Gases

Gases I. comparison of solids, liquids and Gasses A. fluids- things that flow - gases and liquids B. condensed states - liquids and solids C. Vapor - refers to a gas that is formed by evaporation of a liquid or the sublimation of a solid (sublimation - going directly from a solid to a gas)

II some common properties of gases A. Gases can be compressed into smaller volumes; that is, their densities can be increased by applying increased pressure. B. Gases exert pressure on their surroundings; in turn, pressure must be exerted to confine gases. C. Gases expand without limits, and so gas samples completely and uniformly occupy the volume of any container.

D. Gases diffuse into each other, and so samples of gas placed in the same container mix completely. Conversely,different gases in a mixture do not separate on standing. 1. diffusion - going from a high concentration to a low concentration

E. expand to fill container F. take shape of container G. some are invisible

H. some are combustible I. four properties determine the physical behavior of a gas 1. amount of gas 2. volume of the gas 3. temperature of the gas 4. pressure of the gas

Elements that exist as gases at 250C and 1 atmosphere 5.1

5.1

1. Pressure = force = Newtons = pascal Area m2 J. exert Pressure 1. Pressure = force = Newtons = pascal Area m2 a. Named after Blaise Pascal (1625-1662) - father of modern Hydraulics b. units Newtons/m2 = 1 Pascal (Pa) c. 1 Pascal is rather small - more comely used unit is the kilopascal 2. liquid pressure = g x h x d

(force = mass x acceleration) Area Barometer Pressure = (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mm Hg = 760 torr = 101,325 Pa = 14.7 psi = 29.92 in. Hg 5.2

10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm 5.2

4. pressure of gases is due to the billion collision/sec of the gas molecules with the walls of the container 5. Barometer - device used to measure gas pressure a. invented by Evangelista Torricelli

Boyle’s Law P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. P1V1 = P2 V2 Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

Gas Law Problems have 5 steps

e.g. A gas under a pressure of 750. mm of Hg occupies 25 liters. What will its volume be under a pressure of 800. mm of Hg? Temperature remains constant. P1 = 750. mm of Hg = 25 L Step 1 V1 V2 = X P2 = 800. mm or Hg P1V1 = P2V2 - base equation Step 2

P1V1 = V2 - working equation P2 Divide by P2 on both sides P1V1 = P2V2 P2 P2 Make sure your unkown is on the top P1V1 = V2 - working equation P2 Step 3 (750. mm of Hg)( 25 L) = (800. mm of Hg) 23 L Step 5 Step 4

Charles’s Law If n and P are constant, then V α T V and T are directly proportional. V1 V2 = T1 T2 If one temperature goes up, the volume goes up! Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

There are 3 temperature scales remember - temperature is a measure of molecular motions Fahrenheit Celsius Can you have F and C temperatures below 0? If temperature is a measure of molecular movement - How can you have negative temperatures?

We had to come up with a new temperature scale Kelvin K = C + 273 What is 22 C in Kelvin? K = 22 + 273 = 295 K

Absolute zero Really Really Cold 0 K = no molecular motion 0 K = -273 C = -460 F Really Really Cold

A gas occupies a volume of 2.5 liters at a temperature Example of Charles Law A gas occupies a volume of 2.5 liters at a temperature of 22 C. What temperature is required for the gas to occupy 3.0 liters V1 = 2.5 l = 22 C + 273 = 295 K T1 V2 = 3.0 liters T2 = x Step 1

V1 = V2 T1 T2 Step 2 First you must move your unknown to the top Multiply both sides by T2 V1 = V2 T1 T2 T2 T2 T1 T2 V1 = V2 T1 T1 V1 V1

T2 = V2T1 V1 Step 3 = (3.0 liters)(295 K) 2.5 L Step 4 354 K = 350 K

Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P1 P2 = T1 T2 If one temperature goes up, the pressure goes up! Joseph Louis Gay-Lussac (1778-1850)

Combined Gas Law The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P1 V1 P2 V2 = T1 T2

By doing step 1, you will be able to select the right equations. If any on the variables are held constant, disregard them.

e.g. A gas occupies 10.0 L at 700. mm of Hg and a temperature of 30.0 C. What volume will the gas Occupy at 800. mm of Hg and 20.0 C V1=10.0 L P1= 700. mm of Hg T1 = 273 +30.0 = 303.0 K Step 1 V2 = x P2 = 800. mm of Hg T2= 273 + 20.0 C = 293.0 K

Step 2 P1V1 = P2V2 T1 T2 T2P1V1 = V2 Step 3 P2T1 (293.0 K)(700. mm of Hg)(10.0 L) (800. mm of Hg)(303.0 K) Step 4 8.46 L = V2 Step 5

What temperature is required to change 250 ml of a gas at 50.0 C and 860.0 mm of Hg to 150 ml at 1200. mm of Hg V1= 250 P1= 800.0 mm of Hg T1 = 273 +50.0 = 323.0 K V2 = 150 ml P2 = 1200. mm of Hg T2=x

Now just put your numbers in P1V1 = P2V2 T1 T2 T2 T2 P1V1 = P2V2 T1 T2 T1 T2P1V1 T1 = P2V2 T1 P1V1 P1V1 T2 = P2V2T1 P1V1 Now just put your numbers in

on handout - due question 7-15 7-10 are Charles law 11-15 are combined Gas Laws Stop here

Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

Avogadro’s Law V a number of moles (n) V = constant x n V1/n1 = V2/n2 Constant temperature Constant pressure V = constant x n V1/n1 = V2/n2 5.3

Ideal Gas Equation 1 Boyle’s law: V a (at constant n and T) P Charles’ law: V a T (at constant n and P) Avogadro’s law: V a n (at constant P and T) n = number of moles V a nT P V = constant x = R nT P R is the gas constant PV = nRT 5.4

R = 0.082057 L • atm / (mol • K) or .0821 L Atm mole K The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. 22.4 L is standard volume PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K) or .0821 L Atm mole K 5.4

When you use the ideal gas equation Pressure must be in atmospheres 1 atm = 760 mm of Hg Temperature must be in Kelvin volume in liters R is the ideal gas constant = .0821 L Atm mole K

How many liters will 12.0 g of O2 occupy at 780.0 mm of Hg and 30.0 C P = (780.0 mm of Hg) (1atm ) = 1.026 Atm 760 mm of Hg V = X (12.0 g O2)( 1mole ) = .375 moles 32.0 g O2 n = .0821 L atm/mole K R = T = K = 30.0 C + 273 = 303.0 K

PV = nRT V = nRT/P .0821 L atm mole K 303.0 K V= .375 moles 1.026 Atm V = 9.09 L

65 g of a CH4 occupies 30.0 L at a temperature of 10.0 C. What pressure must be exerted on this gas? P = X V= 30.0 L n = (65 g CH4)( 1 mole CH4) = 4.1 moles CH4 (16 g CH4) R = .0821 L Atm Mole K T = 10.0C + 273 = 283.0 K

PV = nRT P = nRT V P = (4.1 mole CH4)(.0821 L Atm ) ( 283.0 K) (30.0 L) ( mole K ) P = 3.1 Atm

What is the molecular weight of a 2.0 g sample of a gas that occupies 5.0 L at a pressure of 0.303 Atm at 22 C. P= 0.303 Atm V= 5.0 L n = x R = 0.0821 L Atm/mole K T = 22 C + 273 ==295 K PV= nRT

remember the units for M.W. is g/mole PV= nRT PV = n RT (0.303 Atm)(5.0 L)(mole K) (295 K) (0.0821 L Atm) n= 0.0625 moles remember the units for M.W. is g/mole M.W. = 2.0 g .0625 moles = 32 g/mole

e.G How many liters will .639 mole of Cl2 occupy under standard conditions

x PV = nRT P = V = n = R = T = 1 Atm V = nRT P .639 moles Cl2 .0821 L Atm Mole K 273 K V = (.639 moles Cl2) (.0821 L Atm) (273 K) 14.3 L = (1 Atm) Mole K

You need your books today On page 358 – do questions 16-20

C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l) Gas Stoichiometry Mass - Volume problems What is the volume of CO2 produced when 5.60 g of glucose are used up in the reaction at STP: C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l) g C6H12O6 mol C6H12O6 mol CO2 V CO2 5.5

C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l) 1 mole 6 moles C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l) x L 5.60 g 22.4 L/mol 180. g/mole molar volume ) ( ) ( ) ( 1 mole C6H12O6 6 moles CO2 (5.60 gC6H12O6) 22.4 L CO2 180. g C6H12O6 1 mole C6H12O6 1 mole CO2 4.18 L

When a problem gives you grams - change it to moles How many liters of Cl2 is required to react with 50.0 g of K to produce KCl at STP? K + Cl 2 -----> KCl 2 mole 1 mole 2 2 50.0 g X L 39.1 g/mole 22.4 L/mole When a problem gives you grams - change it to moles ( 1 mole K ) (1 mole Cl2 ) (50.0 g K) (22.4 L) (39.1g K) ( 2 moles K ) ( 1 mole Cl2 ) = 14.3 L

Volume - mass problems

During a collision, automobile air bags are inflated by N2 gas formed by the explosive decomposition of sodium azide at STP What mass of sodium azide would be needed to inflate 31.0 L 2 NaN3 ====> Na + N2 2 3 x g 31.0 L 65.0 g/mole 22.4 L/mole (2 moles NaN3) (65.0 g NaN3) (1 mole N2) (31.0 L N2) ( 1 mole NaN3) ( 22.4 L N2) ( 3 moles N2)

Try these two How many liters of Hydrogen are produced from 110.0 g of water at STP? H2O => H2 + O2 How many grams of Na are required to produce 200.0 L of H2 when it reacts with water at STP Na + H2O => NaOH + H2

dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M = A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.00C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L 2.21 g L 1 atm x 0.0821 x 300.15 K L•atm mol•K M = M = 54.6 g/mol 5.3

Gas Stoichiometry continued Volume - Volume problems As the name suggest - you start with a volume and end with a volume What volume of O2 is required to completely burn 70.0 L of CH4? CH4 + O2 ====> CO2 + H2O 2 2 As always - make sure you balance all equations

You can see that a balanced chemical equation can CH4 + 2O2 ====> CO2 + 2 H2O (70.0 L of CH4) ( 2 liters of O2 ) = 140. L of O2 ( 1 liter of CH4 ) You can see that a balanced chemical equation can be used for a number of things

1 How many liters of O2 are produced when 50.0 L of H2O2 decomposes H2O2 ===> H2O + O2 2. How many liters of O2 are required to burn 700.0 L of propane? C3H8 + O2 ==> CO2 + H2O

d is the density of the gas in g/L Density (d) Calculations m is the mass of the gas in g m V = PM RT d = M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4

Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal = P1 + P2 5.6

PA = nART V PB = nBRT V PT = PA + PB XA = nA nA + nB XB = nB nA + nB Consider a case in which two gases, A and B, are in a container of volume V. PA = nART V nA is the number of moles of A PB = nBRT V nB is the number of moles of B mole fraction (Xi) = ni nT PT = PA + PB XA = nA nA + nB XB = nB nA + nB Pi = Xi PT PB = XB PT PA = XA PT 5.6

5.6

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH4Cl NH3 17 g/mol HCl 36 g/mol 5.7

GAS DIFFUSION AND EFFUSION diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container.

GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv2 Rate of effusion is inversely proportional to its molar mass. Thomas Graham, 1805-1869. Professor in Glasgow and London.

GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to T inversely proportional to M. Therefore, He effuses more rapidly than O2 at same T. He

Gas Diffusion relation of mass to rate of diffusion HCl and NH3 diffuse from opposite ends of tube. Gases meet to form NH4Cl HCl heavier than NH3 Therefore, NH4Cl forms closer to HCl end of tube.

(8.24 mols CH4 + 0.421mols C2H6 + 0.116 mols C3H8.) Xpropane = A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 mols C3H8. (8.24 mols CH4 + 0.421mols C2H6 + 0.116 mols C3H8.) Xpropane = Xpropane = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm 5.6

PT = PO + PH O 2KClO3 (s) 2KCl (s) + 3O2 (g) 5.6 Bottle full of oxygen gas and water vapor 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO + PH O 2 5.6

Kinetic theory of gases and … Compressibility of Gases Boyle’s Law P a collision rate with wall Collision rate a number density Number density a 1/V P a 1/V Charles’ Law Collision rate a average kinetic energy of gas molecules Average kinetic energy a T P a T 5.7

Kinetic Molecular Theory of Gases A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. Gas molecules exert neither attractive nor repulsive forces on one another. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy http://www.ewellcastle.co.uk/science/pages/P_constantM.html 5.7

Kinetic theory of gases and … Avogadro’s Law P a collision rate with wall Collision rate a number density Number density a n P a n Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas Ptotal = SPi 5.7

What volume of O2 is required to react with 3.00 g of H2 at 22 C and 1.20 atm 2 H2 + O2 ===> H2O 2 ( 1 mole H2) ( 1 mole of O2 ) (3.00g H2) ( 2.02 g H2) ( 2 moles H2 ) = .743 moles O2

P = V = n = R = T = 2.0 Atm PV = nRT 250.0 L PV = n RT x .0821 L Atm Mole K 20.0 C + 273 = 293.0 K (2.0 Atm) (250.0 L) Mole K 21 moles = 0821 L Atm 293.0 K

1. How many grams of Na is required to react with water to produce 250.0 L of H2 at 20.0 C and 2.0 atm? NaOH is the other product. Na + H2O ------> NaOH + H2 2 2 2 Because you are looking for g and start with a volume, it is a volume - mass problem Start with the Ideal gas equation

(21 moles of H2) ( 2 moles of Na ) ( 23 g Na) (1 mole H2) (1 mole of Na) = 970 g Na