12.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law

Learning Objectives

By the end of this section, you will be able to:

  • Identify the mathematical relationships between the various properties of gases
  • Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions

During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure 12.2a), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today’s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas – a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation – the ideal gas law – that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law.

Three illustrations of hot air balloons are pictured. Image labeled ( a ) is a black and white drawing of a hydrogen balloon apparently being deflated by a mob of people. In image ( b ), a blue, gold, and red balloon is being held to the ground with ropes while positioned above a platform from which smoke is rising beneath the balloon. In c, an image is shown in grey on a peach-colored background of an inflated balloon with vertical striping in the air. It appears to have a basket attached to its lower side. A large stately building appears in the background.
Figure 12.2a: In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris. (credit: a) work by unknown, PD; b) work by Claude Louis Desrais, PD; c) work by unknown, digitally enhanced by RawPixel, CC BY-SA 4.0)

Pressure and Temperature: Gay-Lussac’s Law

Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure 12.2b) and the pressure increases.

This figure includes three similar diagrams. In the first diagram to the left, a rigid spherical container of a gas to which a pressure gauge is attached at the top is placed in a large beaker of water, indicated in light blue, atop a hot plate. The needle on the pressure gauge points to the far left on the gauge. The diagram is labeled “low P” above and “hot plate off” below. The second similar diagram also has the rigid spherical container of gas placed in a large beaker from which light blue wavy line segments extend from the top of the liquid in the beaker. The beaker is situated on top of a slightly reddened circular area. The needle on the pressure gauge points straight up, or to the middle on the gauge. The diagram is labeled “medium P” above and “hot plate on medium” below. The third diagram also has the rigid spherical container of gas placed in a large beaker in which bubbles appear near the liquid surface and several wavy light blue line segments extend from the surface out of the beaker. The beaker is situated on top of a bright red circular area. The needle on the pressure gauge points to the far right on the gauge. The diagram is labeled “high P” above and “hot plate on high” below.
Figure 12.2b: The effect of temperature on gas pressure: When the hot plate is off, the pressure of the gas in the sphere is relatively low. As the gas is heated, the pressure of the gas in the sphere increases (credit: Chemistry (OpenStax),CC BY 4.0).

This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure 12.2c. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.

This figure includes a table and a graph. The table has 3 columns and 7 rows. The first row is a header, which labels the columns “Temperature, degrees C,” “Temperature, K,” and “Pressure, k P a.” The first column contains the values from top to bottom negative 150, negative 100, negative 50, 0, 50, and 100. The second column contains the values from top to bottom 173, 223, 273, 323, 373, and 423. The third column contains the values 36.0, 46.4, 56.7, 67.1, 77.5, and 88.0. A graph appears to the right of the table. The horizontal axis is labeled “Temperature ( K ).” with markings and labels provided for multiples of 100 beginning at 0 and ending at 500. The vertical axis is labeled “Pressure ( k P a )” with markings and labels provided for multiples of 10, beginning at 0 and ending at 100. Six data points from the table are plotted on the graph with black dots. These dots are connected with a solid black line. A dashed line extends from the data point furthest to the left to the origin. The graph shows a positive linear trend.
Figure 12.2c: For a constant volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot be made at lower temperatures because of the condensation of the gas.) When this line is extrapolated to lower pressures, it reaches a pressure of 0 at –273 °C, which is 0 on the kelvin scale and the lowest possible temperature, called absolute zero (credit: Chemistry (OpenStax),CC BY 4.0).

Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the P–T relationship for gases is known as either Amontons’ law or Gay-Lussac’s law. Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant. Mathematically, this can be written:

[latex]P\;\propto\;T \;\text{or} \; P = \text{constant} \times T \;\text{or} \; P = k \times T[/latex]

where ∝ means “is proportional to,” and k is a proportionality constant that depends on the identity, amount, and volume of the gas.

For a confined, constant volume of gas, the ratio [latex]\frac{P}{T}[/latex] is therefore constant (i.e., [latex]\frac{P}{T} = k[/latex]). If the gas is initially in “Condition 1” (with P = P1 and T = T1), and then changes to “Condition 2” (with P = P2 and T = T2), we have that [latex]\frac{P_1}{T_1} = k[/latex] and [latex]\frac{P_2}{T_2} = k[/latex], which reduces to [latex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/latex]. This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)

Example 12.2a

Predicting Change in Pressure with Temperature

A can of hair spray is used until it is empty except for the propellant, isobutane gas.

  1. On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why?
  2. The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can?

Solution

A can of hair spray is used until it is empty except for the propellant, isobutane gas.

  1. The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)
  2. We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’/Gay-Lussac’s law. Taking P1 and T1 as the initial values, T2 as the temperature where the pressure is unknown and P2 as the unknown pressure, and converting °C to K, we have:
    [latex]\frac{P_1}{T_1} = \frac{P_2}{T_2} \;\text{which means that} \;\frac{360 \;\text{kPa}}{297 \;\text{K}} = \frac{P_2}{323 \;\text{K}}[/latex]
    Rearranging and solving gives: [latex]P_2 = \frac{360 \;\text{kPa} \times 323 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{K}}{297 \;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{K}} = 390 \;\text{kPa}[/latex]

Exercise 12.2a

A sample of nitrogen, N2, occupies 45.0 mL at 27 °C and 600 torr. What pressure will it have if cooled to –73 °C while the volume remains constant?

Check Your Answer[1]

Volume and Temperature: Charles’ Law

If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.

Watch Liquid Nitrogen Experiments: The Balloon (3 mins)

Video source: Jefferson Lab. (2008, September 2). Liquid Nitrogen Experiments: The Balloon [Video]. YouTube.

These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure 12.2d.

This figure includes a table and a graph. The table has 3 columns and 6 rows. The first row is a header, which labels the columns “Temperature, degrees C,” “Temperature, K,” and “Pressure, k P a.” The first column contains the values from top to bottom negative 100, negative 50, 0, 100, and 200. The second column contains the values from top to bottom 173, 223, 273, 373, and 473. The third column contains the values 14.10, 18.26, 22.40, 30.65, and 38.88. A graph appears to the right of the table. The horizontal axis is labeled “Temperature ( K ).” with markings and labels provided for multiples of 100 beginning at 0 and ending at 300. The vertical axis is labeled “Volume ( L )” with marking and labels provided for multiples of 10, beginning at 0 and ending at 30. Five data points from the table are plotted on the graph with black dots. These dots are connected with a solid black line. The graph shows a positive linear trend.
Figure 12.2d: The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph’s origin, representing a temperature of absolute zero (credit: Chemistry (OpenStax),CC BY 4.0).

The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’ law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. Charles’ law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant.

Mathematically, this can be written as:

[latex]V \propto \; T \;\text{or} \; V = \text{constant} \cdot T \;\text{or} \; V = k \cdot T \;\text{or} \; V_1 / T_1 = V_2 / T_2[/latex]

with k being a proportionality constant that depends on the amount and pressure of the gas.

For a confined, constant pressure gas sample, [latex]\frac{V}{T}[/latex] is constant (i.e., the ratio = k), and as seen with the P–T relationship, this leads to another form of Charles’ law: [latex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/latex].

Example 12.2b

Predicting Change in Volume with Temperature

A sample of carbon dioxide, CO2, occupies 0.300 L at 10 °C and 750 torr. What volume will the gas have at 30 °C and 750 torr?

Solution

Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’ law. Taking V1 and T1 as the initial values, T2 as the temperature at which the volume is unknown and V2 as the unknown volume, and converting °C into K we have:

[latex]\frac{V_1}{T_1} = \frac{V_2}{T_2} \;\text{which menas that} \frac{0.300 \;\text{L}}{283 \;\text{K}} = \frac{V_2}{303 \;\text{K}}[/latex]

Rearranging and solving gives: [latex]V_2 = \frac{0.300 \;\text{L} \times 303 \;\rule[0.25ex]{0.8em}{0.1ex}\hspace{-0.8em}\text{K} }{283 \;\rule[0.25ex]{0.8em}{0.1ex}\hspace{-0.8em}\text{K}} = 0.321 \;\text{L}[/latex]

This answer supports our expectation from Charles’ law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L).

Exercise 12.2b

A sample of oxygen, O2, occupies 32.2 mL at 30 °C and 452 torr. What volume will it occupy at –70 °C and the same pressure?

Check Your Answer[2]

Example 12.2c

Measuring Temperature with a Volume Change

Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm3 when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm3. Find the temperature of boiling ammonia on the kelvin and Celsius scales.

Solution

A volume change caused by a temperature change at constant pressure means we should use Charles’ law. Taking V1 and T1 as the initial values, T2 as the temperature at which the volume is unknown and V2 as the unknown volume, and converting °C into K we have:

[latex]\frac{V_1}{T_1} = \frac{V_2}{T_2} \;\text{which means that} \frac{150.0 \;\text{cm}^3}{273.15 \;\text{K}} = \frac{131.7 \;\text{cm}^3}{T_2}[/latex]

Rearrangement gives [latex]T_2 = \frac{131.7 \;\rule[0.25ex]{0.8em}{0.1ex}\hspace{-0.8em}\text{cm}^3 \times 273.15 \;\text{K} }{150.0 \;\rule[0.25ex]{0.8em}{0.1ex}\hspace{-0.8em}\text{cm}^3} = 239.8 \;\text{K}[/latex]

Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C.

Exercise 12.2c

What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?

Check Your Answer[3]

Volume and Pressure: Boyle’s Law

If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure 12.2e.

This figure contains a diagram and two graphs. The diagram shows a syringe labeled with a scale in m l or c c with multiples of 5 labeled beginning at 5 and ending at 30. The markings halfway between these measurements are also provided. Attached at the top of the syringe is a pressure gauge with a scale marked by fives from 40 on the left to 5 on the right. The gauge needle rests between 10 and 15, slightly closer to 15. The syringe plunger position indicates a volume measurement about halfway between 10 and 15 m l or c c. The first graph is labeled “V ( m L )” on the horizontal axis and “P ( p s i )” on the vertical axis. Points are labeled at 5, 10, 15, 20, and 25 m L with corresponding values of 39.0, 19.5, 13.0, 9.8, and 6.5 p s i. The points are connected with a smooth curve that is declining at a decreasing rate of change. The second graph is labeled “V ( m L )” on the horizontal axis and “1 divided by P ( p s i )” on the vertical axis. The horizontal axis is labeled at multiples of 5, beginning at zero and extending up to 35 m L. The vertical axis is labeled by multiples of 0.02, beginning at 0 and extending up to 0.18. Six points indicated by black dots on this graph are connected with a black line segment showing a positive linear trend.
Figure 12.2e: When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since P and V are inversely proportional, a graph of 1/P vs. V is linear (credit: Chemistry (OpenStax), CC BY 4.0).

Unlike the P–T and V–T relationships, pressure and volume are not directly proportional to each other. Instead, P and V exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:

[latex]P \propto \; 1/V \;\text{or} \; P = k \cdot 1/V \;\text{or} \; P \cdot V = k \;\text{or} \; P_1 V_1 = P_2 V_2[/latex]

with k being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure ([latex]\frac{1}{P}[/latex]) versus the volume (V), or the inverse of volume ([latex]\frac{1}{V}[/latex]) versus the pressure (P). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to “linearize” their data. If we plot P versus V, we obtain a hyperbola (see Figure 12.2f).

This diagram shows two graphs. In a, a graph is shown with volume on the horizontal axis and pressure on the vertical axis. A curved line is shown on the graph showing a decreasing trend with a decreasing rate of change. In b, a graph is shown with volume on the horizontal axis and one divided by pressure on the vertical axis. A line segment, beginning at the origin of the graph, shows a positive, linear trend.
Figure 12.2f: The relationship between pressure and volume is inversely proportional. (a) The graph of P vs. V is a hyperbola, whereas (b) the graph of (1/P) vs. V is linear (credit: Chemistry (OpenStax), CC BY 4.0).

The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle’s law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.

Example 12.2d

Volume of a Gas Sample

The sample of gas in Figure 12.2e has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using:

  1. the P–V graph in Figure 12.2e
  2. the [latex]\frac{1}{p}[/latex] vs. V graph in Figure 12.2e
  3. the Boyle’s law equation

Comment on the likely accuracy of each method.

Solution

  1. Estimating from the P–V graph gives a value for P somewhere around 27 psi.
  2. Estimating from the [latex]\frac{1}{P}[/latex] versus V graph give a value of about 26 psi.
  3. From Boyle’s law, we know that the product of pressure and volume (PV) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have P1V1 = k and P2V2 = k which means that P1V1 = P2V2.
    Using P1 and V1 as the known values 13.0 psi and 15.0 mL, P2 as the pressure at which the volume is unknown, and V2 as the unknown volume, we have:[latex]P_1 V_1 = P_2 V_2 \;\text{or} \; 13.0 \;\text{psi} \times 15.0 \;\text{mL} = P_2 \times 7.5 \;\text{mL}[/latex]Solving:[latex]P_2 = \frac{13.0 \;\text{psi} \times 15.0 \;\rule[0.5ex]{1.2em}{0.1ex}\hspace{-1.2em}\text{mL}}{7.5 \;\rule[0.5ex]{1.2em}{0.1ex}\hspace{-1.2em}\text{mL}} = 26 \;\text{psi}[/latex]

It was more difficult to estimate well from the P–V graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.

Exercise 12.2d

The sample of gas in Figure 12.2e has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 psi, using:

  1. the P–V graph in Figure 12.2e
  2. the [latex]\frac{1}{P}[/latex] vs. V graph in Figure 12.2e
  3. the Boyle’s law equation

Comment on the likely accuracy of each method.

Check Your Answer[4]

Breathing and Boyle’s Law

What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle’s law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle’s law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle’s law cycle for the rest of your life (Figure 12.2g).

This figure contains two diagrams of a cross section of the human head and torso. The first diagram on the left is labeled “Inspiration.” It shows curved arrows in gray proceeding through the nasal passages and mouth to the lungs. An arrow points downward from the diaphragm, which is relatively flat, just beneath the lungs. This arrow is labeled “Diaphragm contracts.” At the entrance to the mouth and nasal passages, a label of P subscript lungs equals 1 dash 3 torr lower” is provided. The second, similar diagram, which is labeled “Expiration,” reverses the direction of both arrows. Arrows extend from the lungs out through the nasal passages and mouth. Similarly, an arrow points up to the diaphragm, showing a curved diaphragm and lungs reduced in size from the previous image. This arrow is labeled “Diaphragm relaxes.” At the entrance to the mouth and nasal passages, a label of P subscript lungs equals 1 dash 3 torr higher” is provided.
Figure 12.2g: Breathing occurs because expanding and contracting lung volume creates small pressure differences between your lungs and your surroundings, causing air to be drawn into and forced out of your lungs (credit: Chemistry (OpenStax), CC BY 4.0).

Moles of Gas and Volume: Avogadro’s Law

The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behaviour of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed byAvogadro’s law: For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant.

In equation form, this is written as:

[latex]V \propto n \;\text{or} \; V = k \times n \;\text{or} \; \frac{V_1}{n_1} = \frac{V_2}{n_2}[/latex]

Mathematical relationships can also be determined for the other variable pairs, such as P versus n, and n versus T.

Exercise 12.2e

Practice using the following PhET simulation: Gas Properties

Activity source: Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC-BY-4.0

The Ideal Gas Law

To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:

  • Boyle’s law: PV = constant at constant T and n
  • Gay-Lussac’s law: [latex]\frac{P}{T}[/latex] = constant at constant V and n
  • Charles’ law: [latex]\frac{V}{T}[/latex] = constant at constant P and n
  • Avogadro’s law: [latex]\frac{V}{n}[/latex] = constant at constant P and T

Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:

[latex]PV = nRT[/latex]

where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol–1 K–1 and 8.314 kPa L mol–1 K–1.

Gases whose properties of P, V, and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behaviour or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behaviour, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behaviour observed for many gases at relatively high pressures and low temperatures.

The ideal gas equation contains five terms, the gas constant R and the variable properties P, V, n, and T. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.

Example 12.2e

Using the Ideal Gas Law

Methane, CH4, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH4. What is the volume of this much methane at 25 °C and 745 torr?

Solution

We must rearrange PV = nRT to solve for V: [latex]V = \frac{nRT}{P}[/latex]

If we choose to use R = 0.08206 L atm mol–1 K–1, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.

Converting into the “right” units:

[latex]n = 655 \;\rule[0.5ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{g CH}_4 \times \frac{1 \;\text{mol}}{16.043 \;\rule[0.5ex]{2.2em}{0.1ex}\hspace{-2.2em}\text{g CH}_4} = 40.8 \;\text{mol}[/latex]
[latex]T = 25 \;^\circ\text{C} + 273 = 298 \;\text{K}[/latex]
[latex]P = 745 \;\rule[0.5ex]{1.8em}{0.1ex}\hspace{-1.8em}\text{torr} \times \frac{1 \;\text{atm}}{760 \;\rule[0.5ex]{1.4em}{0.1ex}\hspace{-1.4em}\text{torr}} = 0.980 \;\text{atm}[/latex]
[latex]V = \frac{nRT}{P} = \frac{(40.8 \;\rule[0.5ex]{1.2em}{0.1ex}\hspace{-1.2em}\text{mol})(0.08206 \text{L}\;\rule[0.5ex]{5.1em}{0.1ex}\hspace{-5.1em}\text{atm mol}^{-1} \;\text{K}^{-1})(298 \;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{K})}{0.980 \;\rule[0.5ex]{1.5em}{0.1ex}\hspace{-1.5em}\text{atm}} = 1.02 \times 10^3 \;\text{L}[/latex]

It would require 1020 L (269 gal) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.

Exercise 12.2f

Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-L storage tank of a modern hydrogen-powered car.

Check Your Answer[5]

If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: [latex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/latex] using units of atm, L, and K. Both sets of conditions are equal to the product of n ×R (where n = the number of moles of the gas and R is the ideal gas law).

Example 12.2f

Using the Combined Gas Law

When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 12.2h). If the water temperature is 27 °C, how many litres of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?

A photograph of four scuba divers underwater with a tank on their back and bubbles ascending from the breathing apparatus.
Figure 12.2h: Scuba divers use compressed air to breathe while underwater (credit: work by Subro89, CC BY-SA 4.0).

Solution

Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have:

[latex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \longrightarrow \frac{(153 \;\text{atm})(13.2 \;\text{L})}{(300 \;\text{K})} = \frac{(3.13 \;\text{atm})(V_2)}{(310 \;\text{K})}[/latex]

Solving for V2:

[latex]V_2 = \frac{(153 \;\rule[0.5ex]{1.2em}{0.1ex}\hspace{-1.2em}\text{atm})(13.2 \;\text{L})(310 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{K})}{(300 \;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{K})(3.13 \;\rule[0.5ex]{1.2em}{0.1ex}\hspace{-1.2em}\text{atm})} = 667 \;\text{L}[/latex]

(Note: Be advised that this particular example is one in which the assumption of ideal gas behaviour is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.)

Exercise 12.2g

A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sample at 0 °C and 1.00 atm.

Check Your Answer[6]

The Interdependence between Ocean Depth and Pressure in Scuba Diving

Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure 12.2i) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.

Colourful underwater corals and anemones in hues of yellow, orange, green, and brown, surrounded by water is pictured.
Figure 12.2i: Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: work by Kyle Taylor, CC BY 2.0)

Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as “atmospheres absolute” or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body’s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in his BCD expands because of lower pressure according to Boyle’s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and she or he begins to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.

Watch The ABC’s of gas: Avogadro, Boyle, Charles – Brian Bennett

Video source: TED-Ed. (2012c, October 9). The ABC’s of gas: Avogadro, Boyle, Charles – Brian Bennett [Video]. YouTube. https://www.youtube.com/watch?v=BY9VGS2eXas

Standard Conditions of Temperature and Pressure

We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa). At STP, an ideal gas has a volume of about 22.4 L—this is referred to as the standard molar volume (Figure 12.2j).

This figure shows three balloons each filled with H e, N H subscript 2, and O subscript 2 respectively. Beneath the first balloon is the label “4 g of He” Beneath the second balloon is the label, “15 g of N H subscript 2.” Beneath the third balloon is the label “32 g of O subscript 2.” Each balloon contains the same number of molecules of their respective gases.
Figure 12.2j: Since the number of moles in a given volume of gas varies with pressure and temperature changes, chemists use standard temperature and pressure (273.15 K and 1 atm or 101.325 kPa) to report properties of gases (credit: Chemistry (OpenStax), CC BY 4.0).

Exercise 12.2h

Check Your Learning Exercise (Text Version)

A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C. If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can?

Check Your Answer[7]

Source: “Exercise 12.2h” is adapted from “Exercise 8.2-10” in General Chemistry 1 & 2, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson, licensed under CC BY 4.0.

Links to Interactive Learning Tools

Explore Volume-Temperature Gas Law from the Physics Classroom.

Explore Pressure-Volume Gas Law from the Physics Classroom.

Explore Pressure-Temperature Gas Law from the Physics Classroom.

Key Equations

  • [latex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/latex]
  • [latex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/latex]
  • [latex]P_1 V_1 = P_2 V_2[/latex]
  • [latex]\; \frac{V_1}{n_1} = \frac{V_2}{n_2}[/latex]
  • [latex]PV = nRT[/latex]
  • [latex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/latex]

Attribution & References

Except where otherwise noted, this page is adapted by JR van Haarlem from “9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. ​Access for free at Chemistry (OpenStax)​
  1. 400 torr ↵
  2. 21.6 mL ↵
  3. 635 mL ↵
  4. (a) about 17–18 mL; (b) ~18 mL; (c) 17.7 mL; it was more difficult to estimate well from the P-V graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow ↵
  5. 350 bar ↵
  6. 0.193 L ↵
  7. 3.4 x 103 torr ↵
definition

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Enhanced Introductory College Chemistry Copyright © 2023 by Gregory Anderson; Caryn Fahey; Jackie MacDonald; Adrienne Richards; Samantha Sullivan Sauer; J.R. van Haarlem; and David Wegman is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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