Without even knowing what gases are made of (OK, they’re made of atoms and molecules) we can understand how they behave on a macroscopic level. Gases are a form of ordinary matter that is much less dense than liquids or solids. Because of this they tend to fill completely any containers they are in and are very compressible.
Everyone has experience working with gases and taking advantage of their properties. If you have ever inflated a balloon, used a straw, inflated a tire, complained about the weather, or taken a deep breath then you already know how this topic relates to “real life”.
There are four mathematical variables used to describe the behavior of gases. Pressure (P), Volume (V), Temperature (T), and amount (n). There are some common units of measurement for each of these variables so your first task will be to become familiar with them.
One more thing: Using these variables will help you to understand the ideal gas laws. Why are they called ideal? It’s not because they are the best possible laws gases could follow. Nor is it because they are the best scientific laws anyone ever found. No, it is because the gases we will discuss are ‘idealized’. That is, they are not real gases and no real gases act the way these equations say they will. But, and this is important, almost all gases come very, very close to acting exactly according to the ideal gas laws. So even though they are not perfect, they are very useful.
At the end of this packet you will find a series of exercises that will allow you to apply the ideas and proportions introduced in the text. Please do these problems showing all work on a separate piece of paper and carefully label and number each problem.
|Cubic Meter||m3||1000 L||a large fish tank|
|a 2 L bottle of soda|
|a drop from an eyedropper|
|The ‘normal’ amount of pressure exerted by Earth’s atmosphere|
|Atmos. pressure can drop to 740 torr during a storm|
A car tire might be
rated for 35 psi
|Fahrenheit||°F||°C × 9/5 + 32||
32°F water freezes
212°F water boils
|Celsius||°C||(°F -32) × 5/9||
0°C water freezes
100°C water boils
|Kelvin||K||°C + 273||
77 K liquid nitrogen boils
273 K water freezes
373 K water boils
1 mol of C =
6.02 x 1023
atoms of C
1 mol of C atoms weighs
exactly 12 g
Now that you know a bit about the units you will be using you are ready to start working with the mathematical laws that we will use to build up the Ideal Gas Law. The first law we will examine is known as Boyle’s Law and was first quantified and mathematically modelled by Robert Boyle is about the year 1662. The law, in words, says the following:
That is, the higher the pressure, the smaller the volume; the lower the pressure, the larger the volume. This relationship can be expressed in the formula:
P · V = constant or more algebraically: P × V = k (both are true only when T and n are constant)
The key to understanding pressure is to think about force per unit area. The molecules of a gas are constantly striking the walls of a gas’s container. Because these collisions are so numerous it all adds up to a steady amount of force applied to every square centimeter of the surface. The SI unit of pressure is the pascal (Pa) which is equal to a force of 1 newton per square meter (1 N/m2). A more familiar unit of force per unit area for us here in the U.S. is pounds per square inch.
Boyle’s Law is a useful proportion that can be put to work to answer questions about changes in volume and pressure. Here is an example of how to apply the law.Example
If a gas’s pressure is reduced by half at constant temperature then its volume doubles. This much should be clear from the work you did in the lab. But how can Boyle’s Law be used to calculate this result without performing an experiment?
First, let’s define the initial pressure and volume as P1 and V1 and the pressure and volume after a change as P2 and V2. According to Boyle’s Law:
P1V1 = k1 and P2V2 = k2
Let’s assume those constants are the same (and they will be as long as we do not add or remove any gas or change the temperature). In that case:
P1V1 = P2V2
Now, here is a question we can answer using this proportional equation: What is the final volume of a gas when its pressure is doubled and its initial pressure is 1.0 atm and its volume is 5.0 L?
P1 = 1.0 atm P2 = 2.0 atm P1V1 = P2V2 V1 = 5.0 L V2 = ? (1.0 atm)(5.0 L) = (2.0 atm)(V2)
Solving for V2 gives the answer 2.5 L. This is exactly what we expected based on the idea that this is an inverse proportion: when one variable is doubled, the other is cut in half. Use this example to help you to answer the questions in the exercises at the end of this packet.
V V — = constant or — = k (only when P and n are constant) T T
This proportion is true when the pressure is kept the same and when no gas is added or taken away. Another important point about this relationship is that a different temperature scale is required for it to work properly. Both the Celsius and the Fahrenheit scales have meaningful values that are lower than zero. Obviously this would make the proportion give nonsense answers about negative volumes so it can’t be allowed. To solve this problem we will use a temperature scale that has no meaningful values below zero: the kelvin scale. Temperatures expressed in kelvin are just the Celsius temperature plus 273. The lowest possible temperature is absolute zero: 0 K or –273°C. For all calculations involving temperature and gases you must use the kelvin temperature scale. Both scales have degrees of the exact same size.
Charles’s Law is a useful proportion that can be put to work to answer questions about changes in volume and temperature. Here is an example of how to apply the law.Example
If a gas’s temperature is reduced by half at constant pressure then its volume is also cut in half. They are directly proportional, after all. But how can Charles’s Law be used to calculate this result without performing an experiment?
First, let’s define the initial temperature and volume as T1 and V1 and the temperature and volume after a change as T2 and V2. According to Charles’s Law:
V1 V2 ------ = k1 and ------ = k2 T1 T2
Let’s assume those constants are the same (and they will be as long as we do not add or remove any gas or change the pressure). In that case:
V1/T1 = V2/T2
Now, here is a question we can answer using this proportional equation: What is the final volume of a gas when its temperature is doubled and its initial temperature is 25°C and its volume is 5.0 L?
A doubling of temperature refers to the absolute temperature expressed in kelvins. So 50°C is not 25°C doubled. Instead, 25°C = 298 K is doubled to 2 × 298 K = 596 K V1 = 5.0 L V2 = ? V1/T1 = V2/T2 T1 = 298 K T2 = 596 K (5.0 L)/(298 K) = (V2)/(596 K)
Solving for V2 gives the answer 10 L. This is exactly what we expected based on the idea that this is an direct proportion: when one variable is doubled, the other is also doubled. Use this example to help you to answer the questions at the end of this packet.
V V — = constant or — = k (only when T and P are constant) n n
This law applies to situations like when you inflate a balloon. The balloon has a fairly constant temperature and pressure but as you add air its volume grows. Practically speaking you cannot reduce the number of moles of a gas by reducing volume. This law just says that a smaller amount of volume holds a smaller number of gas molecules. One useful consequence of this law is that by comparing gases as they would be under a standard set of conditions the volume does tell you something about the amount of gas. One mole of any gas will have a volume of 22.4 L at 273 K (0 °C) and 1 atm. These conditions (1 atm and 273 K) are known as Standard Temperature and Pressure (STP).
Avogadro’s Law is a useful proportion that can be put to work to answer questions about changes in volume and moles of gas. Here is an example of how to apply the law.Example
If the amount of gas in a container is reduced by one third at constant pressure and temperature then its volume is also cut to one third. They are directly proportional, after all. But how can Avogadro’s Law be used to calculate this result without performing an experiment?
First, let’s define the initial volume and amount of gas V1 and n1 and the volume and amount of gas after a change as V2 and n2. According to Avogadro’s Law:
V1 V2 ------ = k1 and ------ = k2 n1 n2
Let’s assume those constants are the same (and they will be as long as we do not change the temperature or change the pressure). In that case:
V1/n1 = V2/n2
Now, here is a question we can answer using this proportional equation: What is the final volume of a gas when the number of moles of gas is cut by one third—say by letting air out of a balloon—when starting with 5.8 mol and the initial volume is 15.0 L?
V1 = 15.0 L V2 = ? V1/n1 = V2/n2 n1 = 5.8 mol n2 = 1.9 mol (15.0 L)/(5.8 mol) = (V2)/(1.9 mol)
Solving for V2 gives the answer 5.0 L. This is exactly what we expected based on the idea that this is an direct proportion: when one variable is cut by one third, the other is also. Use this example to help you to answer the questions at the end of this packet.
PV = nRT is the combined form of the ideal gas law.
In this formula you can see that pressure (P) is still inversely proportional to volume (V). Volume is still directly proportional to the number of moles (n) and the temperature (T). The letter ‘R’ is the constant of the proportion and is called the Universal Gas Constant. In the units we will usually use R = 0.0821 L·atm/K·mol. This formula is useful for calculating a missing value: if you know the number of moles, volume and temperature of a gas you can use this formula to calculate the pressure by re-writing it as P = nRT/V.
There is another form of the combined gas laws that is useful when you have multiple variables changing at once. When a weather balloon rises through the atmosphere to very high levels the external pressure becomes lower and lower but the temperature also becomes less and less. Rather than doing two separate calculations the proportions can be combined as you see here:
P1V1 P2V2 P1V1 P2V2 R = ------ = ------ or ------ = ------ because moles usually don't change n1T1 n2T2 T1 T2 (ex., n1 = n2 = 1)
By solving the combined gas laws equation (also called the ideal gas law equation) for the universal gas constant it is possible to set up a change formula as we did earlier for the simple gas laws. Since most of the time problems will be concerned with closed systems in which the number of moles of a gas is constant, the form of this formula depicted at right above is most commonly used.
The Combined Gas Law is a useful proportion that can be put to work to answer questions about changes in volume, pressure, temperature or moles of gas. Here is two examples of how to apply the law.Example 1
The pressure, volume, number of moles, and temperature are what you need to know to know everything there is to know about a gas. If you know three of these variables then it is easy to calculate the fourth. For example, what is the volume of a gas in a metal helium cylinder given that at a temperature of 20°C it has a pressure of 180 atm and contains 1,500 mol of He gas? Here is how to set up the problem:
T = 20°C = 293 K P = 180 atm nRT (1,500 mol)(0.0821 L·atm/K·mol)(293 K) V = ? V = ----- = --------------------------------------- = 200 L n = 1,500 mol P (180 atm)
Any of the four variables can be found as long as you know the other three. If you keep careful track of your units as you work then you will have a way to confirm that you have done the calculation correctly. The Universal Gas Constant (R) always provides the missing unit by cancelling out with all of the others. This formula is useful for calculations when nothing is changing.Example 2
For situations in which two variables change at the same time the second equation above is useful for calculating the result. Take the example of a balloon full of helium released at sea level where the pressure in the balloon is about equal to the external pressure of 1 atm. The temperature on a warm spring day is about 18°C and the volume of the balloon is 150 L. What will the volume of the balloon be when it reaches a point high above where the temperature is -10°C and the pressure is 0.5 atm? Start by listing all the variables you know and then set up the proportion. Then solve the proportion for the missing variable, plug in your numbers, cancel units, and calculate the answer:
P1 = 1 atm P2 = 0.5 atm V1 = 150 L V2 = ? n1 = ? mol n2 = ? mol (but the same as before) T1 = 18°C = 291 K T2 = -10°C = 263 K
Since n is the same before and after just use the second form of the equation and solve it for V2 P1V1T2 (1 atm)(150 L)(263 K) V2 = ------- so V2 = ---------------------- = 271 L P2T1 (0.5 atm)(291 K)
So the balloon grows a lot in volume as it rises. This makes sense since the pressure is reduced to half: this alone would double the volume. Because the temperature goes down, though, the volume does not become quite as big since there is a direct proportion between temperature and volume.