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Group Activity: Gas Laws

Introduction

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.

Some Units We Will Use

Volume Unit Symbol Convert Example
Cubic Meter m3 1000 L a large fish tank
Liter L 0.001 m3
1000 mL 		a 2 L bottle of soda
Milliliter mL 1/1000 L
0.001 L 		a drop from an eyedropper
Pressure Unit Symbol Convert Example
Atmosphere atm 760 torr
14.7 lb/in2
101.3 kPa 		The ‘normal’ amount of pressure exerted by Earth’s atmosphere
Torr torr 0.00132 atm
0.0193 psi 		Atmos. pressure can drop to 740 torr during a storm
lbs. per
square in. 		psi
lb/in2 		0.0680 atm
51.7 torr 		A car tire might be
rated for 35 psi
Temperature Unit Symbol Convert Example
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
Amount Unit Symbol Convert Example
Mole mol 1 mol of C =
6.02 x 1023
atoms of C 		1 mol of C atoms weighs
exactly 12 g


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Boyle’s Law (1662)

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:

At a given constant temperature (T) and number of moles of gas (n),
the pressure and volume of a gas are inversely proportional.

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.


Charles’s Law (1787)

The next important law you will learn is called Charles’s Law. Mr. Charles was an Englishman and worked on travel by hot air balloon. He never established a mathematical law describing the relationship between temperature and volume but some English historian liked his name better than the name of the man who did. The man who actually figured out this law of nature is Mr. Gay-Lussac, a Frenchman. Whoever figured it out, the law we’re interested in is:
At a given constant pressure (P) and number of moles of gas (n),
the volume and temperature of a gas are directly proportional.
In other words, the more you raise the temperature, the larger the volume of gas (as long at it remains at the same pressure and you don’t add or subtract any molecules). Temperature must be expressed in Kelvins, not degrees Celsius (x°C = x + 273 K). This relationship can be expressed in the formula:


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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.


Avogadro’s Law (1811)

The mole is the chemical unit that is used to count molecules. Like a dozen, it is a word that refers to a certain number. There are 6.02 × 1023 molecules in a mole. The mole is especially necessary for gases because they are so compressible. The volume of a gas can change dramatically depending on temperature and pressure so a method to keep track of the amount of gas apart from volume is necessary. But interestingly enough, if the temperature and pressure of two samples of gas are the same then equal volumes of those gases have equal numbers of moles. The law that Mr. Avogadro proposed is the following:
At a given constant temperature (T) and constant pressure (P),
the volume and the number of moles, n, of a gas are directly proportional.
In other words, the more gas you have, the bigger the volume (as long as you compare volumes at the same pressure and temperature). The mathematical expression of this law is: 
V                    V
— = constant   or    — = k  (only when T and P are constant)
n                    n
Note: you cannot change the number of moles just by changing the volume! For that you have to add or subtract gas particles.

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.


Combined Gas Laws

So what happens if two variables change at once? Say the temperature and the pressure both change and you need to find the volume: are you stuck, unable to find an answer? Fortunately, no. You can use the combined gas laws: 
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.


Exercises

Do all work for this packet on a separate piece of paper. Carefully number each problem on the other sheet.
Part I, Temperatures
  1. Convert 37°C
    to °F and kelvin (K)
  2. Convert 65°F
    to °C and K
  3. Convert 77 K
    to °C and °F
  4. Convert 97°F
    to °C and K

Part II, Boyle’s Law
If you have done the Boyle’s Law lab that includes the following exercises, then skip this next section of exercises. If you have no idea about that lab, then do these exercises:
  1. If a gas in a volume of 25 mL with a pressure of 1 atmosphere (atm) is compressed to 5 mL what is its pressure?
  2. If a gas with a pressure of 2 atm is confined in a volume of 10 L what will its pressure be if the volume is made to be 20 L?
  3. A balloon with an internal pressure of 1.3 atm and a volume of 2.5 L is placed into a vacuum chamber. What is the balloon’s volume if the internal pressure is reduced to 0.17 atm?
  4. What is the new volume of a gas if the pressure of the gas is reduced from 220 kPa to 100 kPa and the initial volume was 1.5 m3?
  1. At the bottom of the Challenger Deep in the Pacific Ocean the pressure due to all that water overhead is 1091 atm. A bubble of gas with a volume of 1 mL is released by an advanced submarine research vessel. What is the volume of the gas bubble when it pops at the surface of the ocean where the pressure is 1 atm?
  2. A gas is confined in a bottle with a pressure of 5.2 atm. The volume of the bottle is 40 L. What volume does the gas have when it is released into the room?
  3. What happens to the pressure of a gas when its volume is changed from 14 mL to 27 mL?
  4. Why does the volume of helium in a weather balloon increase in volume as it rises from the ground to the upper atmosphere?


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Part III, Charles’s Law
  1. A gas is heated so that its volume doubles. If the initial volume is 42 L and its initial temperature is 200 K, what is its final temperature?
  2. A 6.9 L sample of hydrogen gas is cooled from 88°C to -20°C. What is its final volume?
  3. What is the final volume of some air that is heated from 25°C to 100°C if its initial volume was 11 L?
  4. A gas is cooled from a starting temperature of 60°C. Its initial volume was 15 L. Its final volume was 9 L. What was the temperature after cooling was complete?
  5. A sample of carbon monoxide gas occupies 3.20 L at 125°C. Calculate the temperature at which the gas will occupy 1.54 L if the pressure remains constant.
  1. Find the new volume in L when a 452 mL sample of fluorine gas is heated from 22°C to 187°C at constant pressure.
  2. When you drive your car the tires heat up. The volume of the tire does not change very much when this happens. If the tire pressure reads 35 psi when you start driving will the pressure after driving on the highway for an hour be the same, higher, or lower than 35 psi? Justify your answer.
  3. There is a simple proportional relationship between the pressure of a gas and its temperature when volume and the number of moles are held constant. It is, in fact, simply another way to write Charles’s Law. Can you write an equation for it using the example in the text as a guide? Do so.
Part IV, Avogadro’s Law
  1. A sample of 2.0 mol of oxygen gas is confined in a volume of 18.1 L. What volume would 4.0 mol of gas have to have if the temperature and pressure are to remain the same?
  2. If a balloon containing 0.25 mol of air has a volume of 550 mL then how many moles does the balloon contain when its volume is 1.5 L? In both cases the temperature and pressure are the same.
  3. One mole of an ideal gas has a volume of 22.4 L at 1 atm and 273K. (The set of conditions known as STP). At STP what is the volume of 0.50 mol of gas? Of 0.01 mol of gas?
  1. At STP a certain number of moles of gas occupies 11.2 L. Under the same conditions of temperature and pressure what volume does a gas occupy if it has one third as many moles?
  2. A balloon containing 2.1 mol of CO2 has a volume of 36 L at some temperature and pressure. You pump out 1.5 mol. What is the new volume?
Part V, Combined Gas Laws
  1. Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas. It is often used to fill the space between the panes in double-paned windows. Find the pressure (in atm) of 1.82 mol of SF6 in a steel vessel of volume 5.43 L at 320 K.
  2. How many moles of argon (Ar) would be required to fill a 3.5 L container at 0.278 atm and 12°C?
  3. The unit torr for pressure is commonly used in the lab to measure pressure. One atmosphere equals 760 torr (1 atm = 760 torr). What is the pressure in torr of a gas in a 5.0 L vessel in the lab when the temperature is 25°C and there are 0.452 mol of nitrogen gas?
  4. What is the volume (in L) of 3.2 mol of N2 (nitrogen gas) at 3040 torr and 86°C?
  1. A gas starts with V1 = 4,000 mL, P1 = 1.2 atm, and T1 = 66°C. If no gas is added or removed and the V2 = 1.7 L and T2 = 42°C, what is P2? Hint: convert temp. to K.
  2. A gas starts with V1 = 42 L, P1 = 1.7 atm, and T1 = 13°C. If no gas is added or removed and the V2 = 31 L and P2 = 0.85 atm, what is T2?
  3. A bubble rising from the bottom of a deep lake starts with a temperature of 8°C and a pressure of 6.4 atm. When it gets to the surface, just before it bursts, the temperature is 25°C and the pressure is 1 atm. If the initial volume was 2.1 mL what is the final volume?
  4. Gases are usually delivered in large metal cylinders or gas bottles. When carbon monoxide (CO) is compressed as it is pumped into such a gas bottle its volume is reduced from a very large volume at 1 atm and 20°C to a volume of 200 L at 180 atm and a temperature of 60°C. Calculate the initial volume of the gas and report your final result in cubic meters.
Last updated: May 15, 2013