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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.
This link takes you to a discussion of the experiment Boyle performed, the data he gathered and his picture.
His law gives the relationship between pressure and volume if temperature and amount 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.
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.
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.
This is an inverse mathematical relationship. As one quantity goes up in the value, the other goes down.
In the ChemTeam classroom, a student occasionally asked "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.
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 conclude that P1V1 = P2V2.
This equation of P1V1 = P2V2 will be very helpful in solving Boyle's Law problems.
Problem #1: 2.00 L of a gas is at 740.0 mmHg pressure. What is its volume at standard pressure?
This problem is solved by inserting values into P1V1 = P2V2.
(740.0 mmHg) (2.00 L) =(760.0 mmHg) (x)
Multiply the left side and divide (by 760.0 mmHg) to solve for x. 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.
Problem #2: 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L?
Use the same technique.
(1.08 atm) (5.00 L) =(x) (10.0 L)
Problem #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?
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.
(x) (2.50 L) = (101.325 kPa) (8.00 L)
Hopefully you can see that Boyle's Law problems all use 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:
Problem #4: A spherical weather balloon is constructed so that the gas inside can expand as the balloon ascends to higher altitudes where the pressure is lower. If the radius of the spherical balloon is 2.5 m at sea level where the pressure is 1.004 x 105 Pa, what will be the radius at an altitude of about 10 km where the pressure of the gas is 2.799 x 104 Pa? For simplicity, assume the temperature has not changed.
We will use P1V1 = P2V2, but notice something about the volumes:
[(4/3)(3.14)(2.5 m)3] (1.004 x 105 Pa) = [(4/3)(3.14)(x)3] (2.799 x 104 Pa)
The two volumes are shown as the formula for the volume of a sphere: V = (4/3)(π)(r3). Please notice that the (4/3)(3.14) portion will cancel out:
x3 = [(15.625 m3) (1.004 x 105 Pa)] / 2.799 x 104 Pa
x3 = 56.05 m3
x = 3.8 m (to two significant figures)
Problem #5: If we have 6 cm3 of gas at a pressure of 10 N/cm2 and we increase the pressure to 20 N/cm2, what volume will the gas occupy?
Newtons per square centimeter is not a unit you often see, but it doesn't matter what the unit is, just a long as both P1 and P2 are expressed using the same unit.
(10) (6) = (20) (x)
x = 3 cm3
Notice that the pressure doubled, so the volume was cut in half.
Problem #6: A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 liters. If the balloon is filled with 2.0 liters of helium at sea level (101.3 kPa), and rises to an altitude at which the boiling temperature of water is only 88 degrees Celsius, will the balloon burst?
Comment: These is no way of determining the starting temperature of the gas. However, we know something not in the problem: at sea level, the boiling point of water is 100 °C. So:
1) Let us use a ratio and proportion to estimate the pressure required for water to boil at 88 °C:
100 °C is to 101.3 kPa as 88 °C is to x
x = 89.144 kPa
2) Now, we can solve the problem using Boyle's Law:
P1V1 = P2V2
(101.3) (2.0) = (88.144) (x)
x = 2.27 L
The balloon will not burst.
Comment: Boyle's Law assumes that the temperature and amount of gas are constant. Since we never knew the starting temperature, we will assume it never changed as the balloon rose. The temperature would actually change, however, if we decide the gas temperature did change, but by some unknown value, then we cannot solve the problem.
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