Practice problems for these concepts can be found at:
 Gases Multiple Choice Review Questions for AP Chemistry
 Gases FreeResponse Questions for AP Chemistry
The gas laws relate the physical properties of volume, pressure, temperature, and moles (amount) to each other. First we will examine the individual gas law relationships. You will need to know these relations for the AP exam, but the use of the individual equation is not required. Then we will combine the relationships in to a single equation that you will need to be able to apply.
Volume–Pressure Relationship: Boyle's Law
Boyle's law describes the relationship between the volume and the pressure of a gas when the temperature and amount are constant. If you have a container like the one shown in Figure 8.3 and you decrease the volume of the container, the pressure of the gas increases because the number of collisions of gas particles with the container's inside walls increases.
Mathematically this is an inverse relationship, so the product of the pressure and volume is a constant: PV = k_{b}.
If you take a gas at an initial volume (V_{1}) and pressure (P_{1}) (amount and temperature constant) and change the volume (V_{2}) and pressure (P_{2}), you can relate the two sets of conditions to each other by the equation:
 P_{1}V_{1} = P_{2}V_{2}
In this mathematical statement of Boyle's law, if you know any three quantities, you can calculate the fourth.
Volume–Temperature Relationship: Charles's Law
Charles's law describes the volume and temperature relationship of a gas when the pressure and amount are constant. If a sample of gas is heated, the volume must increase for the pressure to remain constant. This is shown in Figure 8.4.
Remember: In any gas law calculation, you must express the temperature in kelvin.
There is a direct relationship between the Kelvin temperature and the volume: as one increases, the other also increases. Mathematically, Charles's law can be represented as:
 V/T = k_{c}
where k_{c} is a constant and the temperature is expressed in kelvin.
Again, if there is a change from one set of volume–temperature conditions to another, Charles's law can be expressed as:
 V_{1}/T_{1} = V_{2}/T_{2}
Pressure–Temperature Relationship: GayLussac's Law
GayLussac's law describes the relationship between the pressure of a gas and its Kelvin temperature if the volume and amount are held constant. Figure 8.5 represents the process of heating a given amount of gas at a constant volume.
As the gas is heated, the particles move with greater kinetic energy, striking the inside walls of the container more often and with greater force. This causes the pressure of the gas to increase. The relationship between the Kelvin temperature and the pressure is a direct one:
Combined Gas Law
In the discussion of Boyle's, Charles's, and GayLussac's laws we held two of the four variables constant, changed the third, and looked at its effect on the fourth variable. If we keep the number of moles of gas constant—that is, no gas can get in or out—then we can combine these three gas laws into one, the combined gas law, which can be expressed as:
 (P_{1}V_{1})/T_{1} = (P_{2}V_{2})/T_{2}
Again, remember: In any gas law calculation, you must express the temperature in kelvin.
In this equation there are six unknowns; given any five, you should be able to solve for the sixth.
For example, suppose a 5.0L bottle of gas with a pressure of 2.50 atm at 20°C is heated to 80°C. We can calculate the new pressure using the combined gas law. Before we start working mathematically, however, let's do some reasoning. The volume of the bottle hasn't changed, and neither has the number of moles of gas inside. Only the temperature and pressure have changed, so this is really a GayLussac's law problem. From GayLussac's law you know that if you increase the temperature, the pressure should increase if the amount and volume are constant. This means that when you calculate the new pressure, it should be greater than 2.50 atm; if it is less, you've made an error. Also, remember that the temperatures must be expressed in kelvin. 20°C = 293 K (K = °C + 273) and 80°C = 353 K.
We will be solving for P_{2}, so we will take the combined gas law and rearrange for P_{2}:
 (T_{2}P_{1}V_{1})/(T_{1}V_{2}) = P_{2}
Substituting in the values:
 (353 K)(2.50 atm)(5.0 L)/(293 K)(5.0 L) = P_{2}
 3.0 atm = P_{2}
The new pressure is greater than the original pressure, making the answer a reasonable one. Note that all the units canceled except atm, which is the unit that you wanted.
Let's look at a situation in which two conditions change. Suppose a balloon has a volume at sea level of 10.0 L at 760.0 torr and 20°C (293 K). The balloon is released and rises to an altitude where the pressure is 450.0 torr and the temperature is –10°C (263 K). You want to calculate the new volume of the balloon. You know that you have to express the temperature in K in the calculations. It is perfectly fine to leave the pressures in torr. It really doesn't matter what pressure and volume units you use, as long as they are consistent in the problem. The pressure is decreasing, so that should cause the volume to increase (Boyle's law). The temperature is decreasing, so that should cause the volume to decrease (Charles's law). Here you have two competing factors, so it is difficult to predict the end result. You'll simply have to do the calculations and see.
Using the combined gas equation, solve for the new volume (V_{2}):
 (P_{1}V_{1})/T_{1} = (P_{2}V_{2})/T_{2}
 (P_{1}V_{1}T_{2})/(P_{2}T_{1}) = V_{2}
Now substitute the known quantities into the equation. (You could substitute the knowns into the combined gas equation first, and then solve for the volume. Do it whichever way is easier for you.)
 (760.0 torr)(10.0 L)(263 K)/(450.0 torr)(293 K) = V_{2}
 15.2 L = V_{2}
Note that the units canceled, leaving the desired volume unit of liters. Overall, the volume did increase, so in this case the pressure decrease had a greater effect than the temperature decrease. This seems reasonable, looking at the numbers. There is a relatively small change in the Kelvin temperature (293 K versus 263 K) compared to a much larger change in the pressure (760.0 torr versus 450.0 torr).

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