GAS LAWS

IMPORTANCE OF GASES  • Airbags fill with nitrogen gas, N 2 , in an accident  • Gas is generated by the decomposition of sodium azide, NaN 3 (s) 2 NaN 3 (s) →2 Na(s) + 3 N 2 (g)

IMPORTANCE OF GASES  Cellular respiration is going on throughout your body. It involves two gases • Oxygen • Carbon Dioxide  C 6 H 12 O 6 + O 2 →C O 2 + H 2 O

PRESSURE  The mercury barometer was developed in 1643 by  Evangelista Torricelli The column height is dependent on atmospheric pressure   1 standard atmosphere = 1 atm = 760 mmHg =760 torr (named after Torricelli) The SI unit of pressure is the PASCAL  1 atm = 101.3 kPa

PRESSURE  P=FA  Pressure = force x area  The force comes from the gas molecules hitting the sides of the container.

 Average atmospheric pressure at sea level is 760 torr or 760 mm Hg or 1 atm or 101, 325 Pa

KINETIC MOLECULAR THEORY (KMT) • Gases consist of large numbers of tiny particles that are far apart relative to their size. The volume each particle occupies is ignored.

KINETIC MOLECULAR THEORY (KMT)  Collisions between gas particles and between particles and the walls of the containers are said to be ELASTIC.

 This means that there is NO loss of kinetic energy…

KINETIC MOLECULAR THEORY (KMT)  Gas molecules are ALWAYS moving. Moving molecules have KINETIC ENERGY  At the same temperature, all gases have the same average KE  As temperature goes up, average kinetic energy goes up

KINETIC MOLECULAR THEORY (KMT) There are NO forces of attraction or repulsion between gas particles When they collide they bounce apart – no energy is exchanged, lost or gained, no damage is done

KINETIC MOLECULAR THEORY (KMT) Velocity of gas molecules is affected by the temperature of the gas.

All gases at the same temperature have the same AVERAGE kinetic energy.

Lighter gas particles travel faster than heavier gas particles.

VELOCITY OF GAS MOLECULES Molecules of any given gas have a range of different speeds

DIFFUSION OF A GAS  Diffusion is the gradual mixing of two or more gases due to their spontaneous random motion

EFFUSION OF A GAS  Effusion is the process by which molecules of a gas confined in a container randomly pass through a tiny opening in the container.

 Think balloons!

 Why does a mylar balloon stay inflated longer than a regular latex balloon?

 Which gas would you expect to effuse faster, hydrogen or carbon dioxide?

 • Why?

GENERAL PROPERTIES OF GASES 

There is a lot of free space between the particles in a gas



Gases can be expanded infinitely



Gases occupy containers uniformly and completely



Gases diffuse and mix rapidly

BOYLE’S LAW 

As Pressure increases, Volume decreases



P↑, V ↓



As Pressure decreases, Volume increases



P ↓, V ↑



PV =k, where k is a constant



Boyle’s Law : P1 V1 = P2 V2

BOYLE’S LAW  At constant temperature, T, and number of moles, n  Volume is inversely proportional to pressure  Go to http://chem.salve.edu/c hemistry/boyle.asp

BOYLE’S LAW EXAMPLE #1  A gas occupies a volume of 4.0 L at 760 torr. What volume will it occupy at 380 torr?



P1 V1 = P2 V2

 Rearrange to solve for the unknown.

BOYLE’S LAW EXAMPLE #2  A gas occupies a volume of 12 L at 1.2 atm. What volume will it occupy at 2.4 atm?



P1 V1 = P2 V2

 Rearrange to solve for the unknown

TEMPERATURE  For gas laws temperature MUST be converted to Kelvin.

 K = o C + 273 o F o C K  H 2 O boils  Room Temp Fahrenheit 212 68  H 2 O freezes  Absolute Zero 32 - 460 Celsius 100 20 0 - 273 Kelvin 373 293 273 0

CHARLES’ LAW  For a sample of gas under constant pressure,  V = kT  The temperature, T must be in Kelvin  As V increases, T increases  As V decreases, T decreases  So for a sample of any gas at constant pressure and number of moles, V is directly proportional to T

CHARLES’ LAW

GAY LUSSAC’S LAW  For a sample of gas under constant volume  P = kT  As T increases, P increases  As T decreases, P decreases  So for a sample of any gas at constant volume and number of moles, T is directly proportional to P  P 1 = P 2  T 1 T 2

COMBINED GAS LAW  changes in temperature, pressure and volume simultaneously.

  When this happens all three variables MUST be dealt with at once By combining Boyle’s Charles and Gay Lussac’s law for a fixed amount of gas the combined gas law can be found.



Temperature must be in Kelvin

CGL EXAMPLE  A gas sample has a volume of 105 L at a pressure of 985 torr and a temperature of 27 o C. What volume will it occupy at a pressure of 760 torr and a temperature of 0 o C?

 Check all units, temperature MUST be in Kelvins  Identify variables  Plug and Chug…  Solve!

CGL EXAMPLE  V1 = 105 L V2 = ?

 P1 = 985 torr P2 = 76 torr  t1 = 27 o C t2 = 0 o C  Convert C into T K  T1 = 27 o C + 273 = 300 K  T2 = 0 o C + 273 = 273 K  Rearrange equation to isolate unknown variable

STANDARD TEMPERATURE AND PRESSURE  Conditions of standard temperature and pressure allow for comparisons between volumes of different gases.

 Standard Pressure = 1 atm  Standard Temperature = 273K = 0 o C

STP PROBLEM  A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL and the temperature is increased to 30.0

o C  What is the new volume in kPa?



Check units; convert o C into K and units of pressure into kPa

 Isolate variable  Plug “n” chug”  Solve… STP PROBLEM

IDEAL GAS LAW  Any gas can be defined by conditions of:  Pressure, P  Volume,V  Temperature, T and  Number of moles, n  It combines the gas laws and can be derived from experiment and theory

PV = NRT

 The ideal gas law holds for all gases and mixtures of gases as long as they are ideal gases.



What is an ideal gases?

 It is an imaginary gas that fits ALL the assumptions of KMT  P =Pressure (atm, kPa, torr, mmHg)  V = Volume (L)  n = Number of moles (mol)  T = Temperature (K)  R = Gas Law Constant

PV = NRT



R is the Gas Law Constant



It’s value depends on the units of pressure

STP PROBLEM  What is the volume of 1 mol of a gas at STP?

 This volume is known as the MOLAR VOLUME.

1 MOLE OF ANY GAS AT STP HAS THE SAME VOLUME:  22.4 Liters 

1 mol H2 at STP = 22.4 L



1 mol N2 at STP = 22.4 L



1 mol Ar at STP = 22.4 L



1 mol He at STP = 22.4 L



1 mol Air at STP = 22.4 L

USING PV = NRT  How much nitrogen gas, N 2 , would fill a small room with a volume of 960. cubic feet, (27 000 L) to a pressure of 745 mmHg at 25 o C?

 R = 0.0821 L•atm/K• .mol

 Solution:  Put all data into correct units  V = 27 000 L  T= 25 o C + 273= 298 K

USING PV = NRT  How much nitrogen gas, N 2 , would fill a small room with a volume of 960. cubic feet, (27 000 L) to a pressure of 745 mmHg at 25 o C?R = 0.0821 L•atm/K• .mol

 Solution:  Rearrange PV = nRT to isolate the unknown variable.



n=PV /RT

USING PV = NRT  How much nitrogen gas, N 2 , would fill a small room with a volume of 960. cubic feet, (27 000 L) to a pressure of 745 mmHg at 25 o C?

 • R = 0.0821 L•atm/K• .mol

 • Solution: n = 1.1 x 103 mol

USING PV = NRT, EXAMPLE #2  What volume would 4.00 mol of He occupy at a pressure of 748 torr and a temperature of 30.0

o C?

 R = 0.0821 L•atm/K• mol V = 101 L

DALTON’S LAW  The partial pressure of a gas in a mixture is simply the fraction of the total pressure which that gas exerts  Gas % atm 0.13

 Other gases  Carbon Dioxide,  Argon,  Oxygen,  Nitrogen, Trace 0.03 0.94 20.99 78.03 59.5

593.0

0.23

7.14

GAS MIXTURES AND PARTIAL PRESSURES 

PTOTAL= P1 + P2 + …

 •

nTOTAL= nA + nB + …

GAS MIXTURES AND PARTIAL PRESSURES  Example…  Calculate the pressure of a 10.0 L container with 0.200 mole methane, 0,300 mole hydrogen and 0.400 mole nitrogen at 298 K.



P = 2.20 atm

4 TOOLS FOR SOLVING GAS STOICHIOMETRY PROBLEMS  1. Use the Molar Mass to convert Mass (g) → Moles or Moles → Mass (g)  2. Use the Balanced Chemical Equation to convert Moles A → Moles B  3. At STP: 1 mole of any gas = 22.4 L  4. If not at STP, use PV = nRT

GAS STOICH- METHOD 1  How many liters of oxygen, O 2 will be formed at STP from the decomposition of 112g potassium chlorate, KClO 3 ?

 Method 1: use 1mol gas at STP = 22.4 L

GAS STOICH- METHOD 2  How many liters of oxygen, O 2 will be formed at STP from the decomposition of 112g potassium  chlorate, KClO 3 ?

• Method 2: using Pv = nRT 

V = 30.7 L