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# 14.2: Gas Laws

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Use Boyle’s law to calculate pressure-volume changes at constant temperature.
• Use Charles’s law to calculate volume-temperature changes at constant pressure.
• Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.
• Use the combined gas law to solve problems in which pressure, volume, and temperature all change.
• Use Avogadro’s law to calculate volume-number of moles changes at constant temperature and pressure.

## Lesson Vocabulary

• Boyle’s law
• Charles’s law
• Combined gas law
• Gay-Lussac’s law

### Recalling Prior Knowledge

• How does the pressure of a confined gas respond when its volume is decreased?
• How does the pressure of a confined gas respond when its temperature is increased?
• How does the pressure of a confined gas respond when more gas is added to the container?

According to the kinetic-molecular theory of gases, the volume of gas particles is negligible and any attractive forces between particles can be ignored. When these simplifying assumptions hold true, a series of simple empirical relationships called the gas laws, which relate pressure, volume, amount, and temperature, can also be mathematically derived.

## Boyle’s Law

Robert Boyle (1627-1691), an English chemist (Figure below), discovered that doubling the pressure of an enclosed sample of gas while keeping its temperature constant caused the volume of the gas to be reduced by half. Boyle’s Law states that the volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant. An inverse relationship means that as one variable increases in value, the other variable decreases.

Robert Boyle is widely considered to be one of the founders of the modern experimental science of chemistry. Among his many contributions is observing the relationship between gas pressure and volume, described by the gas law that bears his name.

Mathematically, Boyle’s Law can be expressed by the following equation:

P × V = k

The k is a constant for a given sample of gas and depends only on the amount of the gas and the temperature. Table below shows pressure and volume data for a set amount of gas at a constant temperature. The third column represents the value of the constant (k) for this data, which is always equal to the pressure multiplied by the volume. As one of the variables changes, the other changes in such a way that the product of P × V always remains the same. In this particular case, that constant is 500 atm•mL.

Pressure-Volume Data
Pressure (atm) Volume (mL) P × V = k (atm•mL)
0.5 1000 500
0.625 800 500
1.0 500 500
2.0 250 500
5.0 100 500
8.0 62.5 500
10.0 50 500

A graph of the data in Table above further illustrates the inverse relationship described by Boyle’s Law (Figure below). Volume is plotted on the x-axis, and the corresponding pressure is indicated on the y-axis.

The pressure of a gas decreases as the volume increases, making Boyle’s Law an inverse relationship.

Boyle’s Law can be used to compare changing conditions for a gas. We use P1 and V1 to stand for the initial pressure and initial volume of a gas. After a change has been made, P2 and V2 stand for the final pressure and volume. The mathematical relationship of Boyle’s Law becomes:

P1 × V1 = P2 × V2

This equation can be used to calculate any one of the four quantities if the other three are known.

Sample Problem 14.1: Boyle’s Law

When a certain sample of oxygen gas is placed in a 425 mL container, its pressure is equal to 387 kPa. The gas is allowed to expand into a 1.75 L container. Calculate the new pressure of the gas.

Step 1: List the known quantities and plan the problem.

Known

• P1 = 387 kPa
• V1 = 425 mL
• V2 = 1.75 L = 1750 mL

Unknown

• P2 = ? kPa

Use Boyle’s Law to solve for the unknown pressure (P2). It is important that the two volumes (V1 and V2) are expressed in the same units, so V2 has been converted to mL.

Step 2: Solve.

First, rearrange the equation algebraically to solve for P2.

$\mathrm{P_2=\dfrac{P_1 \times V_1}{V_2}}$

Now, substitute the known quantities into the equation and solve.

$\mathrm{P_2=\dfrac{387 \ kPa \times 425 \ mL}{1750 \ mL}=94.0 \ kPa}$

The volume has increased to slightly over 4 times its original value, so the pressure is decreased approximately fourfold. The pressure is in kPa, and the correctly rounded value has three significant figures.

Practice Problem
1. What is the new volume of a sample of neon if 5.23 L of the gas, which was originally at 214 mmHg, is compressed until the pressure increases to 796 mmHg?
2. When a balloon containing 635 mL of air is taken from sea level (at standard pressure) to a higher altitude, the balloon expands to 829 mL. What is the air pressure at this new altitude (in atm)?

If you need more information on Boyle's Law, NASA explains it at http://www.grc.nasa.gov/WWW/k-12/airplane/boyle.html.

You can watch an animation of Boyle's Law at http://www.dlt.ncssm.edu/core/Chapter7-Gas_Laws/Chapter7-Animations/BoylesLaw.html.

Boyle's Law has an interesting application - the potato gun. As the volume of air decreases in a PVC pipe, pressure builds and causes a potato plug to launch. Watch this at http://www.youtube.com/watch?v=OolForRzsRk&feature=player_embedded.

A Boyle's Law lab can be viewed at http://www.youtube.com/watch?v=LVE4PUHqPjk&feature=player_embedded. The document that accompanies this lab is found at http://www.dlt.ncssm.edu/core/Chapter7-Gas_Laws/Chapter7-Labs/Boyle%27s_Law_CBL_web_01-02.doc.

Boyle's Law demonstrations can be seen at http://www.youtube.com/watch?v=N5xft2fIqQU&feature=player_embedded.

## Charles’s Law

In the last lesson, “Gas Properties,” you learned that raising the temperature of a gas enclosed in a rigid container results in an increase in pressure. However, what if the container is flexible? A balloon that is heated will expand in response to an increase in the kinetic energy of the enclosed gas molecules. The molecules strike the inside walls of the balloon with more force, pushing them outward. Figure below illustrates this relationship between temperature and volume.

As a closed container of gas is heated, its molecules increase in kinetic energy and push the movable piston outward, resulting in an increase in volume.

French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles’s Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when the pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stoppage of molecular motion.

Mathematically, the direct relationship of Charles’s Law can be represented by the following equation:

$\mathrm{\dfrac{V}{T}=k}$

As with Boyle’s Law, k is constant only for a given gas sample. Table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature.

Temperature-Volume Data
Temperature (K) Volume (mL) V / T = k (mL/K)
50 20 0.40
100 40 0.40
150 60 0.40
200 80 0.40
300 120 0.40
500 200 0.40
1000 400 0.40

When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in Figure below.

The volume of a gas increases as the Kelvin temperature increases, making Charles’s Law a direct relationship.

Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases.

Charles’s Law can also be used to compare changing conditions for a gas. We will use V1 and T1 to stand for the initial volume and temperature of a gas, while V2 and T2 stand for the final volume and temperature. The mathematical relationship of Charles’s Law becomes:

$\mathrm{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}$

This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin; temperatures in degrees Celsius will not work. Recall that the relationship between these two scales is K=°C+273.

Sample Problem 14.2: Charles’s Law

A balloon is filled to a volume of 2.20 L at a temperature of 22°C. The balloon is then heated to a temperature of 71°C. Find the new volume of the balloon.

Step 1: List the known quantities and plan the problem.

Known

• V1 = 2.20 L
• T1 = 22°C = 295 K
• T2 = 71°C = 344 K

Unknown

• V2 = ? L

Use Charles’s law to solve for the unknown volume (V2). The temperatures have first been converted to Kelvin.

Step 2: Solve.

First, rearrange the equation algebraically to solve for V2.

$\mathrm{V_2=\dfrac{V_1 \times T_2}{T_1}}$

Now substitute the known quantities into the equation and solve.

$\mathrm{V_2=\dfrac{2.20 \ L \times 344 \ K}{295 \ K}=2.57 \ L}$

The volume increases as the temperature increases. The result has three significant figures.

Practice Problem
1. A 465 mL sample of gas at 55°C is cooled to standard temperature (0°C). What is its new volume?
2. To what Celsius temperature does 750. mL of gas originally at −12°C need to be heated to bring the volume to 2.10 L?

Watch an animation of Charles's Law at http://www.dlt.ncssm.edu/core/Chapter7-Gas_Laws/Chapter7-Animations/CharlesLaw.html.

A balloon cooled in liquid nitrogen illustrates Charles's Law at http://www.youtube.com/watch?feature=player_embedded&v=Gi5wPnkBEYI.

You can watch an experiment of Charles's Law at http://www.youtube.com/watch?v=5M8GR6_zIps&feature=player_embedded. The accompanying lab document is found at http://www.dlt.ncssm.edu/core/Chapter7-Gas_Laws/Chapter7-Labs/Charles_Law_Lab_web_01-02.doc.

Watch popping film canisters at http://education.jlab.org/frost/canister.html.

## Gay-Lussac’s Law

When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. French chemist Joseph Gay-Lussac (1778-1850) discovered the relationship between the pressure of a gas and its absolute temperature. Gay-Lussac’s Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas when the volume is kept constant. Gay-Lussac’s Law is very similar to Charles’s Law, with the only difference being the type of container. Whereas the container in a Charles’s Law experiment is flexible, it is rigid in a Gay-Lussac’s Law experiment.

The mathematical expressions for Gay-Lussac’s Law are likewise similar to those of Charles’s Law:

$\mathrm{\dfrac{P}{T}=k}$ and $\mathrm{\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}}$

A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume, its pressure continually decreases until the gas condenses to a liquid.

Sample Problem 14.3: Gay-Lussac’s Law

The gas in an aerosol can is under a pressure of 3.00 atm at a temperature of 25°C. It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of 845°C?

Step 1: List the known quantities and plan the problem.

Known

• P1 = 3.00 atm
• T1 = 25°C = 298 K
• T2 = 845°C = 1118 K

Unknown

• P2 = ? atm

Use Gay-Lussac’s law to solve for the unknown pressure (P2). The temperatures have first been converted to Kelvin.

Step 2: Solve.

First, rearrange the equation algebraically to solve for P2.

$\mathrm{P_2=\dfrac{P_1 \times T_2}{T_1}}$

Now, substitute the known quantities into the equation and solve.

$\mathrm{P_2=\dfrac{3.00 \ atm \times 1118 \ K}{298 \ K}=11.3 \ atm}$

The pressure increases dramatically due to the large increase in temperature.

Practice Problem
1. The pressure in a car tire is 217 kPa at 24°C. After being driven on a hot summer day, the pressure in the tire increases to 258 kPa. What is the Celsius temperature of the air in the tire?
2. A gas sample originally at −39°C is heated to standard temperature (0°C), at which point its pressure is measured to be 808 mmHg. What was the original pressure of the gas?

Watch a simulation of Gay-Lussac's Law at http://www.dlt.ncssm.edu/core/Chapter7-Gas_Laws/Chapter7-Animations/Gay-Lussac%27sLaw.html.

A video of a Gay-Lussac's Law lab is at http://www.youtube.com/watch?v=1pVVZGOBIVg&feature=player_embedded. The accompanying document for this lab is found at http://www.dlt.ncssm.edu/core/Chapter7-Gas_Laws/Chapter7-Labs/Gay_Lussac's_Law_web_01-02.doc.

## The Combined Gas Law

To this point, we have examined the relationships between any two of the variables P, V, and T, while the third variable is held constant. However, situations can also arise where all three variables change. The combined gas law expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas. For a combined gas law problem, only the amount of gas is held constant.

$\mathrm{\dfrac{P \times V}{T}=k}$ and $\mathrm{\dfrac{P_1 \times V_1}{T_1}=\dfrac{P_2 \times V_2}{T_2}}$

Sample Problem 14.4: Combined Gas Law

2.00 L of a gas at 35°C and 0.833 atm is brought to standard temperature and pressure (STP). What will be the new gas volume?

Step 1: List the known quantities and plan the problem.

Known

• P1 = 0.833 atm
• V1 = 2.00 L
• T1 = 35°C = 308 K
• P2 = 1.00 atm
• T2 = 0°C = 273 K

Unknown

• V2 = ? L

Use the combined gas law to solve for the unknown volume (V2). STP is 273 K and 1 atm. The temperatures have been converted to Kelvin.

Step 2: Solve.

First, rearrange the equation algebraically to solve for V2.

$\mathrm{V_2=\dfrac{P_1 \times V_1 \times T_2}{P_2 \times T_1}}$

Now substitute the known quantities into the equation and solve.

$\mathrm{V_2=\dfrac{0.833 \ atm \times 2.00 \ L \times 273 \ K}{1.00 \ atm \times 308 \ K}=1.48 \ L}$

Both the increase in pressure and the decrease in temperature cause the volume of the gas sample to decrease. Since both changes are relatively small, the volume does not decrease dramatically.

Practice Problem
1. A 400. mL sample of gas at 15°C and 113 kPa is heated to 227°C while expanding into a 900. mL container. What is the new pressure of the gas in kPa?

It may seem challenging to remember all the different gas laws introduced so far. Fortunately, Boyle’s, Charles’s, and Gay-Lussac’s Laws can all be easily derived from the combined gas law. For example, consider a situation where a change occurs in the volume and pressure of a gas while the temperature is being held constant. In that case, it can be said that T1 = T2. Look at the combined gas law and cancel the T variable out from both sides of the equation. What is left over is Boyle’s Law: P1 × V1 = P2 × V2. Likewise, if the pressure is constant, then P1 = P2, and canceling P out of the equation leaves Charles’s Law. If the volume is constant, then V1 = V2, and canceling V out of the equation leaves Gay-Lussac’s Law.

You have learned about Avogadro’s hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. Avogadro’s Law states that the volume of a gas is directly proportional to the number of moles of gas when the temperature and pressure are held constant. The mathematical expression of Avogadro’s Law is

$\mathrm{V=k \times n}$ and $\mathrm{\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2}}$

where n is the number of moles of gas and k is a constant. Avogadro’s Law can be observed whenever you blow up a balloon. The volume of the balloon increases as you add more gas particles to the balloon by blowing it up.

If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro’s Law. Adding gas to a rigid container makes the pressure increase.

A balloon has been filled to a volume of 1.90 L with 0.0920 mol of helium gas. If an additional 0.0210 mol of helium is added to the balloon while the temperature and pressure are held constant, what is the new volume of the balloon?

Step 1: List the known quantities and plan the problem.

Known

• V1 = 1.90 L
• n1 = 0.0920 mol
• n2 = 0.0920 + 0.0210 = 0.1130 mol

Unknown

• V2 = ? L

Note that the final number of moles has to be calculated by adding the original number of moles to the moles of added helium. Use Avogadro’s Law to solve for the final volume.

Step 2: Solve.

First, rearrange the equation algebraically to solve for V2.

$\mathrm{V_2=\dfrac{V_1 \times n_2}{n_1}}$

Now, substitute the known quantities into the equation and solve.

$\mathrm{V_2=\dfrac{1.90 \ L \times 0.1130 \ mol}{0.0920 \ mol}=2.33 \ L}$

Since a relatively small amount of additional helium was added to the balloon, its volume increases slightly.

Practice Problem
1. 2.50 g of CO2 gas is confined in a rigid cylinder at a pressure of 4.65 atm. If 0.42 g of gas is released from the cylinder, what is the new pressure?

## Lesson Summary

• As the pressure of a gas increases, its volume decreases if the temperature is held constant.
• As the absolute temperature of a gas increases, its volume also increases if the pressure is held constant.
• As the absolute temperature of a gas increases, its pressure also increases if the volume is held constant.
• The combined gas law is used when the pressure, volume, and temperature of a gas all change.
• As the number of moles of gas increases, its volume increases if the pressure and temperature are held constant.

## Lesson Review Questions

### Reviewing Concepts

1. Why is it necessary to use Kelvin temperatures in all gas law calculations that involve temperature?
2. Explain the difference between a direct and an inverse relationship.

### Problems

1. 2.20 L of air at 1.10 atm is allowed to expand to fill a 6.30 L container. Find the final pressure.
2. A balloon is inflated to a volume of 1.25 L. If the temperature of the air inside is cooled from 35°C to 15°C, what will be the final volume of the balloon?
3. A sample of nitrogen in a rigid container that is originally at 25.0°C and 98.0 kPa is heated until the pressure is 177 kPa. What is the final temperature in Kelvin? In degrees Celsius?
4. A 5.25 L sample of oxygen gas at 720. torr and 35°C is compressed to a volume of 3.80 L while being heated to 78°C. What is the new pressure?
5. 3.30 moles of a gas occupies 95.0 L. If the gas is allowed to escape until the volume is 41.0 L, how many moles of the gas remain?
6. 400.0 mL of helium is held in a container at 1.1 atm and a temperature of –5.0°C. If the gas then escapes into a 2.0 L container while the pressure drops to 0.90 atm, what is the final temperature of the helium in degrees Celsius?
7. A 250. mL sample of air at STP is injected into an evacuated sphere with a radius of 7.00 cm. The temperature is measured to be 20.0°C. What is the pressure inside the sphere, measured in atm? (The volume of a sphere is (4/3)πr3).
8. 2.0 liters of hydrogen is held in a rigid container at STP. If the temperature is dropped to –65.0°C, what is the resulting pressure in mmHg?
9. 870 mL of gas at a pressure of 47.4 kPa is brought to standard pressure. If the temperature remains constant, what is the final volume?

## Points to Consider

There is one equation that relates the pressure to the volume, temperature, and number of moles when the gas is assumed to be ideal.

• In what situations is the ideal gas law more useful than the other gas laws?
• Under what conditions can a gas be assumed to be ideal?

Feb 23, 2012

Mar 26, 2015