Gases appear to us as material of very low density that must be enclosed to keep together. Unlike solids, gases have no definite shape. Unlike liquids, gases have no definite volume, but they completely fill a container. The volume of the container is the volume of the gas in it. A gas exerts a pressure on all sides of the container that holds it. Gas can be compressed by pressures greater than the pressure the gas on its container. The words vapor, fume, air, or miasma also describe a gas. Air describes the common mixture of gases in the atmosphere. A miasma is usually a bad-smelling or poisonous gas. The words vapor and fume suggest that the gas came from a particular liquid.
- THE IDEAL GAS LAW FORMULA
- VARIATIONS ON THE IDEAL GAS LAW FORMULA
- THE COMBINED GAS LAW FORMULA
- BOYLE'S LAW
- CHARLES'S LAW
- THE THIRD LAW
- GAS STOICHIOMETRY MATH
- POINTERS ON GAS LAW MATH PROBLEMS
- AVOGADRO'S LAW
- DALTON'S LAW OF PARTIAL PRESSURES
- GRAHAM'S LAW OF DIFFUSION (OR EFFUSION)
- GAS LAW MATH PROBLEMS
In the gaseous state matter is made of particles (atoms or molecules) that are not attached to each other. The intermolecular or interatomic forces that hold solids and liquids have been overcome by the motion of the molecules. The particles of a gas have too much thermal energy to stay attached to each other. The motion and vibration of the atoms pull the individual molecules apart from each other.
Liquid air (with all of the molecules touching each other) has a density of 0.875 grams per milliliter. By Avogadro's law, a mol of any gas occupies 22.4 liters at standard temperature and pressure (STP).
1 mol of any gas at STP = 22.4 liters
Air in the gas phase at standard temperature and pressure ( 1 atmosphere of pressure and 0°C.) has a mol of it (28.96 g) in 22.4 liters, coming to about 1.29 grams per liter. Liquid air is over 680 times denser than the air at one atmosphere. As an estimate, each molecule of gas in the air has 680 times its own volume to rattle around in. Gases are mostly unoccupied space. Each molecule of a gas can travel for a long distance before it encounters another molecule. We can think of a gas as having a 'point source of mass', that is, the volume of the molecule is negligible compared to the space it occupies.
When a gas molecule hits another one, they bounce off each other, ideally in a completely elastic encounter. There is pressure within the gas that is caused by the gas molecules in motion striking each other and anything else in the gas. The pressure that a gas exerts on its container comes from the molecules of gas hitting the inside of the container and bouncing off.
There are some materials that do not appear in the form of a gas because the amount of molecular motion necessary to pull a molecule away from its neighbors is enough to pull the molecule apart. For this reason you are not likely to see large biological molecules such as proteins, fats, or DNA in the form of a gas.
THE IDEAL GAS LAW FORMULA
A gas may be completely described by its makeup, pressure, temperature, and volume. Where P is the pressure, V is the volume, n is the number of mols of gas, T is the absolute temperature, and R is the Universal Gas Constant,
P V = n R T
This formula is the "Ideal Gas Law Formula." The formula is pretty accurate for all gases as we assume that the gas molecules are point masses and the collisions of the molecules are totally elastic. (A completely elastic collision means that the energy of the molecules before a collision equals the energy of the molecules after a collision, or, to put it another way, there is no attraction among the molecules.) The formula becomes less accurate as the gas becomes very compressed and as the temperature decreases. There are some correction factors for both of these factors for each gas to convert it to a Real Gas Law Formula, but the Ideal Gas Law is a good estimation of the way gases act. We will consider only the Ideal Gas Law Formula here. The Universal Gas Constant, R, can be expressed in several ways, depending upon the units of P, V, and T. One common R is 0.0821 liter - atmospheres per mol - degree. It is highly recommended that you know this value for R and the Ideal Gas Law Formula.
VARIATIONS ON THE IDEAL GAS LAW FORMULA
The Ideal Gas Law Formula is a wonderful place to begin learning almost all of the formulas for gases. You are not likely to get out of a chemistry class without a question like: What is the mass of neon in a neon light at 0.00545 Atmospheres at 24 degrees Celsius if the inside volume is 0.279 liters?
GIVEN: P = 0.00545 Atmospheres T = 24°C + 273° = 297K V = 0.279 liters
FIND: mass (m) of neon
m/Fw can now substitute for n and P V = (m/Fw) R T or Fw P V = m R T. When you solve for m, you almost have the problem completely done.
THE COMBINED GAS LAW FORMULA
The Combined Gas Law Formula is the relationship of changing pressure, temperature, and volume of an ideal gas. The same amount of the same gas is given at two different sets of conditions. Let's call the first set of measurements, 'condition #1,' and the second set of measurements, 'condition #2.' We could label the pressure, temperature and volume symbols each with the subscripted number of the condition it represents. P1 is the pressure at condition #1. P2 is the pressure at condition #2. V1 is the volume at condition #1, etc. The gas laws apply to both conditions, so P1 V1 = n R T1 and P2 V2 = n R T2. R is always the same Universal Gas Constant. If we are considering the same gas only at two different conditions, then n1 = n2. Since they are both equations, we could divide one equation by the other to get:
|R = R||n1 = n2|
|P1V1||n1 R T1||P1V1||n1 T1||P1V1||T1|
|P2V2||n2 R T2||P2V2||n2 T2||P2V2||T2|
The last form can be a very useful one. This is the form of the Combined Gas Law Formula that Chemtutor finds easiest to remember. The formulas that most books call the Gas Laws are all contained in the Combined Gas Law. The Combined Law Formula is the one to use if you have any doubt about which of the Gas Laws to use.
Boyle's Law is useful when we compare two conditions of the same gas with no change in temperature. (Remember, "Always Boyle's at the same temperature!") No change in temperature means T1 = T2, so we can cancel the two temperatures in the Complete Gas Law Formula and get:
P1 V1 = 1 or P1 V1 = P2 V2
P2 V2 the usual Boyle's Law
P1 V1 = P2 V2
The usual expression of Boyle's Law was lurking right there in the Combined Gas Law Formula. As you can see, Boyle's Law is in the classic form of, "P is inversely proportional to V." We could predict that from the P and V being together in the numerator of the same side of the equation.
To get a feel for Boyle's Law, visualize a small balloon between your hands. The balloon is so small that you can push all sides of it together between your hands without any of the balloon pouching out at any point. When you push your hands together the volume of the gas in the balloon decreases as the pressure increases. When you let up on the pressure, the volume increases as the pressure decreases.
Again we start with the Combined Law to get Charles's Law, but now there is no change in the pressure volume, so P1 = P2.
P1 V1 = T1
P2 V2 T2
If you cancel out the two pressures, you get a form of Charles's Law that I consider easiest to remember. You can still see the P V = n R T in it if you look hard enough.
V1 = T1
You may have seen this written differently, as in the following form:
V1 = V2
These two expressions are mathematically exactly the same, but the first one shows its origin in the Combined Law. Remember it by, "Charles is under constant pressure."
To get a better feeling for Charles's Law, consider a child's toy balloon. At points between the beginning of filling of a balloon and the maximum stretching of a balloon, the change in internal pressure of a balloon is negligible as the balloon increases in size. A balloon is partially filled at room temperature and placed in the sun inside a car on a hot day in summer. The balloon expands in proportion to the Kelvin temperature. When the same balloon is take out of the car and put into a home freezer, the volume of the balloon decreases.
THE THIRD LAW
The third gas law from the Combined Gas Law has been named for Gay-Lussac in some books, Amonton in others, and not named in a large number of books. It is sometimes amusing to read a book that does not name the third law and needs to refer to it. The third law is the relationship of pressure and temperature with constant volume (V1 = V2.) the pressure and absolute temperature of a gas are directly proportional.
P1 V1 = T1
P2 V2 T2
And so we get the third law, the relationship between the pressure and temperature of a gas.
P1 = T1
Similarly to Charles's Law, it can be arranged so that it appears in the same form you see in most books.
P1 = P2
To get a feel for the third Law, consider an automobile tire. With a tire gauge measure the pressure of the tire before and immediately after a long trip. When cool, the tire has a lower pressure. As the tire turns on the pavement, it alters its shape and becomes hot. There is some expansion of the air in the tire, as seen by the tire riding slightly higher, but we can ignore that small effect. If you were to plot the temperature versus pressure of a car tire, would zero pressure extrapolate out to absolute zero? Remember what you are measuring. The pressure of a car tire is actually the air pressure above atmospheric pressure. If you add atmospheric pressure to your tire gauge, you would certainly come closer to extrapolating to absolute zero.
GAS STOICHIOMETRY MATH
As you know from the Mols, percents, and stoichiometry section, stoichiometry is the calculation of an unknown material in a chemical reaction from the information given about another of the materials in the same chemical reaction. What if either the given material or the material you are asked to find is a gas? In stoichiometry you need to know the amount of one material. For gases not at STP, you must know the pressure, temperature, and volume to know the amount of material given. If you are given a gas not at STP, you will be able to substitute P V = n R T for the given side and plug it directly into the mols place by solving the equation for 'n'. Here is a sample problem using a gas not at STP as the given.
What mass of ammonia would you get from enough nitrogen with 689 liters of hydrogen gas at 350°C and 4587 mmHg?
Given: 689 l H2 = V T = 350°C + 273° = 623K P = 4587 mmHg (change to Atm)
Notice we have all three of the bits of data to know the amount of hydrogen.
Find: Mass (m) of NH3
3 H2 + N2 2 NH3
The outline plan of direction from the stoichiometry roadmap is:
(gas laws) (mols given) (mol ratio) (formula weight find) (mass find)
The ideal gas law ( P v = n R T ) must be solved for 'n' so it can be used as the 'given' of the outline.
|(||P V||)||(||mols NH3||)||(||Fw NH3||)||=||ammonia mass|
|R T||mols H2||mols NH3|
|given||mol ratio||Fw find||mass find|
Things are a bit different when you need to find the volume, pressure, or temperature of a gas not at STP. You will need to solve P V = n R T for the dimension you need to find and attach it to the end of the sequence using the roadmap to find 'n' for the gas. Let's take another problem based on the same chemical equation to explore how to set up finding a gas not at STP.
What volume of ammonia at 7.8 atmospheres and 265°C would you get from 533 grams of nitrogen?
Given: m H2 = 533 g (Now hydrogen is the known material.)
Find: Volume of ammonia at P = 7.8 Atm and T = 265°C + 273° = 538K
The outline plan is now: (mass given) (Fw given) (mols given) (mol ratio) (gas laws)
|(||mass of H2||)||(||mols H2||)||(||mols NH3||)||=||mols of ammonia|
|1||Fw of H2||mols H2|
|given||Fw given||mol ratio||mols find|
Now the result of the stoichiometry is the number of mols of ammonia, 'n' in the ideal gas formula. We solve for the volume we want to find.
V = n R T/P and insert the numbers with 'n' coming from the stoichiometry, or we can tack ( RT/P ) onto the end of the stoichiometry.
|(||mass of H2||)||(||mols H2||)||(||mols NH3||)||(||R T||)||=||V of NH3|
|1||Fw of H2||mols H2||P|
|given||Fw given||mol ratio||gas law||volume|
POINTERS ON GAS LAW MATH PROBLEMS
1. Know the units and dimensions of pressure, volume and temperature and how to convert them to what you want.
2. The gas laws require an absolute temperature, usually Kelvin, in the formulas. Know how to convert any temperature measurement you are given to Kelvin.
3. Know the number and units of 'R' to use in the gas equations. Remember to convert all the units to the units of the 'R' you use to cancel the units.
4. Carefully label the dimension and condition of each variable. The dimensions of the same condition must be labeled with the same subscript.
A 20.6 liter tire at 23°C and 3.21 Atmospheres inside pressure is run on the Interstate for four hours. The tire is now 20.8 liters at 235°C. What is the pressure in the hot tire? You must group the V = 20.6 liters, P = 3.21 Atmospheres, and T = 296 K as one condition. Each of these measurements must have the same subscript, whatever you choose. For instance, V1 = 20.6 liters, P1 = 3.21 Atmospheres, and T1 = 296 K The second condition has a missing component. You are given the volume and temperature, but not the pressure. V2 = 20.8 Liters, T2 = 508 K, and you need to find P2.
5. You can use the Combined Gas Law Formula for any of these problems, but you must carefully cancel any dimensions that are the same in both conditions.
6. Solve for the unknown, insert the given quantities, and cancel the units to make sure your answer will come out right.
There is even more we can do with good old P V = n R T. The first part of this section introduced you to Avogadro's Law. One mole of any gas takes up a volume of 22.4 liters at standard temperature and pressure (STP). If we go back to the comparison of two formulas of the Ideal Gas Law, we have:
P1 V1 = n1 R T1
P2 V2 = n2 R T2
The R's are the same, so they can be cancelled. At standard temperature, T1 = T2 = 273K, and the T's can be cancelled. At standard pressure, P1 = P2 = 1 atmosphere, and the P's can be cancelled. When all the canceling has been done,
V1 = n1
If the volume is proportional to the number of mols of a gas, there is a constant, k, that we can use in the formula, V = k n, to express the proportionality of V and n. What is that proportionality constant? At standard temperature and pressure, the pressure is one atmosphere and the temperature is 273K. The Universal Gas Constant is still 0.0821 Liter - atmospheres per mol - degree. Let's set n at one to find out what k is.
P V = n R T and V = n R T/P
V = (1 mol) (0.0821 L - A/ mol - K) (237 K) / (1 A)
Cancel the mols, the A's (for Atmosphere) and the K's. Do the math.
V = 22.4 Liters
We have seen this number before in Avogadro's Law, and this is where it comes from. When n is one mol and V is 22.4 Liters, k is 22.4 Liters/mol.
1 mol of any gas at STP = 22.4 liters
DALTON'S LAW OF PARTIAL PRESSURES
Similarly to the way we derived V = k n for Avogadro's Law above when the pressure is constant, we can derive P = k n for conditions when the volume does not change. This time there is no notable significance to the k, so we will just say that P is proportional to n when the temperature and pressure are constant. In conditions when more than one gas is mixed, we could number and add the pressures and mols. If we were to have P1 of gas #1 due to n1 mols of it and P2 of another gas (#2) due to n2 mols of it, those two gases in the same volume (They must be at the same temperature.) can be added together. PT is the total pressure and nT is the total number of mols.
n1 + n2 = nT and P1 + P2 = PT
This has nothing to do with whether gas #1 is the same as gas #2. Dalton's Law of Partial Pressures says that, "The sum of all the partial pressures of the gases in a volume is equal to the total pressure." Where PT is the total pressure, P1 is the partial pressure of 'gas #1', P2 is the partial pressure of 'gas #2', Pn is the pressure of the last gas, whatever number (n) it is.
PT = P1 + P2 + ..... + Pn
GRAHAM'S LAW OF DIFFUSION (OR EFFUSION)
Gases under no change of pressure that either diffuse in all directions from an original concentration or effuse through a small hole move into mixture at a rate that is inversely proportional to the square root of the formula weight of the gas particle.
The mental picture of diffusion could be the drop of ink (with the same specific gravity as water) being carefully placed in the center of a glass of water. The ink will diffuse from the original point where it was deposited with no mixing of the glass of water. The mixing of diffusion is due to the movement of the molecules. Gases diffuse more quickly than liquids because the energy of motion is higher and the available path for unobstructed straight movement is much greater in gases.
Temperature is a type of energy. Temperature is the way we feel the motion of the molecules. E = 1/2 m v2 is the formula for energy of motion. This very motion of the molecules is the operating motion of the mixing action of diffusion. The mass of the molecule is the formula weight or molecular weight of the gas particle.
From the formula for energy of motion we can see that the mass of the particle (the formula weight) is inversely proportional to the square of the velocity of the particle. This is the easiest way to remember Graham's Law.
(v1)2 = Fw2
Notice in the above formula that 'v1' is over 'v2' and that 'Fw2' is over 'Fw1'. This is so that the inverse relationship can be expressed in the formula.
If you are solving for the effusion velocity of a particle, you might take the square root of both sides to get the other useful Graham's Law formula.
GAS LAW MATH PROBLEMS
- Helium takes up 5.71 liters at O°C and 3.95 atmospheres. What is the volume of the same helium at 32°F and 800 mmHg?
- 257 mL of oxygen in a gas tube goes from 17°C to 42°C from being out in the sun. The pressure in the tube is 39 #/in2, but it does not change as the temperature increases. What is the volume of the tube after it has heated?
- An enormous (57,400 cubic meter) expandable helium balloon at 22°C is heated up by a fire under it and the action of the sun on the dark plastic covering on top. There will be a small increase in pressure from 785 mmHg to 790 mmHg, but the major effect wanted is an increase in volume so the balloon can lift its cargo. To what temperature must the balloon get in order to fill out to 60,500 cubic meters?
- What volume of air at standard pressure gets packed into an 11 ft3 SCUBA tank at the same temperature at 15.8 atmospheres?
- Air is 20% oxygen and 80% nitrogen. What is the mass of air in an automobile tire of 19.7 L and internal pressure of 46.7 PSI at 24°C? (That pressure is the same as the 32 PSI difference you usually measure as the tire pressure 32 PSI + 14.7 PSI. You will have to use a weighted average for the molar mass of air.)
- A constant pressure tank of gas at 1.01 Atm has propane in it at 15°C when it is at 255 cubic meters. What is its volume at 48°C?
- A SCUBA tank is filled with air at 16.7 Atm at 24°C, but someone leaves it out in the sun to warm to 65°C. What is the tank pressure?
- The usual partial pressure of oxygen that people get at sea level is 0.20 Atm., that is, a fifth of the usual sea level air pressure. People used to 1 Atm. air pressure begin to become "light-headed" at about 0.10 Atm oxygen. As a rule of thumb, the air pressure decreases one inch of mercury each thousand feet of altitude above sea level. At what altitude should airplane cabins be pressurized? Up to about what altitude should you be able to use unpressurized pure oxygen? (Express your answer in feet above Mean Sea Level, or MSL.)
- Which diffuses faster, the bad smell from a cat-pan due to ammonia or an expensive French perfume with an average molecular weight of 170 g/mol? "How much faster does the faster one diffuse?
- What is the mass of neon in a 625 mL neon tube at 357 mmHg & 25°C?
- What is the mass of 15 liters of chlorine gas at STP?
- How many liters of ammonia at STP are produced when 10 g of hydrogen is combined with nitrogen?
- How many milliliters of hydrogen at 0°C and 1400 mmHg are produced if 15g of magnesium reacts with sulfuric acid?
- What is the mass of 25 liters of fluorine gas at 2.85 atm, 450°C?
- A nine liter tank has 150 atmospheres of bromine in it at 27°C. What is the added mass of the tank due to the gas?
- A 250 Kg tank of liquid butane (C4H1O) burns to produce carbon dioxide at 120°C. What volume of carbon dioxide is produced at 1 Atm?
- How many liters of product at 950 mmHg and O°C is produced by the burning of three liters of acetylene (C2H2) at 5 atm and 20°C?
- Five grams of octane (C8H18) and enough oxygen to burn it are in an automobile cylinder compressed to 20 atm at 28°C. The mixture explodes and heats the cylinder to 150°C. What is the pressure in the (same sized) cylinder after the explosion?
- If 0.515g of magnesium is added to HCl, it makes hydrogen gas and magnesium chloride. The hydrogen is collected at 23°C and 735mmHg. What is the volume of hydrogen?
- What is the mass of 150 liters of propane gas (C3H8) at 37°C and 245 inHg?
- Isopropyl alcohol, C3H7OH, makes a good fuel for cars. What volume of oxygen at 735 mmHg and 23°C is needed to burn one kilogram of isopropyl alcohol?
- What volume does 4 Kg of nitrogen gas take up at 27°C and 3 atm?
- The dirigible Hindenburg had 3.7E6 m3 of hydrogen in its gas bags at 1.1 atm and 7°C. What was the weight of the hydrogen in pounds?
|ANSWERS TO GAS LAW MATH PROBLEMS|
|1. 21.4 L||2. 279 ml||3. 39.9°C||4. 174 ft3|
|5. 73.9 g||6. 284 cubic meters||7. 19.0 Atm||8a. 15,000 ft. MSL|
|8b. 27,000 ft. MSL||9. Ammonia diffuses 3.16 times faster (Wouldn't you KNOW it?)|
|10. 0.242 g||11. 47.5 g||12. 74.7 L||13. 7.51 L|
|14. 45.6 g||15. 8.76 Kg||16. 5.56 E5 L||17. 33.5 L|
|18. 35.4 Atm.||19. 532 ml||20. 2.12 Kg||21. 209 KL|
|22. 1.17 KL||23. 7.80 E5 L|
- Dr. Davis Wilner (http://chemtutor.com)