Go to Boyle's Law Problems #1 - 15

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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.

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.

Why?

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 (P or V) increases in its value, the other value (P or V) goes down. The constant K does not change in value.

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

Suppose P_{1} and V_{1} 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 V_{2}, then the pressure will spontaneously change to P_{2}. 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:

P_{1}V_{1}= k

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

P_{2}V_{2}= k

Since k = k, we can create this equality:

P_{1}V_{1}= P_{2}V_{2}

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:

P_{1}V_{1}= P_{2}V_{2}

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 P_{1}V_{1} = P_{2}V_{2} 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:

- 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 haven't yet noticed.

Here's an example of 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, question authors over the years have come up with some extra spice to season the sause, so to speak.

**Example #4:** If we have 6 cm^{3} of gas at a pressure of 10 N/cm^{2} and we increase the pressure to 20 N/cm^{2}, what volume will the gas occupy?

**Solution:**

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 P_{1} and P_{2} are expressed using the same unit.

(10) (6) = (20) (x)x = 3 cm

^{3}

Notice that, when the pressure was doubled, the volume was cut in half.

**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)

**Bonus Example #1:** 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?

**Solution:**

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 xx = 89.144 kPa

2) Now, we can solve the problem using Boyle's Law:

P_{1}V_{1}= P_{2}V_{2}(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.

**Bonus Example #2:** 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 10^{5} Pa, what will be the radius at an altitude of about 10 km where the pressure of the gas is 2.799 x 10^{4} Pa? For simplicity, assume the temperature has not changed.

**Solution:**

We will use P_{1}V_{1} = P_{2}V_{2}, but notice something about the volumes:

[(4/3)(3.14)(2.5 m)^{3}] (1.004 x 10^{5}Pa) = [(4/3)(3.14)(x)^{3}] (2.799 x 10^{4}Pa)

The two volumes are shown as the formula for the volume of a sphere: V = (4/3)(π)(r^{3}). Please notice that the (4/3)(3.14) portion will cancel out:

x^{3}= [(15.625 m^{3}) (1.004 x 10^{5}Pa)] / 2.799 x 10^{4}Pax

^{3}= 56.05 m^{3}x = 3.8 m (to two significant figures)

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