### Boyle's LawDiscussion and Ten Examples

Discovered by Robert Boyle in 1662. On the continent of Europe, this law is attributed to Edme Mariotte, therefore those counties tend to call this law by his name. Mariotte, however, did not publish his work until 1676.

Boyle's Law gives the relationship between pressure and volume if temperature and amount are held constant. In words, Boyle found these to be true:

1) If the volume of a container is increased, the pressure decreases.
2) If the volume of a container is decreased, the pressure increases.

What makes them true? We can make brief reference to the ideas of kinetic-molecular theory (KMT), which Boyle did not have access to in the 1600's. KMT was developed in its modern form about 200 years after Boyle.

1) Suppose the volume is increased. This means gas molecules have farther to go and they will impact the container walls less often per unit time. This means the gas pressure will be less because there are less molecule impacts per unit time.

2) 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 molecule 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. By the way, what Boyle made is referred to as an empirical discovery. His data told him what was true (PV = k), but he had no idea why it was true.

Boyle's Law is an inverse mathematical relationship. As one quantity (P or V) increases in its value, the other value (P or V) decreases. The constant k does not change in value.

A student might occasionally ask "What is the value for k?"

Suppose P1 and V1 are a 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. The ChemTeam does 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?" The ChemTeam believes it is best right now to ignore that question even though it is a perfectly valid one. The answer lies in the area of kinetic-molecular theory, a topic for another day.

So we know this:

P1V1 = k

And we know that the second data pair equals the same constant:

P2V2 = k

Since k = k, we can create this equality:

P1V1 = P2V2

The equation just above will be very helpful in solving Boyle's Law problems.

By the way, PV = k is Boyle's Law, not the one just above. The one above is just an equation derived from Boyle's Law.

Example #1: 2.00 L of a gas is at 740.0 mmHg pressure. What is its volume at standard pressure?

Solution:

1) Use this equation:

P1V1 = P2V2

2) Insert values:

(740.0 mmHg) (2.00 L) = (760.0 mmHg) (x)

3) Multiply the left side and divide (by 760.0 mmHg) to solve for x.

x = 1.95 L (to three significant figures)

Note that the units of mmHg 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. Don't put a unit on the unknown.

Also, you need to know what the standard value are for pressure (and for temperature).

Example #2: 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L?

Solution:

Use the same technique as in Example #1:

(1.08 atm) (5.00 L) = (x) (10.0 L)

x = 0.540 L (to three sig figs)

Example #3: 9.48 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?

Solution:

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 pressure units were used above.

Once again, insert into P1V1 = P2V2 for the solution.

(x) (9.48 L) = (101.325 kPa) (8.00 L)

x = 85.5 kPa

Hopefully you can see that Boyle's Law problems all use basically the same solution technique. It's just a question of where the x is located. Two problems will arise during the gas laws unit in your classroom:

1. How to match a given problem with what law it is, so you can solve it.
2. Watching out for questions worded in a slightly confusing manner or with unnecessary information. Teachers like to do these sorts of things, if you haven't yet noticed.

Here's an example of asking a question in a confusing manner: give the pressure in the problem in one unit (say, mmHg) but ask for the answer to be in a different unit (say, atm.). You have to either (a) convert the mmHg to atm before the calculation or (b) convert the mmHg answer to atm after the calculation. Believe me, a lot of students get trapped by this technique.

A variant of the above is to give two pressure values in the problem (thus making it be volume that you are calculating). However, the two different pressures are provided using different units (say, atm and mmHg). You MUST convert one unit to the other unit (either conversion direction is OK) before doing the calculation.

Personally, I think this attempt to make the problems confusing stems from the fact that, more or less, all Boyle's Law problems are the same. So, people have come up with some extra spice to season the sauce, so to speak.

Example #4: If we have 6.00 cm3 of gas at a pressure of 10.0 N/cm2 and we increase the pressure to 20.0 N/cm2, what volume will the gas occupy?

Solution:

Newtons per square centimeter is not a unit you often see in chemistry, but it doesn't matter what the unit is, just a long as both P1 and P2 are expressed using the same unit.

(10.0) (6.00) = (20.0) (x)

x = 3.00 cm3

Notice that, when the pressure was doubled, the volume was cut in half. Also, be careful. Your teacher may want you to include the units in the problem, like this:

(10.0 N/cm2) (6.00 cm3) = (20.0 N/cm2) (x)

Example #5: What pressure is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder whose volume is 26.0 liters?

Solution:

(1.00 atm) (196.0 L) = (x) (26.0 L)

x = 7.54 atm (to three sig figs)

Example #6: An evacuated flask A, which has a volume of 30 mL, is attached to a second flask B containing an ideal gas at a pressure of 5 atm. When the two flasks are connected the pressure in the system drops to 2 atm. Calculate the volume of flask B.

Solution:

1) Here's the set-up to solve the problem:

P1V1 = P2V2

(5 atm) (x) = (2 atm) (x + 30)

2) Left side of equation:

Flask B has all the pressure since A is evacuated. We do not know the volume of Flask B. That's this part of the above equation:

(5 atm) (x)

3) Right side of equation:

The two flasks are now connected and the total volume goes up by 30 mL and the total pressure drops to 2 atm. That's this part:

(2 atm) (x + 30)

4) We solve:

5x = 2x + 60

3x = 60

x = 20 mL

5) Does it work?

(5 atm) (20 mL) = (2 atm) (50 mL)

Example #7: Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 39.0 mL. The first bulb has a volume of 56.0 mL and contains 5.92 atm of argon, the second bulb has a volume of 250.0 mL and contains 1.28 atm of neon, and the third bulb has a volume of 37.0 mL and contains 8.50 atm of hydrogen. If the stopcocks (valves) that isolate all three bulbs are opened, what is the final pressure of the whole system in atm?

Solution:

1) Get the total volume of the system:

39.0 + 56.0 + 250.0 + 37.0 = 382 mL

2) You will use Boyle's Law three times, then add up the three results for the final pressure.

argon ---> (5.92 atm) (56.0 mL) = (x) (382.0 mL) ---> x = 0.8678534 atm
neon ---> (1.28 atm) (250.0 mL) = (y) (382.0 mL) ---> y = 0.8376963 atm
hydrogen ---> (8.50 atm) (37.0 mL) = (z) (382.0 mL) ---> z = 0.8232984 atm

3) Add 'em up and round off:

2.5288481 atm

2.53 atm (to three sig figs)

4) You could also do this:

P1V1 + P2V2 + P3V3 = P4V4

And do the left-hand side on the calculator, then divide by V4 (which is the 382.0 value)

Example #8: In order to measure the volume of a piece of apparatus, a chemist filled a 750. mL flask with 46.65 kPa pressure of gas, then expanded it into the apparatus. The final pressure was 14.95 kPa. Calculate the total volume occupied by the gas.

Solution:

P1 = 46.65 kPa
V1 = 750. mL

P2 = 14.95 kPa
V2 = x + 750. mL

(46.65 kPa) (750. mL) = (14.95 kPa) (x + 750. mL)

34987.5 = 14.95x + 11212.5

14.95x = 23775

x = 1590 mL (to three sig figs)

Example #9: Boyle's Law deals with the relationship between two of the variables (of four) that describe gas behavior. Which two variables are held constant in Boyle's Law problems?

(a) pressure/moles
(b) temperature/volume
(c) pressure/volume
(d) temperature/moles
(e) volume/moles

Solution:

The correct answer is (d). Boyle's Law deals with the relationship between pressure and volume (two of the four variables). For Boyle's Law to be valid, the other two variables must be held constant. Those two variables are temperature and amount of gas (the last one being measured in moles).

Example #10: A balloon contains 7.20 L of He. The pressure is reduced to 2.00 atm and the balloon expands to occupy a volume of 25.2 L. What was the initial pressure exerted on the balloon?

Solution:

1) Write the equation used to solve Boyle's Law problems and manipulate it to isolate P1:

P1V1 = P2V2

P1 = (P2V2) / V1

2) Insert values and solve:

P1 = [(2.00 atm) (25.2 L)] / 7.20 L

P1 = 7.00 atm