Chapter 14: Gases
Chapter 14: Gases
Getting started with gas calculations:
Before we can start talking about how gases behave in numerical terms, we need to define some of the quantitative properties that are characteristic of gases:
• Pressure (P): The force of gas molecules as they hit the sides of the container in which they are placed.
o Common units of pressure:
▪ atmospheres (atm): The average air pressure at sea level.
▪ kilopascals (kPa): The SI unit for pressure; 101.325 kPa = 1 atm.
▪ mm Hg (Torr): 760 Torr = 1 atm.
• Volume (V): The amount of space in which a gas is enclosed.
o The only commonly used unit of volume is liters (L).
• Temperature (T): A measurement of the amount of energy that molecules have. The higher the energy, the higher the temperature.
o Common units of pressure:
▪ Kelvin (K): The only units that can be used when doing numerical problems with gases.
▪ Degrees Celsius (0C): Must be converted to Kelvin before doing problems (by adding 273).
Other terms frequently used:
• STP: Stands for “standard temperature and pressure”, namely 273 K (00 C) and 1.00 atm.
• “Room temperature”: 298 K (250 C)
A whirlwind tour through the early gas laws:
Boyle’s Law: P1V1 = P2V2
• For any gas, the product of the pressure and the volume before a change is equal to the product of the pressure and the volume after a change.
• In plain English, what this means is that if you put pressure on a gas, it gets smaller. If you decrease pressure on a gas, it gets larger.
• Demonstration: Have them sit on balloons to demonstrate that decreasing the volume increases the pressure inside the balloon so much that it pops!
o Another demonstration (if the equipment is available): Use a vacuum pump to decrease the pressure around a balloon – the balloon will get bigger (until it pops, probably).
• Sample problems:
o If I have 10 L of gas at a pressure of 1 atm and double the pressure, what will the new volume of the gas be? 5 L
o If 250 L of a gas is in a sealed container at a pressure of 1.5 atm and I decrease the volume of the container to 100 L, what will the gas pressure inside the container be? 3.75 atm.
Charles’s Law: [pic]
• If you increase the temperature of a gas, the volume also increases. (Note: The temperature must be in Kelvin, NOT degrees centigrade)
• Why? The KMT tells us that the amount of energy that a gas has is determined by the temperature of the gas. The more energy a gas has, the faster the gas molecules move away from each other, causing more space between the molecules and a larger overall volume.
• Demonstration: Show the students two balloons, both of which originally had the same volume. One goes into a freezer and comes out smaller than the one at room temperature.
o Alternate demonstration: Do a mini lab where students pop the rubber from a rubber glove out of the end of a test tube by heating it. Will need rubber gloves, large test tubes, and rubber bands to do this demo.
• Examples:
o If you heat a 1.25 L balloon from a temperature of 250 C to 400 C, what will the new volume of the balloon be? 1.31 L
o What temperature will be required to raise the volume of a 1.0 L balloon to 1.25 L if the initial temperature is 250 C? 373 K
Gay-Lussac’s Law: [pic]
• When you increase the temperature of an enclosed gas, the pressure of the gas goes up.
• This is why it’s a bad idea to put a spray can into a campfire – eventually the pressure rises so much that the sides of the can split and the can explodes.
• Example:
o If you have a spray can at a pressure of 20 atm at room temperature and put it into a campfire at a temperature of 12000 C, what will the pressure in the canister be right before it explodes? 98.9 atm
The combined gas law:
• If we put the last three gas laws together, we can devise another law that encompasses all three of them (making it unnecessary to memorize the three):
[pic]
• How to use this law: Whenever you have a problem in which you change the pressure, volume, and/or temperature, just plug the values into it.
o If one of the variables isn’t mentioned, we can assume that it’s kept constant and we can just cross it out of the equation.
o Demonstrate how the three gas laws can be easily derived from this law.
• Examples:
o If I have 25 mL of a gas at a pressure of 2.1 atm and a temperature of 300 K, what will the pressure become if I raise the temperature to 400 K and decrease the volume to 10 mL? 7 atm
o If I have a container with an internal pressure of 1.5 atm and temperature of 250 C, what will the pressure be if I heat the container to 1500 C? 2.13 atm
• Worksheet like this
• Tin can lab – have them determine if the implosion was caused by the cooling of the steam or by something else. If something else, what?
• Homework like this
Ideal Gases:
Now that we know how gases behave when we manipulate P, V, and T, it’s time to start thinking about how to deal with things like moles and grams. After all, if we’re going to do chemical reactions with gases, we’ll need to know how to calculate these!
Avogadro’s law: One mole of every gas has the same volume.
• This law assumes that all gases behave perfectly and identically according to the rules of the kinetic molecular theory. Though not precisely true, it gives us very good answers under most conditions.
• Ideal gas: A gas that behaves according to the kinetic molecular theory.
o No intermolecular forces, infinitely small, etc.
o There is no ideal gas in the real world, but some gases come closer than others:
▪ The gas molecules are small.
▪ The gas molecules have very weak intermolecular forces.
▪ The gas molecules are very hot, so they move quickly around and don’t interact with each other much.
▪ The gas is at low pressure, so the molecules have a lot of space between them.
Assuming that all gases are ideal, we can use an equation to relate the number of moles to the pressure (P), volume (V), and temperature (T), giving us the…
Ideal gas law: PV = nRT
P = pressure (in atm of kPa)
V = volume (L)
n = number of moles
T = temperature (Kelvin)
R = ideal gas constant (depends on the unit of pressure used)
• 8.314 L kPa/mol K
• 0.08206 L atm/mol K
Examples:
• If I have 10 liters of a gas at a pressure of 1.5 atm and a temperature of 250 C, how many moles of gas do I have? 0.61 mol.
• If I have 3.5 moles of a gas at a pressure of 895 kPa and they take up a volume of 50 L, what’s the temperature? 1538 K
• Practice worksheet
• Moles of air in the classroom lab
• Homework (honors and general chem have different worksheets)
Dalton’s Law of Partial Pressures: Honors Chemistry
So far, we’ve assumed that all the gases we’ve been working with are pure. However, this isn’t true most of the time. Most gases we use are actually mixtures of several different elements or compounds.
To deal with this, we’ve come up with the idea of partial pressure:
Partial pressure: The pressure that each of the components in a mixture of gases would exert if the other gases in the mixture were removed.
To work with mixtures, we usually treat each gas as if it were alone, rather than in a mixture. This is a reasonable assumption because gases don’t interact with each other much anyway. As a result, the total pressure of the mixture of gases would be the same as the sums of all the partial pressures.
Ptot = P1 + P2 + P3 + … Dalton’s Law of Partial Pressures
Example: If I have a container that contains both oxygen and nitrogen, what’s the total pressure inside the container of the partial pressure of oxygen is 0.50 atm and the partial pressure of nitrogen is 0.75 atm?
• Answer: 0.50 atm + 0.75 atm = 1.25 atm
Example: What’s the pressure in atmospheres if I put 2 moles of N2 and 3 moles of H2 in a 20 L container at 310 K?
• Answer:
o Ptot = Pnitrogen + Phydrogen
o To solve for each of the partial pressures, we need to use the equation PV=nRT to find the partial pressures of each gas, then add them up.
▪ Pnitrogen(20 L) = (2 mol)(0.08206 Latm/molK)(310 K) = 2.54 atm
▪ Phydrogen(20 L) = (3 mol)(0.08206 Latm/molK)(310 K) = 3.82 atm
o Ptot = 2.54 atm + 3.82 atm = 6.36 atm
Example: A 3 L flask containing hydrogen gas at a pressure of 1.5 atm is connected to a 5 L flask containing nitrogen gas at a pressure of 2.5 atm. When the tube between the two flasks is opened, what will the total pressure inside the system be?
• Answer:
o Use the combined gas law to solve the partial pressure of each gas when you open the connector between the flasks, then add the partial pressures together.
▪ Hydrogen: P1V1 = P2V2; P2 = 0.563 atm.
▪ Nitrogen: P1V1 = P2V2; P2 = 1.56 atm.
o Overall, Ptot = 0.563 atm + 1.56 atm = 2.12 atm
Honors practice worksheet: Dalton’s Law of Partial Pressures
Hissing gas tank lab from “17”
Honors: Dalton’s Law Homework
Dalton’s Law of Partial Pressures: General Chemistry
So far, we’ve assumed that all the gases we’ve been working with are pure. However, this isn’t true most of the time. Most gases we use are actually mixtures of several different elements or compounds.
To deal with this, we’ve come up with the idea of partial pressure:
Partial pressure: The pressure that each of the components in a mixture of gases would exert if the other gases in the mixture were removed.
To work with mixtures, we usually treat each gas as if it were alone, rather than in a mixture. This is a reasonable assumption because gases don’t interact with each other much anyway. As a result, the total pressure of the mixture of gases would be the same as the sums of all the partial pressures.
Ptot = P1 + P2 + P3 + … Dalton’s Law of Partial Pressures
Example: A container contains three gases. If the partial pressure of carbon dioxide is 0.50 atm, the partial pressure of nitrogen is 0.30 atm, and the partial pressure of methane is 0.40 atm, what’s the total pressure inside the container?
• Ptot = 0.50 atm + 0.30 atm + 0.40 atm = 1.20 atm
Example: A container has a total internal pressure of 1.50 atm. This container contains two gases, nitrogen (which has a partial pressure of 0.80 atm) and oxygen (with unknown partial pressure). Given this information, what’s the partial pressure of oxygen in the container?
Ptot = Pnitrogen + Poxygen
1.50 atm = 0.80 atm + X
X = 0.70 atm
Practice sheet
Gas tank lab
Homework sheet
Gas Stoichiometry – Honors Chemistry
Explain to the honors chemistry classes that you need to convert between liters ( moles in both cases using PV = nRT.
• At STP, one mole has a volume of 22.4 liters.
If the compounds you’re working with on both sides of the equation are measured in liters, you can use the mole ratio conversion directly between them.
[pic]
Example: Using the equation 2 H2 + O2 ( 2 H2O, how many liters of water can be made from 25 liters of oxygen? Answer: 50 L
Example (same equation): If we do the reaction above using electrochemistry, we can make liquid water from H2 and O2 gas. If I use 20 L of oxygen gas at a pressure of 1.0 atm and a temperature of -350 C, how many grams of water will I make? Answer: 36.9 grams
Honors gas stoichiometry practice problem
Honors homework on gas stoichiometry
Gas Stoichiometry – General Chemistry
• Explain to the general chemistry classes that at STP, the conversion factor between liters and moles is 22.4 L = 1 mole.
What this means is that, whatever method of stoichiometry we use, we need only replace the molar masses of the gases with the number 22.4 – for compounds that are solids, we still use the molar mass.
Example: Using the equation 2 H2 + O2 ( 2 H2O, how many liters of water can be made from 25 liters of oxygen at STP? Answer: 50 L
Example (same equation): How many liters of water can be made from 240 grams of hydrogen? Answer: 2688 L
General chem practice in class practice worksheet
General chem gas stoichiometry practice problem
General homework on gas stoichiometry
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related searches
- dod 7000.14 r volume 2a chapter 1
- dod 7000.14 r volume 7a chapter 57
- dod 7000.14 r chapter 12
- dod 7000.14 r volume 10 chapter 13
- dod 7000.14 r volume 7a chapter 26
- dod 7000.14 r volume 8 chapter 5
- dod 7000.14 r volume 12 chapter 7
- dod 7000.14 r volume 4 chapter 6
- fmr 7000.14 r volume 2a chapter 1
- chapter 14 lesson 4 cells and eneeryy gt
- chapter 14 alcohols exam style questions
- chapter 14 alcohol ms