5 Six Gas Laws and Solved Problems

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The Six Gas Laws Content of Lecture 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

The six formulas of gas laws are summarized (as well as the combined gas law, the ideal gas law, Dalton law of partial pressure, Graham’s law.) Boyle's Law Charles' Law Guy-Lussac's Law Avogadro's Law Diver Law “No-name” Law The Combined Gas Law The Ideal Gas Law Dalton's Law of Partial Pressures Graham’s Law

1. Boyle's Law

PV

= k1

2. Charles' Law

T/V

= k2

3. Guy-Lussac's Law

P/T

= k3

4. Avogadro's Law

V/n

= k4

5. Diver Law

P/n

= k5

1/nT

=1/k

6. “No-name” Law 6 -----------------------------------------------------------------------------------------------------------Combined Gas Law

P1 V1 / n1 T1 PV

= P2 V2 / n2 T2 =nRT

Ideal Gas Law -----------------------------------------------------------------------------------------------------------Dalton's Law of Partial Pressures

Ptotal = P1 + P2 + P3 + ... + Pn 1/2

Graham’s Law (m1/m2) = v2/v1 ------------------------------------------------------------------------------------------------------------

1. Boyle's Law Discovered by the British Robert Boyle in 1662. However, on the continent of Europe, this law is attributed to the French Edme Mariotte, therefore those countries tend to call this law by his name. Mariotte, however, did not publish his work until 1676.

Discussion of Boyle's experiment His law gives the relationship between pressure and volume if temperature and amount of gas are held constant. If the volume of a container is increased, the pressure decreases. If the volume of a container is decreased, the pressure increases. Why should that take place? Suppose the volume is increased. This means gas molecules have farther to go and they will affect the container walls less often collisions per unit time. This means the gas pressure will be less because there are less molecule impacts per unit time. If the volume is decreased, the gas molecules have a shorter distance to go, thus striking the walls more often per unit time. This results in pressure being increased because there are more molecules’ impacts per unit time. The mathematical form of Boyle's Law is

PV = k

This means that the pressure-volume product will always be the same value if the temperature and amount remain constant. This relationship was what Boyle discovered. This is an inverse mathematical relationship. As one quantity goes up in the value, the other goes down. In the classroom, a student will occasionally ask, "What is k?" Suppose that P1 and V1 are the pressure-volume pair of data at the start of an experiment. In other words, some container of gas is created and the volume and pressure of that container is measured. Keep in mind that the amount of gas and the temperature DOES NOT CHANGE. We do not care what the exact numbers are, just that there are two numbers. When you multiply P and V together, you get a number that is called k, we don't care what the exact value is. Now, if the volume is changed to a new value called V2, then the pressure will spontaneously change to P2. It will do so because the PV product must always equal k. The PV product CANNOT just change to any old value. It MUST go to k (If the temperature and amount remain the same.) Of course, you now want to ask, "Why does it have to stay at k?" We believe it is best right now to ignore that question even though it is a perfectly valid one. So, we know this P1 V1 = k And we know that the second data pair equals the same constant P2 V2 = k Since k = k, we can conclude that P1 V1 = P2 V2. This equation of P1 V1 = P2 V2 will be very helpful in solving Boyle's Law problems.

Example #1: 2.00 L of a gas is at 740.0 mm Hg pressure. What is its volume at standard pressure? Answer: this problem is solved by inserting values into P1 V1 = P2 V2. (2.00 L) (740.0 mm Hg) = (x) (760.0 mm Hg)

Cross-multiply and divide to solve for x. Note that the units of mm Hg will cancel. x is a symbol for an unknown and, technically, does not carry units. So do not write x L for x liters. Just keep checking to see you are using the proper equation and you have all the right values and units. Do not put a unit on the unknown.

Example #2: 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L? Answer: use the same technique. (5.00 L) (1.08 atm) = (10.0 L) (x)

Example #3: 2.50 L of a gas was at an unknown pressure. However, at standard pressure, its volume was measured to be 8.00 L. What was the unknown pressure? Answer: notice the units of the pressure were not specified, so any can be used. If this were a test question, you might want to inquire of the teacher as to a possible omission of desired units. Let's use kPa since the other two units were used above. Once again, insert into P1V1 = P2V2 for the solution. (2.50 L) (x) = (8.00 L) (101.325 kPa) You can see that Boyle's Law problems all use the same solution technique. It is just a question of where the x is located. Two problems will arise during the gas laws unit in your classroom: How to match a given problem with what law it is, so you can solve it. Watching out for questions worded in a slightly confusing manner or with unnecessary information. Teachers like to do these sorts of things, if you have noticed. ------------------------------------------------------------------------------------------------------------

Robert Boyle and His Data Regarding the picture of Robert Boyle, be safe to note that this is not his real hair. Wigs were the fashion in his days. The table below shows the values Boyle collected. The titles of each column are rather wordy and so are given below the table. All measurements are in inches. It was published in "A Defense of the Doctrine Touching the Spring And Weight of the Air . . . .," published in 1662.

A

B

C

D

48 46 44 42 40 38 36 34 32 30 28 26 24 23 22 21 20 19 18 17 16 15 14 13 12

00 29 1/8 29 2/16 01 7/16 29 1/8 30 9/16 02 13/16 29 1/8 31 15/16 04 6/16 29 1/8 33 8/16 06 3/16 29 1/8 35 5/16 07 14/16 29 1/8 37 10 2/16 29 1/8 39 5/16 12 8/16 29 1/8 41 10/16 15 1/16 29 1/8 44 3/16 17 15/16 29 1/8 47 1/16 21 3/16 29 1/8 50 5/16 25 3/16 29 1/8 54 5/16 29 11/16 29 1/8 58 13/16 32 3/16 29 1/8 61 5/16 34 15/16 29 1/8 64 1/16 37 15/16 29 1/8 67 1/16 41 9/16 29 1/8 70 11/16 45 29 1/8 74 2/16 48 12/16 29 1/8 77 14/16 53 11/16 29 1/8 82 12/16 58 2/16 29 1/8 87 14/16 63 15/16 29 1/8 93 1/16 71 5/16 29 1/8 100 7/16 78 11/16 29 1/8 107 13/16 88 7/16 29 1/8 117 9/16 All entries in C are 29 1/8

E 29 2/16 30 6/16 31 12/16 33 1/7 35 36 15/19 38 7/8 41 2/17 43 11/16 46 3/5 50 53 10/13 58 2/8 60 18/23 63 6/11 66 4/7 70 73 11/19 77 2/3 82 4/17 87 3/8 93 1/5 99 6/7 107 7/13 116 4/8

Titles of each column A - The number of equal spaces in the shorter leg, that contained the same parcel of air diversely extended. B - The height of the mercurial cylinder in the longer leg, that compressed the air into those dimensions. C. The height of the mercurial cylinder, that counter-balanced the pressure of the atmosphere. D - The Aggregate of the last columns B and C, exhibiting the pressure sustained by the included air. E - What that pressure should be according to the hypothesis, that supposes the pressures and expansions to be in reciprocal proportions. For the better understanding of this experiment, it may not be a miss to take notice of the following particulars: 1. That the tube being so tall, that we could not conveniently make use of it in a chamber, we were fain to use it on a pair of stairs, which yet were very lightsome, the tube being for preservation's sake by strings so suspended, that it did scarce touch the box presently to be mentioned. 2. The lower and crooked part of the pipe was placed in a square wooden box, of a good largeness and depth, to prevent the loss of the quicksilver, that might fall aside in the

transfusion from the vessel into the pipe, and to receive the whole quicksilver in case the tube should break. 3. That we were two to make the observation together, the one to take notice at the bottom, how the quicksilver rose in the shorter cylinder, and the other to pour in at the top of the longer; it being very hard and troublesome for one man alone to do both accurately. 4. That the quicksilver was poured in but by little and little, according to the direction of him that observed below; it being far easier to pour in more, than to take out any, in case too much at once had been poured in. The graph just below is of Robert Boyle's data from above.

The following figure is from an article in the May 1992 issue of the Journal of College Science Teaching. Pages 363-365 is an excellent article by Frank Fazio titled "Using Robert Boyle's Original Data in the Physics and Chemistry Classrooms."

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2. Charles' Law Discovered by Joseph Louis Guy-Lussac in 1802. He made reference in his paper to unpublished work done by Jacques Charles about 1787 (he died in 1850). Charles had found that oxygen, nitrogen, hydrogen, carbon dioxide, and air expand to the same extent over the same 80 degree interval. Charles did invent the hydrogen-filled balloon and on December 1, 1783, he ascended into the air and became possibly the first man in history to witness a double sunset. Guy-Lussac was no slouch in the area of ballooning. On September 16, 1804, he ascended to an altitude of 7016 meters (just over 23,000 feet - about 4.3 miles). This remained the world altitude record for almost 50 years when that record was broken by only a few meters. Because of Guy-Lussac's reference to Charles' work, many people have come to call the law by the name of Charles' Law. There are some books which call the temperature-volume relationship by the name of Guy-Lussac's Law, and there are some which call it the Law of Charles and Guy-Lussac. Needless to say, there are some confused people out there. Most textbooks call it Charles' Law. The same year a 23-year-old Guy-Lussac discovered this law, he had occasion to walk into a linen draper's shop in Paris and there he made a wonderous discovery. He found the 17-years old showgirl reading a chemistry textbook while waiting for customers. Needless to say, he was intrigued by this, and made more visits to the shop. In 1808, he and Josephine were married and over the years, five little Guy-Lussac ankle-biters were added to the scene. This law gives the relationship between volume and temperature if pressure and amount are held constant. If the temperature of a container is increased, the volume increases. If the temperature of a container is decreased, the volume decreases. Why? Suppose the temperature is increased. This means gas molecules will move faster and they will impact the container walls more often. This means the gas pressure inside the container will increase (but only for an instant. Think of a short span of time.). The greater pressure on the inside of the container walls will push them outward, thus increasing the volume. When this happens, the gas molecules will now have farther to go, thereby lowering the number of impacts and dropping the pressure back to its constant value. It is important to note that this momentary increase in pressure lasts for only a very, very small fraction of a second. You would need a very fast, accurate pressure-sensing device to measure this momentary change. Charles' Law is a direct mathematical relationship. This means there are two connected values and when one goes up, the other also increases. The mathematical form of Charles' Law is:

T÷V=k

This means that the temperature/volume fraction will always be the same value if the pressure and amount remain constant. Let T1 and V1 be a temperature-volume pair of data at the start of an experiment. If the temperature is changed to a new value called T2, then the volume will change to V2. The new temperature-volume data pair will preserve the value of k. We do not care what the actual value of k is, only that two different temperature-volume data pairs equal the same value and that value is called k. So we know this: T1 ÷ V1 = k And we know this: T2 ÷ V2 = k Since k = k, we can conclude that T1 ÷ V1 = T2 ÷ V2. This equation of T1 ÷ V1 = T2 ÷ V2 will be very helpful in solving Charles' Law problems.

Notice that the right-hand equation results from cross-multiplying the first one. Some people remember one better than the other, so both are provided. Before going to some sample problems, let's be very clear: EVERY TEMPERATURE USED IN A CALCULATION MUST BE IN KELVIN, NOT DEGREES CELSIUS. DON'T YOU DARE USE CELSIUS IN A NUMERICAL CALCULATION. USE KELVIN EVERY TIME.

Example # 1: A gas is collected and found to fill 2.85 L at 25.0°C, what will be its volume at standard temperature? Answer: convert 25.0°C to Kelvin and you get 298 K. Standard temperature is 273 K. We plug into our equation like this:

Remember that you have to plug into the equation in a very specific way. The temperatures and volumes come in connected pairs and you must put them in the proper place.

Example # 2: 4.40 L of a gas is collected at 50.0°C, what will be its volume upon cooling to 25.0°C?

First of all, 2.20 L is the wrong answer. Sometimes a student will look at the temperature being cut in half and reason that the volume must also be cut in half. That would be true if the temperature was in Kelvin. However, in this problem the Celsius is cut in half, not the Kelvin. Answer: convert 50.0°C to 323 K and 25.0°C to 298 K. Then plug into the equation and solve for x, like this:

Example # 3: 5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure (as required by Charles' Law)? Answer:

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3. Guy-Lussac's Law Discovered by Joseph Louis Guy-Lussac in the early 1800's. It gives the relationship between pressure and temperature when volume and amount are held constant. If the temperature of a container is increased, the pressure increases. If the temperature of a container is decreased, the pressure decreases. Why? Suppose the temperature is increased. This means gas molecules will move faster and they will impact the container walls more often. This means the gas pressure inside the container will increase, since the container has rigid walls (volume stays constant). Guy -Lussac's Law is a direct mathematical relationship. This means there are two connected values and when one goes up, the other also increases. The mathematical form of Gay-Lussac's Law is:

P÷T=k

This means that the pressure/temperature fraction will always be the same value if the volume and amount remain constant. Let P1 and T1 be a pressure-temperature pair of data at the start of an experiment. If the temperature is changed to a new value called T2, then the pressure will change to P2. Keep in mind that when volume is not discussed (as in this law), it is constant. That means a container with rigid walls. As with the other laws, the exact value of k is unimportant in our context. It is important to know the PT data pairs obey a constant relationship, but it is not important for us what the exact value of the constant is. Besides which, the value of k would shift based on what pressure units (atm, mmHg, or kPa) you were using. We know this: P1 ÷ T1 = k And we know this: P2 ÷ T2 = k Since k = k, we can conclude that P1 ÷ T1 = P2 ÷ T2. This equation of P1 ÷ T1 = P2 ÷ T2 will be very helpful in solving Guy-Lussac's Law problems.

Notice the similarities to the Charles' Law. This is because both laws are direct relationships. Make sure to convert any Celsius temperature to Kelvin before using it in your calculation.

Example # 1:

10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure? Answer: change 25.0°C to 298.0 K and remember that standard pressure in kPa is 101.325. Insert values into the equation and get:

The answer is 311.3 K, but the question asks for Celsius, so you subtract 273 to get the final answer of 38.3°C, but then you knew that. Right?

Example #2: 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure? Answer: convert to Kelvin and insert:

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4. Avogadro's Law Discovered by Amedo Avogadro, of Avogadro's Hypothesis fame, probably sometime in the early 1800s. It gives the relationship between volume and amount when pressure and temperature are held constant. Remember amount is measured in moles. Also, since volume is one of the variables, that means the container holding the gas is flexible in some way and can expand or contract. If the amount of gas in a container is increased, the volume increases. If the amount of gas in a container is decreased, the volume decreases. Why? Suppose the amount is increased. This means there are more gas molecules and this will increase the number of impacts on the container walls. This means the gas pressure inside the container will increase (for an instant), becoming greater than the pressure on the outside of the walls. This causes the walls to move outward. Since there is more wall space the impacts will lessen and the pressure will return to its original value. The mathematical form of Avogadro's Law is:

V÷n=k

This means that the volume/amount fraction will always be the same value if the pressure and temperature remain constant. Let V1 and n1 be a volume-amount pair of data at the start of an experiment. If the amount is changed to a new value called n2, then the volume will change to V2. We know this: V1 ÷ n1 = k And we know this: V2 ÷ n2 = k Since k = k, we can conclude that V1 ÷ n1 = V2 ÷ n2. This equation of V1 ÷ n1 = V2 ÷ n2 will be very helpful in solving Avogadro's Law problems.

Avogadro's Law is a direct mathematical relationship. You prove Avogadro's Law every time you blow up a balloon.

Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? Answer: this time I'll use V1n2 = V2n1 (5.00 L) (1.80 mol) = (x) (0.965 mol) ------------------------------------------------------------------------------------------------------------

Combined Gas Law To derive the Combined Gas Law, do the following: Step 1: Write Boyle's Law: P1V1 = P2V2 Step 2: Multiply by Charles Law: P1 V12 / T1 = P2 V22 / T2 Step 3: Multiply by Gay-Lussac's Law: P12 V12 / T12 = P22 V22 / T22 Step 4: Take the square root to get the combined gas law: P1 V1 / T1 = P2 V2 / T2 If all six gas laws are included (the three above as well as Avogadro, Diver, and "no-name"), we would get the following: P1 V1 / n1 T1 = P2 V2 / n2 T2 However, this more complete combined gas law is rarely, if ever, discussed. Consequently, we will ignore it in future discussions and use only the law given in step 4 above.

Problem The following type of combined gas law problem (where everything goes to STP) is VERY common: 2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP? (Note: STP is a common abbreviation for "standard temperature and pressure.") You have to recognize that five values are given in the problem and the sixth is an unknown x. Remember to change the Celsius temperatures into Kelvin. To solve the problems, write it as a matrix, like this:

and fill it in with data from the problem. Here is the right-hand side filled in with the STP values:

Here's the solution matrix completely filled in:

Insert the values in their proper places in the combined gas law equation: P1V1 / T1 = P2V2 / T2 and solve for x.

Problem The next problem uses two gas laws in sequence. It involves using Dalton's Law of Partial Pressures first, then use of the Combined Gas Law. The explanation will assume you understand Dalton's Law. These two laws occurring together in a problem is VERY COMMON. 1.85 L of a gas is collected over water at 98.0 kPa and 22.0°C., what is the volume of the dry gas at STP? The key phrase is "over water." Another phrase to look for is "wet gas." This means the gas was collected by bubbling it into an inverted bottle filled with water that is sitting in a water bath. The gas bubbles-in, and is trapped. It displaces the water that flows out into the water bath. The problem is that the trapped gas now has water vapor mixed in with it. This is a consequence of the technique and cannot be avoided. However, we can use a preparatory calculation technique based on Dalton's Law. That calculation should allow us to remove the effect of the water vapor and treat the gas as "dry." For this example, we write Dalton's Law like this: We need to know the vapor pressure of water at 22.0°C and to do this we must look it up in a reference source. It is important to recognize the P tot is the 98.0 value. P tot is the combined pressure of the dry gas AND the water vapor. We want the water vapor pressure OUT. We solve the problem for P gas and get 95.3553 kPa. Notice that it is not rounded off. The only rounding off done is at the FINAL answer, which this is not. Placing all the values into the solution matrix yields this:

Solve for x in the usual manner of cross-multiplying and dividing.

Derivation of the Ideal Gas Law PV = n R T

Ideal Gas Law

Emil Clapeyron first wrote the Ideal Gas Law in 1834. This is just one way to derive the Ideal Gas Law: For a static sample of gas, we can write each of the six gas laws as follows: PV = k1 V/T = k2 P/T = k3 V/n = k4 P/n = k5 1/nT= 1 / k6 Note that the last law is written in reciprocal form. The subscripts on k indicate that six different values would be obtained. When you multiply them all together, you get: P3 V3 / n3 T3 = k1 k2 k3 k4 k5 / k6 Let the cube root of k1 k2 k3 k4 k5 / k6 be called R. The units work out: k1 = atm-L k2 = L / K k3 = atm / K k4 = L / mol k5 = atm / mol 1 / k6 = 1 / mol-K Each unit occurs three times and the cube root yields L-atm / mol-K, the classic units for R when used in a gas law context. Resuming, we have: PV / n T = R or, more commonly: PV = n R T R is called the gas constant. Sometimes it is referred to as the universal gas constant. If you wind up taking enough chemistry, you will see it showing up over and over and over. The Numerical Value for R R's value can be determined many ways. This is just one way:

We will assume we have 1.000 mol of a gas at STP. The volume of this amount of gas under the conditions of STP is known to a high degree of precision. We will use the value of 22.414 L. By the way, 22.414 L at STP has a name. It is called "molar volume." It is the volume of ANY ideal gas at standard temperature and pressure. Let's plug our numbers into the equation: (1.000 atm) (22.414 L) = (1.000 mol) (R) (273.15 K) Notice how atmospheres were used as well as the exact value for standard temperature. Solving for R gives 0.08206 L atm / mol K, when rounded to four significant figures. This is usually enough. Remember the value. You'll need it for problem solving. Notice the weird unit on R: say out loud "liter atmospheres per mole Kelvin." This is not the only value of R that can exist. It depends on which units you select. Those who take more chemistry will meet up with 8.3145 Joules per mole Kelvin, but we will only use the 0.08206 value in gas-related problems. A sample of dry gas weighing 2.1025 grams is found to occupy 2.850 L at 22.0°C and 740.0 mmHg. How many moles of the gas are present? Notice that the units for pressure MUST be in atm., so the 740.0 mm Hg must be converted first. 740.0 mm Hg ÷ 760.0 mm Hg/atm = 0.9737 atm However, the unrounded-off value should be used in the calculation just below. Now, plug into the equation: (0.9737 atm) (2.850 L) = (n) (0.08206 L atm / mol K) (295.0 K) and solve for n. When using the problem above, what is the molar mass of the gas? This is a very common use of this law and the odds are very good you will see this type of question on a test. The key is to remember the units on molar mass: grams per mole. We know from the problem statement that 2.1025 grams of the gas is involved and we also know how many moles that is. We know that from doing the calculation above and getting 0.1146 mol. So all we have to do is divide the grams of gas by how many moles it is: 2.1025 g ÷ 0.1146 mol = 18.34 g/mol ------------------------------------------------------------------------------------------------------------

Dalton's Law of Partial Pressures John Dalton discovered this law in 1801. For any pure gas (let's use helium), PV = n R T holds true. Therefore, P is directly proportional to n if V and T remain constant. As n goes up, so would P. Or the reverse.

Example Suppose you were to double the moles of helium gas present. What would happen? Answer: the gas pressure doubles. However, suppose the new quantity of gas added was a DIFFERENT gas. Suppose that, instead of helium, you added neon. What would happen to the pressure? Answer: the pressure doubles, same as before.

Derivation Dalton's Law immediately follows from this example since each gas is causing 50% of the pressure. Summing their two pressures gives the total pressure. Written as an equation, it looks like this: P He + P Ne = P total Dalton's Law of Partial Pressures: each gas in a mixture creates pressure as if the other gases were not present. The total pressure is the sum of the pressures created by the gases in the mixture. P total = P 1 + P 2 + P 3 + .... + P n Where n is the total number of gases in the mixture. The only necessity is that the two gases do not interact in some chemical fashion, such as reacting with each other. The pressure each gas exerts in mixture is called its partial pressure. The most common use of Dalton's Law seen is with water vapor. A common method of collecting gas during an experiment is by trapping it "over water." An inverted bottle filled with water sits in a water bath. A tube from the reaction vessel conducts the gas into the bottle where it bubbles to the top and displaces water, which runs out the mouth of the bottle into the water bath. However, there is an unavoidable problem. The gas saturates with water vapor and now the total pressure inside the bottle is the sum of two pressures - the gas itself and the added water vapor. WE DO NOT WANT THE WATER VAPOR PRESSURE. So we get rid of it by subtraction. This means we must get the water vapor pressure from somewhere. We get it from a table because the water vapor pressure depends only on the temperature, NOT how big the container is or the pressure of the other gas. Usually the textbook will have an abbreviated table with more complete tables in reference manuals like "The Handbook of Chemistry and Physics."

So finally, here is the example problem: 0.750 L of a gas is collected over water at 23.0°C with a total pressure of 99.75 kPa. What is the pressure of the dry gas? Look up the vapor pressure data here. Work this problem one out!

Example Another common concept that crops up in a Dalton's Law context is the mole fraction. Suppose you had equal moles of two different gases in a mixture. Then the mole fraction for each would be 0.50. The mole fraction for each gas is simply the moles of that gas divided by the total moles in the mixture. Seems simple enough. How does it relate to Dalton's Law?

Answer: The mole fraction also gives the fraction of the total pressure each gas contributes. So if the mole fraction for a gas was 0.50, then it would contribute 50% of the total pressure. If the mole fraction of a gas was 0.15, then its partial pressure would be 0.15 times the total pressure. The reverse is also true. If you divided the partial pressure of a gas by the total pressure, you would get the mole fraction for that gas. (I hope you know enough by now that the two pressures would have to be in the same units!) By the way, mole fractions are unit-less numbers. The mole (or pressure) units cancel out.

Graham's Law Discovered by Thomas Graham of Scotland in the 1830s (I think!). Consider samples of two different gases at the same Kelvin temperature. Since temperature is proportional to the kinetic energy of the gas molecules, the kinetic energy of the two gas samples is also the same. In equation form, we can write: KE1 = KE2 Since KE = (1/2) m v2, we can write the following equation: m1 v1 2 = m2 v2 2 Note that the value of one-half cancels. The equation above can be rearranged algebraically into the following: (m1 / m2)1/2 = v2 / v1 This last equation is the modern way of stating Graham's law. Graham's Law Problems Abbreviations

Conversions

atm – atmosphere mm Hg - millimeters of mercury torr - another name for mm Hg Pa - Pascal (kPa = kilo Pascal) K – Kelvin °C - degrees Celsius

K = °C + 273 1 cm3 (cubic centimeter) = 1 mL (milliliter) 1 dm3 (cubic decimeter) = 1 L (liter) = 1000 mL Standard Conditions 0.00 °C = 273 K 1.00 atm = 760.0 mm Hg = 101.325 kPa = 101,325 Pa

Problems 136. If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster? 137. What is the molecular weight of a gas which diffuses 1/50 as fast as hydrogen? 138. Two porous containers are filled with hydrogen and neon respectively. Under identical conditions, 2/3 of the hydrogen escapes in 6 hours. How long will it take for half the neon to escape? 139. If the density of hydrogen is 0.090 g/L and its rate of diffusion is 6 times that of chlorine, what is the density of chlorine? 140. How much faster does hydrogen escape through a porous container than sulfur dioxide?