Transcript Gas Laws - Moreau Catholic High School
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