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15.4: Gas Laws

Difficulty Level: At Grade Created by: CK-12

Lesson Objectives

The student will:

• state Boyle's law, Charles's law, and Gay-Lussac's law.
• solve problems using Boyle's law, Charles's law, and Gay-Lussac's law.
• state the combined gas law.
• solve problems using the combined gas law.

Vocabulary

• Boyle's law
• Charles's law
• combined gas law
• Gay-Lussac's law

Introduction

The gas laws are mathematical expressions that relate the volume, pressure, temperature, and quantity of gas present. They were determined from the results of over 100 years of experimentation. They can also be derived logically by examining the present day definitions of pressure, volume, and temperature.

Boyle’s Law

Gases are often characterized by their volume, temperature, and pressure. These characteristics, however, are not independent of each other. Gas pressure is dependent on the force exerted by the molecular collisions and the area over which the force is exerted. In turn, the force exerted by these molecular collisions is dependent on the absolute temperature. The relationships between these characteristics can be determined both experimentally and logically from their mathematical definitions.

The relationship between the pressure and volume of a gas was first determined experimentally by an Irish chemist named Robert Boyle (1627-1691). The relationship between the pressure and volume of a gas is commonly referred to as Boyle’s law.

When we wish to observe the relationship between two variables, it is absolutely necessary to keep all other variables constant so that the change in one variable can be directly related to the change in the other. Therefore, when the relationship between the volume and pressure of a gas is investigated, the quantity and temperature of the gas must be held constant so that these factors do not contribute to any observed changes.

You may have noticed that when you try to squeeze a balloon, the resistance to squeezing becomes greater as the balloon becomes smaller. That is, the pressure inside the balloon becomes greater when the volume is reduced. This phenomenon can be studied more carefully with an apparatus like the one shown below. This device is a cylinder with a tightly fitted piston that can be raised or lowered. There is also a pressure gauge fitted to the cylinder so that the gas pressure inside the cylinder can be measured. The amount of gas of gas inside the cylinder cannot change, and the temperature of the gas is not allowed to change.

In the picture on the left, a 4.0-liter volume of gas is exerts a pressure of \begin{align*}2.0 \ \mathrm{atm}\end{align*}. If the piston is pushed down to decrease the volume of the gas to \begin{align*}2.0 \ \mathrm{liters}\end{align*}, the pressure of the gas is found to be \begin{align*}4.0 \ \mathrm{atm}\end{align*}. The piston can be moved up and down to positions for several different volumes, and the pressure of the gas can be for each of the volumes. Several trials would generate a data set like that shown in Table below.

PV Data
Trial Volume Pressure
1 \begin{align*}8.0\;\mathrm{liters}\end{align*} \begin{align*}1.0\;\mathrm{atm}\end{align*}
2 \begin{align*}4.0\;\mathrm{liters}\end{align*} \begin{align*}2.0\;\mathrm{atm}\end{align*}
3 \begin{align*}2.0\;\mathrm{liters}\end{align*} \begin{align*}4.0\;\mathrm{atm}\end{align*}
4 \begin{align*}1.0\;\mathrm{liters}\end{align*} \begin{align*}8.0\;\mathrm{atm}\end{align*}

We might note from casual observation of the data that doubling volume is associated with the pressure being reduced by half. Likewise, if we move the piston to cause the pressure to double, the volume is halved. We can analyze this data mathematically by adding a fourth column to our table – namely, a column showing the product of multiplying pressure times volume for each trial (see Table below).

PV Data
Trial Volume Pressure Pressure \begin{align*}\times \end{align*} Volume
1 \begin{align*}8.0\;\mathrm{liters}\end{align*} \begin{align*}1.0\;\mathrm{atm}\end{align*} \begin{align*}8.0\;\mathrm{liters} \cdot \mathrm{atm}\end{align*}
2 \begin{align*}4.0\;\mathrm{liters}\end{align*} \begin{align*}2.0\;\mathrm{atm}\end{align*} \begin{align*}8.0\;\mathrm{liters} \cdot \mathrm{atm}\end{align*}
3 \begin{align*}2.0\;\mathrm{liters}\end{align*} \begin{align*}4.0\;\mathrm{atm}\end{align*} \begin{align*}8.0\;\mathrm{liters} \cdot \mathrm{atm}\end{align*}
4 \begin{align*}1.0\;\mathrm{liters}\end{align*} \begin{align*}8.0\;\mathrm{atm}\end{align*} \begin{align*}8.0\;\mathrm{liters} \cdot \mathrm{atm}\end{align*}

The data in the last column shows that with constant temperature and quantity of gas, the pressure times the volume for this sample of gas yields a constant. A mathematical constant (often represented by \begin{align*}k\end{align*}) is a number that does not change even when other quantities in the formula do change. The value of \begin{align*}k_1\end{align*} will change if a different quantity of gas is used or if the trials are carried out at a different temperature, but for a particular mass of a particular gas at a particular temperature, the value of \begin{align*}k_1\end{align*} will always be the same. A subscript 1 is used to distinguish this constant from the constants of other gas laws. This relationship can be shown in a mathematical equation.

\begin{align*}PV = k_1\end{align*}

This equation is a mathematical statement of Boyle’s law. This particular equation demonstrates what is called an inverse proportionality. When one of the variables is increased, the other variable will decrease by exactly the same factor. This relationship can be easily seen in a graph like the one shown below.

This result matches our logic intuition. If the pressure a gas exerts is equal to the force divided by the area over which it is exerted, we would expect the pressure to increase when we decrease the area but keep the force constant. Similarly, if we maintain the same number of molecules of gas and we keep the same temperature, we expect the total force exerted by the molecules to be the same. As a result, if we expand the volume of the gas, which increases the area over which the force is exerted, we would expect the pressure to decrease.

This video is a laboratory demonstration of Boyle's Law (4c): http://www.youtube.com/watch?v=J_I8Y-i4Axc (1:38).

Charles’s Law

The relationship between the volume and temperature of a gas was investigated by a French physicist, Jacques Charles (1746-1823). The relationship between the volume and temperature of a gas is often referred to as Charles’s law.

An apparatus that can be used to study the relationship between the temperature and volume of a gas is shown below. Once again, the sample of gas trapped inside a cylinder so that no gas can get in or out. Thus, we would have a constant mass of gas inside the cylinder. In this setup, we would also place a mass on top of a movable piston to keep a constant force pushing against the gas. This guarantees that the gas pressure in the cylinder will be constant. If the pressure inside increases, the piston will be pushed up until the inside pressure becomes equal to the outside pressure. Similarly, if the inside pressure decreases, the outside pressure will push the cylinder down, decreasing the volume until the two pressures again become the same. This system guarantees constant gas pressure inside the cylinder.

With this set up, we can adjust the temperature and measure the volume at each temperature to produce a data table similar to the one we created for comparing pressure and volume. The picture on the left in the diagram above shows the volume of a sample of gas at \begin{align*}250\;\mathrm{K}\end{align*}, while the picture on the right shows the volume when the temperature has been raised to \begin{align*}500\;\mathrm{K}\end{align*}. After two more trials, the collected data is shown in Table below.

Charles's Law Data
Trial Volume Temperature Volume/Temp
1 \begin{align*}1000.\;\mathrm{mL}\end{align*} \begin{align*}250.\;\mathrm{K}\end{align*} \begin{align*}4.00\;\mathrm{mL/K}\end{align*}
2 \begin{align*}1200.\;\mathrm{mL}\end{align*} \begin{align*}300.\;\mathrm{K}\end{align*} \begin{align*}4.00\;\mathrm{mL/K}\end{align*}
3 \begin{align*}2000.\;\mathrm{mL}\end{align*} \begin{align*}500.\;\mathrm{K}\end{align*} \begin{align*}4.00\;\mathrm{mL/K}\end{align*}
4 \begin{align*}2400.\;\mathrm{mL}\end{align*} \begin{align*}600.\;\mathrm{K}\end{align*} \begin{align*}4.00\;\mathrm{mL/K}\end{align*}

In order to find a constant from this data, it was necessary to divide each volume with the corresponding Kelvin temperature. The mathematical expression for Charles’s Law is:

\begin{align*} \frac {V} {T} = k_2\end{align*}

This relationship is to be expected if we recognize that we are increasing molecular collisions with the walls by raising the temperature. The only way to keep the pressure from increasing is to increase the area over which that force is exerted. This mathematical relationship is known as a direct proportionality. When one variable is increased, the other variable also increases by exactly the same factor.

Historical note: In addition to exploring the relationship between volume and temperature for gases, Jacques Charles was also the first person to fill a large balloon with hydrogen gas and take a solo balloon flight.

This video is a laboratory demonstration of Charle's Law (4c): http://www.youtube.com/watch?v=IkRIKGN3i0k (4:02).

Gay-Lussac’s Law

The relationship between temperature and pressure was investigated by the French chemist, Joseph Gay-Lussac (1778-1850). An apparatus that could be used for this investigation is shown below. In this case, the cylinder does not have a movable piston because it is necessary to hold the volume, as well as the quantify of gas, constant. This apparatus allows us to alter the temperature of a gas and measure the pressure exerted by the gas at each temperature.

After a series of temperatures and pressures have been measured, Table below can be produced. Like Charles’s Law, in order to produce a mathematical constant when operating on the data in the table, we must divide pressure by temperature. The relationship, again like Charles’s Law, is a direct proportionality.

Pressure vs. Temperature Data
Trial Temperature Pressure Pressure/Temp
1 \begin{align*}200.\;\mathrm{K}\end{align*} \begin{align*}600.\;\mathrm{torr}\end{align*} \begin{align*}3.00\;\mathrm{torr/K}\end{align*}
2 \begin{align*}300.\;\mathrm{K}\end{align*} \begin{align*}900.\;\mathrm{torr}\end{align*} \begin{align*}3.00\;\mathrm{torr/K}\end{align*}
3 \begin{align*}400.\;\mathrm{K}\end{align*} \begin{align*}1200.\;\mathrm{torr}\end{align*} \begin{align*}3.00\;\mathrm{torr/K}\end{align*}
4 \begin{align*}500.\;\mathrm{K}\end{align*} \begin{align*}1500.\;\mathrm{torr}\end{align*} \begin{align*}3.00\;\mathrm{torr/K}\end{align*}

The mathematical form of Gay-Lussac’s Law is:

\begin{align*} \frac {P} {T} =k_3K\end{align*}

This relationship demonstrates that when the temperature is increased, the pressure must also increase to maintain the value of the constant, \begin{align*}k_3\end{align*}. If the area the molecules are occupying is kept the same, the collisions of the molecules with the surroundings will increases as the temperature increases. This results in a higher pressure.

Standard Conditions for Temperature and Pressure (STP)

It should be apparent by now that expressing a quantity of gas simply by stating its volume is inadequate. Ten liters of oxygen gas could contain any mass of oxygen from \begin{align*}4,000\;\mathrm{grams}\end{align*} to \begin{align*}0.50\;\mathrm{grams}\end{align*} depending on the temperature and pressure of the gas. Chemists have found it useful to choose a standard temperature and pressure with which to express gas volume. The standard conditions for temperature and pressure (STP) were chosen to be \begin{align*}0^\circ \text{C} \ (273\;\mathrm{K})\end{align*} and \begin{align*}1.00\;\mathrm{atm}\end{align*} (\begin{align*}760\;\mathrm{mm \ of \ Hg}\end{align*}). You will commonly see gas volumes expressed as \begin{align*}1.5\;\mathrm{liters}\end{align*} of gas under standard conditions or \begin{align*}1.5\;\mathrm{liters}\end{align*} of gas at STP. Once you know the temperature and pressure conditions of a volume of gas, you can calculate the volume under other conditions.

This video is a black board presentation of some ideal gas law calculations and it includes the definition of standard temperature and pressure (4d): http://www.youtube.com/watch?v=GwoX_BemwHs (13:01).

The Combined Gas Law

Boyle’s law shows how the volume of a gas changes when its pressure is changed with the temperature held constant, while Charles’s law shows how the volume of a gas changes when the temperature is changed with the pressure held constant. Is there a formula we can use to calculate the change in volume of a gas if both pressure and temperature change? The answer is yes, as we can use a formula that combines both Boyle’s law and Charles’s law.

Boyle's law states that for a sample of gas at constant temperature, every volume times pressure trial will yield the same constant. We use the subscript 1 to represent one set of conditions, and the subscript 2 to represent a second set of conditions,

\begin{align*}P_1 V_1= P_2 V_2 \ \ \ \ \ \text{so} \ \ \ \ \ V_2 = V_1 \cdot \frac {P_1} {P_2}\end{align*}

We can find a similar expression for Charles's law:

\begin{align*}\frac{V_1} {T_1} = \frac{V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ V_2 = V_1 \cdot \frac{T_2} {T_1}\end{align*}

Combining the two equations yields:

\begin{align*}V_2 = V_1 \cdot \frac {P_1} {P_2} \cdot \frac {T_2} {T_1}\end{align*}

The terms in this equation are rearranged and are commonly written in the form shown below. This equation is also known as the combined gas law.

\begin{align*} \frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2}\end{align*}

When solving problems with the combined gas law, temperatures must always be in Kelvin. The units for pressure and volume may be any appropriate units, but the units of pressure must be the same for \begin{align*}P_1\end{align*} and \begin{align*}P_2\end{align*}, and the units of volume for \begin{align*}V_1\end{align*} and \begin{align*}V_2\end{align*} must also be the same.

Example:

A sample of gas has a volume of \begin{align*}400\end{align*}. liters when its temperature is \begin{align*}20.^\circ\mathrm{C}\end{align*} and its pressure is \begin{align*}300.\;\mathrm{mm \ of \ Hg}\end{align*}. What volume will the gas occupy at STP?

Solution:

Step 1: Assign known values to the appropriate variable.

\begin{align*} \begin{array}{ll} P_1 = 300.\ \text{mm\ of\ Hg} & P_2 = 760.\ \text{mm of Hg (standard pressure)}\\ V_1 = 400.\ \text{liters} & V_2 = x\ \text{(the unknown)}\\ T_1 = 20.^\circ \text{C} + 273 = 293\ \text{K} & T_2 = 0^\circ \text{C} + 273 = 273\ \text{K} \end{array} \end{align*}

Step 2: Solve the combined gas law for the unknown variable.

\begin{align*} \frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ V_2 = \frac {P_1V_1T_2} {P_2T_1}\end{align*}

Step 3: Substitute the known values into the formula and solve for the unknown.

\begin{align*}V_2 = \frac {(300.\ \text{mm of Hg}) \cdot (400.\ \text{L}) \cdot (273\ \text{K})} {(760.\ \text{mm of Hg}) \cdot (293\ \text{K})} = 147 \ \text{liters}\end{align*}

Example:

A sample of gas occupies \begin{align*}1.00\;\mathrm{liter}\end{align*} under standard conditions. What temperature would be required for this sample of gas to occupy \begin{align*}1.50\;\mathrm{liters}\end{align*} and exert a pressure of \begin{align*}2.00\;\mathrm{atm}\end{align*}?

Step 1: Assign known values to the appropriate variable.

\begin{align*} \begin{array}{ll} P_1 = 1.00\ \text{atm (standard pressure)} & P_2 = 2.0\ \text{atm} \\ V_1 = 1.00\ \text{liter} & V_2 = 1.50\ \text{liters} \\ T_1 = 273\ \text{K (standard temperature)} & T_2 = x \ \text{(unknown)} \end{array} \end{align*}

Step 2: Solve the combined gas law for the unknown variable.

\begin{align*} \frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ T_2 = \frac {P_2V_2T_1} {P_1V_1}\end{align*}

Step 3: Substitute the known values into the formula and solve for the unknown.

\begin{align*}T_2 = \frac {(2.00\ \text{atm}) \cdot (1.50\ \text{L}) \cdot (273\ \text{K})} {(1.00\ \text{atm}) \cdot (1.00\ \text{L})} = 819\ \text{K}\end{align*}

Example:

A \begin{align*}1.00\;\mathrm{liter}\end{align*} sample of oxygen gas under standard conditions has a density of \begin{align*}1.43\;\mathrm{g/L}\end{align*}. What is the density of oxygen gas at \begin{align*}500.\;\mathrm{K}\end{align*} and \begin{align*}760.\;\mathrm{torr}\end{align*}?

Solution:

You can find the mass of oxygen in the \begin{align*}1.00\;\mathrm{liter}\end{align*} sample by multiplying volume times density, which yields a mass of \begin{align*}1.43\;\mathrm{grams.}\end{align*} Changing the temperature and/or pressure of a sample of gas changes its volume and therefore its density, but it does not change the mass. Therefore, when the new volume of the gas is found, the mass of oxygen gas in it will still be \begin{align*}1.43\;\mathrm{grams.}\end{align*} The density under the new conditions can be found by dividing the mass by the volume the gas now occupies.

Step 1: Assign known values to the appropriate variable.

\begin{align*} \begin{array}{ll} P_1 = 760.\ \text{torr} & P_2 = 760.\ \text{torr} \\ V_1 = 1.00\ \text{L} & V_2 = x \ \text{unknown} \\ T_1 = 273\ \text{K}& T_2 = 500.\ \text{K} \end{array} \end{align*}

Step 2: Solve the combined gas law for the unknown variable.

\begin{align*}V_2 = \frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ \frac {P_1V_1T_2} {P_2T_1} \end{align*}

Step 3: Substitute the known values into the formula and solve for the unknown.

\begin{align*}\frac {(760.\ \text{torr}) \cdot (1.00\ \text{L}) \cdot (500.\ \text{K})} {(760.\ \text{torr}) \cdot (273\ \text{K})} = 1.83\ \text{liters}\end{align*}

\begin{align*}D_{O_2@500\ \text{K}} = \frac {\text{mass}} {\text{volume}} = \frac {1.43\ \text{g}} {1.83\ \text{L}} = 0.781\ \text{g/L}\end{align*}

Lesson Summary

• Boyle's law states that for a gas at constant temperature, volume is inversely proportional to pressure.
• Charles's law states that for a gas at constant pressure, volume in directly proportional to temperature.
• Gay-Lussac's law states that for a gas at constant volume, pressure is directly proportional to temperature.
• The volume of a mass of gas is dependent on the temperature and pressure. Therefore, these conditions must be given along with the volume of a gas.
• Standard conditions for temperature and pressure are \begin{align*}0^\circ \text{C}\end{align*} and \begin{align*}1.0\;\mathrm{atm}\end{align*}.
• The combined gas law relates the temperature, pressure, and volume of a gas.

Review Questions

1. When a sample of gas is placed in a larger container at the same temperature, what happens to the total force of the molecules hitting the walls?
2. When a sample of gas is placed in a larger container at the same temperature, what happens to the pressure exerted by the gas?
3. If \begin{align*}X\end{align*} and \begin{align*}Y\end{align*} are quantities that are related to each other by inverse proportion, what will the value of \begin{align*}Y\end{align*} become when the value of \begin{align*}X\end{align*} is increased by a factor of five?
4. Under what conditions will the value for the constant, \begin{align*}K\end{align*}, change in the equation for Boyle’s Law, \begin{align*}PV = K\end{align*}.
5. A sample of gas has a volume of \begin{align*}500.\ \mathrm{mL}\end{align*} under a pressure of \begin{align*}500.\ \mathrm{mm \ of \ Hg}\end{align*}. What will be the new volume of the gas if the pressure is reduced to \begin{align*}300.\ \mathrm{mm \ of \ Hg}\end{align*} at constant temperature?
6. A graph is made illustrating Charles’s Law. Which line would be appropriate assuming temperature is measured in Kelvin?
7. At constant pressure, the temperature of a sample of gas is decreased. Will the volume of the sample
1. increase
2. decrease
3. remain the same?
8. A sample of gas has its temperature increased from \begin{align*}-43^\circ\mathrm{C}\end{align*} to \begin{align*}47^\circ\mathrm{C}\end{align*} at constant pressure. If its volume at \begin{align*}-43^\circ\mathrm{C}\end{align*} was \begin{align*}500.\;\mathrm{mL}\end{align*}, what is its volume at \begin{align*}47^\circ\mathrm{C}\end{align*}?
9. A gas is confined in a rigid container and exerts a pressure of \begin{align*}250.\ \mathrm{mm \ of \ Hg}\end{align*} at a temperature of \begin{align*}17^\circ \text{C}\end{align*}. To what temperature must this gas be cooled in order for its pressure to become \begin{align*}216\ \mathrm{mm \ of \ Hg}\end{align*}? Express this temperature in \begin{align*}^\circ \text{C}\end{align*}.
10. What is the abbreviation used to indicate standard conditions for temperature and pressure?
11. A sample of gas has a volume of \begin{align*}500.\ \mathrm{mL}\end{align*} at standard conditions. Find its volume at \begin{align*}47^\circ \text{C}\end{align*} and \begin{align*}800.\ \mathrm{torr}\end{align*}.
12. A sample of gas has a volume of \begin{align*}100.\ \mathrm{L}\end{align*} at \begin{align*}17^\circ \text{C}\end{align*} and \begin{align*}800.\ \mathrm{torr}\end{align*}. To what temperature must the gas be cooled in order for its volume to become \begin{align*}50.0\ \mathrm{L}\end{align*} at a pressure of \begin{align*}600. \ \mathrm{torr}\end{align*}?

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