 0 0 and 1 atm Gas Molar Volume (L) Oxygen (O 2 ) 22.397 Nitrogen (N 2 ) 22.402 Hydrogen (H 2 ) 22.433 Helium (He) 22.434 Argon (Ar) 22.397 Carbon dioxide (CO 2 ) 22.260 Ammonia (NH 3 ) 22.079

Gas Law: Solving for Pressure Problem: Calculate the pressure in a container whose Volume is 87.5 L and it is filled with 5.038kg of Xenon at a temperature of 18.8 o C. Plan: Convert all information into the units required, and substitute into the Ideal Gas Equation ( PV=nRT ). Solution: n Xe = 5038 g Xe = 38.37014471 mol Xe 131.3 g Xe / mol T = 18.8 o C + 273.15 K = 291.95 K

Gas Law: Solving for Pressure Problem: Calculate the pressure in a container whose Volume is 87.5 L and it is filled with 5.038kg of Xenon at a temperature of 18.8 o C. Plan: Convert all information into the units required, and substitute into the Ideal Gas equation ( PV=nRT ). Solution: n Xe = 5038 g Xe = 38.37014471 mol Xe 131.3 g Xe / mol T = 18.8 o C + 273.15 K = 291.95 K PV = nrt P = so P = nrt V (38.37 mol )(0.0821 L atm)(291.95 K) 87.5 L (mol K) = 10.5108 atm = 10.5 atm

Ideal Gas Calculation - Nitrogen Calculate the pressure in a container holding 375 g of nitrogen gas. The volume of the container is 0.150 m 3 and the temperature is 36.0 o C. n = 375 g N 2 / (28.0 g N 2 / mol) = 13.4 mol N 2 V = 0.150 m 3 x (1000 L / m 3 )= 150 L T = 36.0 o C + 273.15 = 309.2 K

Ideal Gas Calculation Nitrogen- part 2 Calculate the pressure in a container holding 375 g of Nitrogen gas. The volume of the container is 0.150 m 3 and the temperature is 36.0 o C. n = 375 g N 2 / 28.0 g N 2 / mol = 13.4 mol N 2 V = 0.150 m 3 x 1000 L / m 3 = 150 L T = 36.0 o C + 273.15 = 309.2 K PV=nRT so P= nrt/v ( 13.4 mol) ( 0.08206 L atm/mol K) ( 309.2 K) P = 150 L P = 2.27 atm

Mass of Air in a Hot Air Balloon - Part I Calculate the mass of air in a spherical hot air balloon that has a volume of 14,100 cubic feet when the temperature of the gas is 86 o F and the pressure is 748 mm Hg? (Avg. molar mass of gases in air is ~29 g/mol) P = 748 mm Hg x 1atm / 760 mm Hg= 0.984 atm V = 1.41 x 10 4 ft 3 x T = 86 o F (convert to K) n = PV/RT Then convert to grams (29 g/mol).

Mass of Air in a Hot Air Balloon - Part II Calculate the mass of air in a spherical hot air balloon that has a volume of 14,100 cubic feet when the temperature of the gas is 86 o F and the pressure is 748 mm Hg? (Avg. molar mass of gases in air is ~29 g/mol) P = 748 mm Hg x 1atm / 760 mm Hg= 0.984 atm V = 1.41 x 10 4 ft 3 x (12 in/1 ft) 3 x (2.54 cm/1 in) 3 x x (1mL/1 cm 3 ) x ( 1L / 1000 ml) =3.99 x 10 5 L T = o C =(86-32)5/9 = 30 o C 30 o C + 273 = 303 K

Mass of Air in a Hot Air Balloon - Part III PV = nrt n = PV / RT ( 0.984 atm) ( 3.99 x 10 n = 5 L) = 1.58 x 10 4 mol ( 0.08206 L atm/mol K) ( 303 K ) mass = 1.58 x 10 4 mol air x 29 g air/mol air = 4.58 x 10 5 g Air x (1 kg / 1000 g) = 458 kg Air

Ammonia Density Problem Calculate the density of ammonia gas (NH 3 ) in grams per liter at 752 mm Hg and 55 o C. P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm T = 55 o C + 273 = 328 K PV = nrt so n/v = = P R x T ( 0.989 atm) ( 0.08206 L atm / mol K) ( 328 K) n/v = 0.03674 mol/l density = ρ = mass per unit volume density = 0.03674 mol/l x (17.03 g/mol NH 3 ) ρ = 0.626 g / L (more than 1000x less dense than water!)

Ammonia Density Problem by another method Calculate the density of ammonia gas (NH 3 ) in grams per liter at 752 mm Hg and 55 o C. Density = ρ = mass per unit volume = m / L P = 752 mm Hg x (1 atm/ 760 mm Hg) =0.989 atm T = P 55 x M o C + 273 ( 0.989 = 328 atm) K( 17.03 g/mol) R x T ( 0.08206 L atm/mol K) ( 328 K) n = mass / Molar mass = m / M PV = nrt so m/(mv) = n/v = P/(RT)

Applying the Temperature - Pressure Relationship (constant n and V) Problem: A copper tank is filled with compressed gas to a pressure of 4.28 atm at a temperature of 0.185 o F. What will be the pressure if the temperature is raised to 95.6 o C? Plan: The volume of the tank is not changed, and we only have to deal with the temperature change, so convert to SI units, and calculate the pressure ratio from the T ratio. Solution: T 1 = (0.185 o F - 32.0 o F)x 5/9 = -17.68 o C T 1 = -17.68 o C + 273.15 K = 255.47 K T 2 = 95.6 o C + 273.15 K = 368.8 K P 2 = 4.28 atm x = 6.18 atm 368.8 K 255.47 K P 1 = P 2 = nr/v T 1 T 2 P 2 = P 1 x T 2 =? T 1

Change of Conditions :Problem -I A gas sample in the laboratory has a volume of 45.9 L at 25 o C and a pressure of 743 mm Hg. If the temperature is increased to 155 o C by pumping (compressing) the gas to a new volume of 3.10 ml, what is the pressure? P 1 = 743 mm Hg x1 atm/ 760 mm Hg=0.978 atm P 2 =? V 1 = 45.9 L V 2 = 3.10 ml = 0.00310 L T 1 = 25 o C + 273 = 298 K T 2 = 155 o C + 273 = 428 K

Change of Condition Problem I : continued P 1 x V 1 P 2 x V 2 = = nr = constant T 1 T 2 ( 0.978 atm) ( 45.9 L) P 2 (0.00310 L) = ( 298 K) ( 428 K) P 2 = ( 428 K) ( 0.978 atm) ( 45.9 L) = 2.08 x 10 4 atm ( 298 K) ( 0.00310 L)

Change of 2 Conditions : Problem II A weather balloon is released at the surface of the earth. If the volume was 100 m 3 at the surface ( T = 25 o C, P = 1 atm ), what will its volume be at its peak altitude of 90,000 ft where the temperature is - 90 o C and the pressure is 15 mm Hg? Initial Conditions Final Conditions V 1 = 100 m 3 V 2 =? T 1 = 25 o C + 273.15 T 2 = -90 o C +273.15 = 298 K = 183 K P 1 = 1.0 atm P 2 = 15 mm Hg 760 mm Hg/ atm PV/T = nr = constant = 0.0197 atm

Change of 2 Conditions Problem II: continued P 1 x V P 1 2 x V 2 T = 1 T 2 V 2 = P 1 V 1 T 2 T 1 P 2 ( 1.0 atm) ( 100 m 3 ) (183 K) V 2 = = (298 K) (0.0197 atm) V 2 = 3,117.23 m 3 = 3,120 m 3 or 31 times the volume!!!

Example Problem: Molar Mass of a Gas from its weight and P,V,T Problem: A sample of natural gas is collected at 25.0 o C in a 250.0 ml flask. If the sample had a mass of 0.118 g at a pressure of 550.0 Torr, what is the molecular weight of the gas? Plan: Use the Ideal gas law to calculate n, then calculate the molar mass. Solution: P = 550.0 Torr x 1mm Hg 1.00 atm x = 0.724 atm 1 Torr 760 mm Hg V = 250.0 ml x 1.00 L = 0.250 L 1000 ml T = 25.0 o C + 273.15 K = 298.2 K n = M = mass / n = n = P V R T

Example problem 2: Molar Mass of a gas Problem: A volatile liquid is placed in a flask whose volume is 590.0 ml and allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of 100.0 o C and 736 mm Hg pressure. If the mass of the flask empty and after the experiment was 148.375g and 149.457 g, what is the molar mass of the liquid? Plan: Use the gas law to calculate the moles of gas and its mass to get M Solution: Pressure = 736 mm Hg x 1 atm = 0.9684 atm 760 mm Hg mass = 149.457g - 148.375g = 1.082 g Molar Mass = mass / n n = PV/RT Molar Mass =

Example problem 2: Molar Mass of a gas Problem: A volatile liquid is placed in a flask whose volume is 590.0 ml and allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of 100.0 o C and 736 mm Hg pressure. If the mass of the flask empty was 148.375g and after the experiment 149.457 g, what is the molar mass of the liquid? Plan: Use the gas law to calculate the moles, then its molar mass. Solution: Pressure = 736 mm Hg x 1 atm = 0.9684 atm 760 mm Hg mass = 149.457g - 148.375g = 1.082 g ( 0.9684 atm)(0.590 L) n = (PV/RT) = (0.0821 L atm/mol K)(373.2 K) = 0.0186 mol M = mass / n = 1.082 g / 0.0186 mol = 58.02 g/mol note: the compound is acetone C 3 H 6 O = MM = 58g mol!

Chemical Equation Calc - III Atoms (Molecules) Avogadro s Number Reactants 6.02 x 10 23 Molecules Moles Molecular Weight Mass g/mol Products Molarity moles / liter PV = nrt Solutions Gases

Sodium Azide Decomposition-I Sodium Azide (NaN 3 ) is used in some air bags in automobiles. Calculate the volume of Nitrogen gas generated at 21 o C and 823 mm Hg by the decomposition of 60.0 g of NaN 3. 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) mol NaN 3 = mol N 2 = mol NaN 3 x (3 mol N 2 / 2 mol NaN 3 ) V = nrt/p

Sodium Azide Decomposition-I Sodium Azide (NaN 3 ) is used in some air bags in automobiles. Calculate the volume of Nitrogen gas generated at 21 o C and 823 mm Hg by the decomposition of 60.0 g of NaN 3. 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) mol NaN 3 = 60.0 g NaN 3 / (65.02 g NaN 3 / mol NaN 3 ) = = 0.9228 mol NaN 3 mol N 2 = 0.9228 mol NaN 3 x (3 mol N 2 / 2 mol NaN 3 ) = 1.38 mol N 2

Sodium Azide Calc: - II PV = nrt V = nrt/p V = ( 1.38 mol) (0.08206 L atm / mol K) (294 K) 823 mm Hg x ( 1 atm / 760 mmhg ) V = 30.8 liters

Like Example 5.6 (P 151) Problem: Calculate the volume of carbon dioxide produced by the thermal decomposition of 36.8 g of calcium carbonate at 715 mm Hg, and 37 0 C according to the reaction below: Solution: CaCO 3(s) CaO (s) + CO 2 (g) 1 mol CaCO 36.8 g x 3 = 0.368 mol CaCO 100.1 g CaCO 3 3 Moles of CO 2 = mol CaCO 3 x (1 mol CaCO 3 / 1 mol CO 2 ) V = = nrt P

Like Example 5.6 (P 151) Problem: Calculate the volume of carbon dioxide produced by the thermal decomposition of 36.8 g of calcium carbonate at 715 mm Hg, and 37 0 C according to the reaction below: Solution: CaCO 3(s) CaO (s) + CO 2 (g) 1 mol CaCO 36.8 g x 3 100.1g CaCO = 0.368 mol CaCO 3 3 Moles of CO 2 = mol CaCO 3 x (1 mol CaCO 3 / 1 mol CO 2 ) nrt V = = P (0.368 mol CO 2 )(0.08206 L atm/mol K)(310.15K) (715mmHg/(760mm Hg/atm) V = 9.96 L

Partial Pressures P i

Figure 5.9: Partial pressure of each gas in a mixture of gases depends on the number of moles n i n total of that gas (n i ). Defined as P i = x P total

Ideal Gas Mixtures Ideal gas equation applies to the mixture as a whole and to each gas species (i) individually: P total V= n total RT P total is what is measured by pressure guages. Σ n i = n 1 +n 2 +n 3 = n total i P i V= n i RT for all species i. P i harder to measure. Note: Σ P i = P 1 +P 2 +P 3 = P total i

Partial Pressures - Prob #1 A 2.00 L flask contains 3.00 g of CO 2 and 0.10 g of Helium at a temperature of 17.0 o C. What are the Partial Pressures of each gas, and the total Pressure? T = 17 o C + 273 = 290 K n CO2 = 3.00 g CO 2 / 44.01 g CO 2 / mol CO 2 = 0.0682 mol CO 2 P CO2 = n CO2 RT/V P CO2 = P CO2 = 0.811 atm ( 0.0682 mol CO 2 ) ( 0.08206 L atm/mol K) ( 290 K) (2.00 L)

Partial Pressures Problem - #1 cont. n He = 0.10 g He / 4.003 g He / mol He = 0.025 mol He P He = n He RT/V P He = P He = 0.30 atm (0.025 mol) ( 0.08206 L atm / mol K) ( 290 K ) ( 2.00 L ) P Total = P CO2 + P He = 0.812 atm + 0.30 atm P Total = 1.11 atm

Mole fraction of species i: x i = n i /n total P i V= n i RT for all i, and Σ P i = P 1 +P 2 +P 3 = P total Σ n i = n 1 +n 2 +n 3 = n total Σ x i = x 1 +x 2 +x 3 = 1.00000 P i = n i RT/V = (ni/n total )n total RT/V = x i. P total P i = (ni/n total )P total = x i. P total

Example Problem using mole fractions A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm? Total # moles = 4.46 + 0.74 + 2.15 = 7.35 mol = n total x Ne = n Ne / n total = 4.46 mol Ne / 7.35 mol = 0.607 P Ne = x Ne P Total = 0.607 ( 2.00 atm) = 1.21 atm Ne x Ar = 0.74 mol Ar / 7.35 mol = 0.10 P Ar = x Ar P Total = 0.10 (2.00 atm) = 0.20 atm Ar x Xe = 2.15 mol Xe / 7.35 mol = 0.293 P Xe = x Xe P Total = 0.293 (2.00 atm) = 0.586 atm Xe

Saturation.....................

Figure 5.10: Production of O 2 (g) by thermal decomposition of KCIO3: It is mixed with vapor pressure of water.

Collection of Hydrogen gas over Water - Vapor pressure - I 2 HCl (aq) + Zn (s) ZnCl 2 (aq) + H 2 (g) Calculate the mass of Hydrogen gas collected over water if 156 ml of gas is collected at 20 o C and 769 mm Hg. P Total = P H2 + P H2O P H2 = P Total -P H2O FROM TABLE P H2 = T = 20 o C + 273 = 293 K P H2 = V = 0.156 L n H2 =P H2 V/RT = Convert to mass

Collection of Hydrogen gas over Water - Vapor pressure - I 2 HCl (aq) + Zn (s) ZnCl 2 (aq) + H 2 (g) Calculate the mass of Hydrogen gas collected over water if 156 ml of gas is collected at 20 o C and 769 mm Hg. P Total = P H2 + P H2O P H2 = P Total -P H2O FROM TABLE P H2 = 769 mm Hg - 17.5 mm Hg = 752 mm Hg T = 20 o C + 273 = 293 K P H2 = 752 mm Hg /760 mm Hg /1 atm = 0.989 atm V = 0.156 L

COLLECTION OVER WATER CONT.-II P H2 V = n H2 RT n H2 = P H2 V / RT n H2 = (0.989 atm) (0.156 L) (0.0821 L atm/mol K) (293 K) n H2 = 0.00641 mol Mass H 2 = 0.00641 mol x 2.01 g H 2 / mol H 2 = 0.0129 g H 2

Chemical Equation Calc - III Atoms (Molecules) Avogadro s Number Reactants 6.02 x 10 23 Molecules Moles Molecular Weight Mass g/mol Products Molarity moles / liter P i V= n i RT Solutions Gases

Gas Law Stoichiometry - I - NH 3 + HCl Problem: A slide separating two containers is removed, and the gases are allowed to mix and react. The first container with a volume of 2.79 L contains Ammonia gas at a pressure of 0.776 atm and temperature = 18.7 o C. The second with a volume of 1.16 L contains HCl gas at a pressure of 0.932 atm and a temperature of 18.7 o C. What mass of solid ammonium chloride will be formed, and what will be remaining in the container, and what is the pressure? Plan: This is a limiting reactant problem, so we must calculate the moles of each reactant using the gas law to determine the limiting reagent. Then we can calculate the mass of product, and determine what is left in the combined volume of the container, and the conditions. Solution: NH 3 HCl Reaction: NH 3 (g) + HCl (g) NH 4 Cl (s) T NH3 = 18.7 o C + 273.15 = 291.9 K

Gas Law Stoichiometry - II - NH 3 + HCl n i = P i V RT NH 3 (g) + HCl (g) NH 4 Cl (s) (0.776 atm) (2.79 L) n NH3 = = 0.0903 mol NH 3 1 (0.0821 L atm/mol K) (291.9 K) (0.932 atm) (1.16 L) n HCl = = 0.0451 mol HCl 1 (0.0821 L atm/mol K) (291.9 K) limiting reactant Therefore the product will be 0.0451 mol NH 4 Cl or 2.28 g NH 4 Cl Ammonia remaining = 0.0903 mol - 0.0451 mol = 0.0452 mol NH 3 V = both volumes added = 1.16 L + 2.79 L = 3.95 L n P NH3 = NH3 RT (0.0452 mol) (0.0821 L atm/mol K) (291.9 K) = = 0.274 atm V (3.95 L)

Kinetic Molecular Theory The kinetic energy associated with the molecules velocity: KE = (1/2) mass x u 2 The temperature of matter is a measure of its thermal energy, such that T is proportional to the average KE. so that their velocity (u) is proportional to T. Gas molecules make collisions with the walls, and on average simply reverse their momentum in the direction perpendicular to the wall. This gives rise to the pressure (force per unit area) on the wall.

Figure 5.11: An ideal gas particle in a cube whose sides are of length L (in meters).

Figure 5.12: (b) The velocity u of any gas particle can be broken down into three mutually perpendicular components, u x, u y, and u z. Momentum = mass x velocity = m. u

Figure 5.13: (a) In collision: exact reversal of the x component of the velocity, so that momentum change on impact is 2 x mass x u x. (b) The # impacts on the shaded walls per unit time can be calculated from the velocity and the number density. Proportional to both. (c) Force on wall = the average momentum change per unit time = (momentum change per impact) x (# impacts per unit time)

Velocity and Energy Kinetic Energy for one gas molecule = E K = ½ mu 2 Average Kinetic Energy of a mole of gas = (KE) avg = 3/2 RT independent of gas identity (use R in J/mol K) Average Kinetic Energy of one molecule = E K = (KE) avg / N A = 3/2 RT / N A independent of mass, identity = ½ mu 2 (mean value of u 2 ) ½mu 2 = 3/2 RT/N A u 2 = 3RT/(mN A ) = 3RT/M since m = M/N A (u 2 ) = (3RT/M) Root mean square velocity Must use R in J/mol K, and remember that 1 J = 1 kg m 2 /s 2

Figure 5.16: A plot of the relative number of N 2 molecules that have a given velocity at three temperatures. Also: At same T, heavier gases go slower, as 1/ M: Average u 2 /u 1 = (M 1 /M 2 ) 1/2 Kinetic energy per mole = ½ Mu 2 = 3/2 RT, independent of M HEAT!

Velocity, Energy and HEAT Kinetic Energy for one gas molecule = E K = ½ mu 2 Average Kinetic Energy of a mole of gas = (KE) avg = 3/2 RT independent of gas identity (use R in J/mol K) Average Kinetic Energy of one molecule = E K,avg = (KE) avg / N A = 3/2 RT / N A independent of mass, identity = ½ mu 2 (mean value of u 2 ) ½mu 2 = 3/2 RT/N A u 2 = 3RT/(mN A ) = 3RT/M since m = M/N A (u 2 ) = (3RT/M) Root mean square velocity Must use R in J/mol K, and remember that 1 J = 1 kg m 2 /s 2 Molecular picture of HEAT!

Molecular Mass and Molecular Speeds - I Problem: Calculate the molecular speeds of the molecules of hydrogen, methane, and carbon dioxide at 300K! Plan: Use the equations for the average kinetic energy of a molecule, and the relationship between kinetic energy and velocity to calculate the average speeds of the three molecules. Solution: for Hydrogen, H 2 = 2.016 g/mol E K,avg = 3 R x T = 1.5 x 8.314 J/mol K x 300K = 2 N A 6.022 x 10 23 molecules/mol E K,avg = 6.213x10-21 J/molecule = 1/2 mu 2 m = M/N A = 2.016 x 10-3 kg/mole = 3.348x10-27 kg/molecule 6.022 x 10 23 molecules/mole E K,avg = 6.213x10-21 J/molecule = 1/2 mu 2 =1.674x10-27 kg/molecule (u 2 ) u = 1.926 x 10 3 [J/kg x (1 kg m 2 /s 2 )/J] 1/2 = 1.926 x 10 3 m/s = 1,926 m/s

Molecular Mass and Molecular Speeds - II For methane, CH 4 : M = 16.04 g/mole = 16.04 x 10-3 kg/mole (u 2 ) = (3RT/M) R = 8.314 J/mol K = 8.314 kg m 2 /(s 2 mol K) (u 2 ) =

Molecular Mass and Molecular Speeds - II For methane, CH 4 : M = 16.04 g/mole = 16.04 x 10-3 kg/mole (u 2 ) = (3RT/M) R = 8.314 J/mol K = 8.314 kg m 2 /(s 2 mol K) (u 2 ) = [(3x 8.314 kg m 2 /(s 2 mol K) x 300 K)/ 16.04 x 10-3 kg/mole] Note: 1 J = 1 kg m 2 /s 2 (u 2 ) = 6.838 x 10 2 m/s = 683.8 m/s Root mean square velocity

Molecular Mass and Molecular Speeds - III for Carbon dioxide CO 2 = 44.01 g/mole E K,avg = 3 R x T = 1.5 x 8.314 J/mol K x 300K = 2 N A 6.022 x 10 23 molecules/mole E K,avg = 6.213 x 10-21 J/molecule = 1/2mu 2 m = 44.01 x 10-3 kg/mole = 7.308 x 10-26 kg/molecule 6.022 x 10 23 molecules/mole 6.213 x 10-21 J/molecule = 3.654 x 10-26 kg/molecule (u 2 ) u = 4.124 x 10 2 (J/kg) 1/2 = 412.4 m/s

Molecular Mass and Molecular Speeds - IV Molecule H 2 CH 4 CO 2 Molecular Mass (g/mol) Kinetic Energy (J/molecule) Velocity at 300 K (m/s) 1/ M 2.016 16.04 44.01 6.213 x 10-21 6.213 x 10-21 6.213 x 10-21 1,926 683.8 412.4

Diffusion vs Effusion (Rates go as gas velocity) Diffusion - One gas mixing into another gas, or gases, of which the molecules are colliding with each other, and exchanging energy between molecules. Effusion - A gas escaping from a container into a vacuum (where there are very few molecules to block it s molecules paths)

Figure 5.19: (a) demonstration of the relative diffusion rates of NH 3 and HCL molecules through air.

Diffusion Rates: proportional to average velocity or to 1/ M Rate 1 / Rate 2 = u 1 /u 2 = (M 2 /M 1 ) 1/2 HCl = 36.46 g/mol NH 3 = 17.03 g/mol Rate NH3 = Rate HCl x ( 36.46 / 17.03 ) 1/ 2 Rate NH3 = Rate HCl x 1.463

Figure 5.18: Effusion rate of a gas into an evacuated chamber = collision rate with imaginary surface.

Example 5.9 (P168) Calculate the impact rate on a 1.00 cm 2 section of wall on a vessel containing oxygen gas at a pressure of 1.00 atm and 27 0 C. Solution: Calculate Z A : Z A = A N RT V 2πM 1.00 atm A = 1.00 x 10-4 m 2 N = P = ( V RT 0.08206 L atm) (300. K) K mol N = 4.06 x 10 V -2 Mol x 6.022 x 10 23 molecules x 1000 L = L mol m 3 = 2.44 x 10 25 molecules/m 3 M = 32.0 g/mol x 1 kg = 3.2 x 10-2 kg/mol 1000g Z A = (1.00 x 10-4 m 2 ) (2.44 x 10 25 m -3 ) x 8.3145 J (300. K) K mol Z 2(3.14)( 3.20 x 10-2 kg/mol) A = 2.72 x 10 23 collisions / s

Figure 5.21: The cylinder swept out by a gas molecule of diameter d: It hits those molecules inside the cylinder. Z = Collision rate of one molecule with other molecules in container = # collisions/s = Volume swept per unit time x (N/V) = 4d 2 (πrt/m) 1/2 x (N/V)

Like Example 5.10 (P 170) Problem: Calculate the collision frequency for a Nitrogen molecule in a sample of pure nitrogen gas at 28 0 C and 2.0 atm. Assume that the diameter of an N 2 molecule is 290 pm. Z = 4 (N/V) d 2 (πrt/m) 1/2 Solution: n V = P 2.0 atm = = 8.097 x 10-2 mol/l RT ( 0.08206 L atm )(301. K) K mol N V = (8.097 x 10-2 mol/l)( 6.022 x 10 23 molecules/mol)(1000l/m 3 ) N V = 4.876 x 10 25 molecules/m 3 d= 290 pm = 2.90 x 10-10 m For N 2, M = 2.80 x 10-2 kg/mol Z = 4 (N/V) d 2 (πrt/m) 1/2 Z = 4(4.876 x 10 25 m -3 )(1.45 x 10-10 m) 2 x π (8.3145 J /mol K)(301 K) Z = 1.38 x 10 19 collisions / sec 2.80 x 10-2 kg/mol