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# 14.3: Ideal Gases

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Know the ideal gas law, and know which of the different values for the ideal gas constant to use in a given situation.
• Use the ideal gas law to calculate the pressure, volume, temperature, or number of moles of an ideal gas when the other three quantities are known.
• Use the ideal gas law to calculate the molar mass or the density of a gas.
• Use the ideal gas law in stoichiometry problems involving gases that are not at STP.
• Explain the conditions under which real gases are most ideal or least ideal.

## Lesson Vocabulary

• ideal gas constant
• ideal gas law
• real gas

### Recalling Prior Knowledge

• What assumptions from the kinetic-molecular theory of gases are not capable of being precisely followed?
• How is molar volume used in stoichiometry and gas density calculations?

The combined gas law is useful for analyzing situations where a gas at a certain initial set of conditions is changed to a new set of conditions. In this lesson, you are introduced to the ideal gas law, which is useful for studying a quantity of gas under any given conditions of temperature and pressure.

## Ideal Gas Law

A number of relationships between the four variables that are used to describe gases—P, V, T, and n—have been established with the gas laws from the previous lesson. The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro’s Law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation.

P1×V1T1×n1=P2×V2T2×n2\begin{align*}\mathrm{\dfrac{P_1 \times V_1}{T_1 \times n_1}=\dfrac{P_2 \times V_2}{T_2 \times n_2}}\end{align*}

As with the other gas laws, we can also say that (P × V) / (T × n) is equal to a constant. The constant can be evaluated, provided that the gas being described is considered to be ideal.

The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles for any ideal gas. If we substitute in the variable R for the constant, the equation becomes:

P×VT×n=R\begin{align*}\mathrm{\dfrac{P \times V}{T \times n}=R}\end{align*}

The ideal gas law is conventionally rearranged to look this way, with the multiplication signs omitted:

PV=nRT

The variable R in the equation is called the ideal gas constant.

### Evaluating the Ideal Gas Constant

The value of R, the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature, and it is conventional to use units of liters for the volume. However, pressure is commonly measured in one of three units: kPa, atm, or mmHg. The numerical value of R depends on the units in which pressure is expressed.

We will demonstrate how R is calculated when the pressure is measured in kPa. Recall from the chapter, “The Mole,” that the volume of 1.00 mol of any gas at STP is measured to be 22.414 L. We can substitute 101.325 kPa for pressure, 22.414 L for volume, and 273.15 K for temperature into the ideal gas equation and solve for R.

R=PVnT=101.325 kPa×22.414 L1.000 mol×273.15 K=8.314 kPaL/Kmol\begin{align*}\mathrm{R=\dfrac{PV}{nT}=\dfrac{101.325 \ kPa \times 22.414 \ L}{1.000 \ mol \times 273.15 \ K} = 8.314 \ kPa \cdot L / K \cdot mol}\end{align*}

This is the value of R that should be used in the ideal gas equation when the pressure is given in kPa. The table below (Table below) shows a summary of this and the other possible values of R. It is important to choose the correct value of R to use for a given problem.

Values of the Ideal Gas Constant
Unit of P Unit of V Unit of n Unit of T Value and unit of R
kPa L mol K 8.314 J/K•mol
atm L mol K 0.08206 L•atm/K•mol
mmHg L mol K 62.36 L•mmHg/K•mol

Notice that the unit for R when the pressure is in kPa has been changed to J/K•mol. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule (J).

Sample Problem 14.6: Ideal Gas Law

What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4 kPa and a temperature of 19°C? Assume the oxygen acts as an ideal gas.

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

Known

• P = 88.4 kPa
• T = 19°C = 292 K
• mass of O2 = 3.760 g
• molar mass of O2 = 32.00 g/mol
• R = 8.314 J/K•mol

Unknown

• V = ? L

In order to use the ideal gas law, the number of moles of O2 (n) must be found from the given mass and the molar mass. Then, use PV = nRT to solve for the volume of the sample.

Step 2: Solve.

3.760g×1molO232.00gO2=0.1175molO2\begin{align*}\mathrm{3.760\:g \times \dfrac{1\:mol\:O_2}{32.00\:g\:O_2}=0.1175\:mol\:O_2}\end{align*}

Rearrange the ideal gas law and solve for V.

V=nRTP=0.1175 mol×8.314 J/Kmol×292 K88.4 kPa=3.23 L O2\begin{align*}\mathrm{V=\dfrac{nRT}{P}=\dfrac{0.1175 \ mol \times 8.314 \ J/K \cdot mol \times 292 \ K}{88.4 \ kPa}=3.23 \ L \ O_2}\end{align*}

The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume (22.4 L/mol) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for T and P. Since a joule (J) = kPa•L, the units cancel correctly, leaving a volume in liters.

Practice Problem
1. At what Celsius temperature does 0.815 mol of an ideal gas occupy 33.7 L at a pressure of 108.1 kPa?
2. What is the pressure (in atm) in a 850. mL flask that holds 9.44 g of chlorine gas (Cl2) at 11°C?

### Finding Molar Mass and Density

A chemical reaction is performed that produces a gas, which is then collected. After determining the mass and volume of the collected sample, the molar mass of the unknown gas can be found using the ideal gas law, provided the temperature and pressure of the gas are also known.

Sample Problem 14.6: Molar Mass and the Ideal Gas Law

A certain reaction occurs, producing a gaseous oxide of nitrogen. The gas has a mass of 1.211 g and occupies a volume of 677 mL. The temperature in the laboratory is 23°C and the air pressure is 0.987 atm. Calculate the molar mass of the gas and deduce its formula. Assume the gas is ideal.

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

Known

• mass = 1.211 g
• V = 677 mL = 0.677 L
• T = 23°C = 296 K
• P = 0.987 atm
• R = 0.08206 L•atm/K•mol

Unknown

• n = ? mol
• molar mass = ? g/mol

First, the ideal gas law will be used to solve for the moles of unknown gas (n). Then, the mass of the gas divided by the moles will give the molar mass.

Step 2: Solve.

n=PVRT=0.987 atm×0.677 L0.08206 Latm/Kmol×296 K=0.0275 mol\begin{align*}\mathrm{n=\dfrac{PV}{RT}=\dfrac{0.987 \ atm \times 0.677 \ L}{0.08206 \ L \cdot atm/K \cdot mol \times 296 \ K}=0.0275 \ mol}\end{align*}

Now divide the mass of the sample (in g) by the amount (in mol) to get the molar mass.

molar mass=1.211 g0.0275 mol=44.0 g/mol\begin{align*}\mathrm{molar \ mass=\dfrac{1.211 \ g}{0.0275 \ mol}=44.0 \ g/mol}\end{align*}

Since nitrogen has a molar mass of 14 g/mol, and oxygen has a molar mass of 16 g/mol, the formula N2O would produce the correct molar mass.

The R value that includes units of atm was chosen for this problem. The calculated molar mass corresponds to a reasonable molecular formula for a compound composed of nitrogen and oxygen.

Practice Problem
1. Determine the molar mass of a gas if 4.91 g of the gas occupies a volume of 2.85 L at a pressure of 812 mmHg and a temperature of 29°C. Which of the noble gases would be consistent with this data?

The ideal gas law can be used to find the density of a gas at conditions that are not standard. For example, we will determine the density of ammonia (NH3) at 0.913 atm and 20°C, assuming the ammonia acts as an ideal gas. First, the molar mass of ammonia is calculated from its formula to be 17.04 g/mol. Next, assume exactly 1 mol of ammonia (n = 1) and calculate the volume that such an amount would occupy at the given temperature and pressure.

V=nRTP=1.00 mol×0.08206 Latm/Kmol×293 K0.913 atm=26.3 L\begin{align*}\mathrm{V=\dfrac{nRT}{P}=\dfrac{1.00 \ mol \times 0.08206 \ L \cdot atm/K \cdot mol \times 293 \ K}{0.913 \ atm}=26.3 \ L}\end{align*}

Now, the density can be calculated by dividing the mass of one mole of ammonia by the volume above.

density=17.04 g26.3 L=0.647 g/L\begin{align*}\mathrm{density=\dfrac{17.04 \ g}{26.3 \ L}=0.647 \ g/L}\end{align*}

As a point of comparison, this density is slightly less than the density of ammonia at STP, which is equal to (17.04 g/mol)/(22.4 L/mol) = 0.761 g/L. It makes sense that the density should be lower than at STP, since both the increase in temperature (from 0°C to 20°C) and the decrease in pressure (from 1 atm to 0.913 atm) would cause the NH3 molecules to spread out a bit farther from one another.

Video lectures are extremely helpful in learning about ideal gases. You can find five successive lectures at:

## Gas Stoichiometry

In the chapter Stoichiometry, you learned how to use molar volume to solve stoichiometry problems for chemical reactions involving one or more gases at STP. Now, we can use the ideal gas law to expand our treatment of chemical reactions to solve stoichiometry problems for reactions that occur at any temperature and pressure.

Sample Problem 14.7: Gas Stoichiometry and the Ideal Gas Law

What volume of carbon dioxide is produced by the combustion of 25.21 g of ethanol (C2H5OH) at 54°C and 728 mmHg? Assume the gas is ideal.

Before using the ideal gas law, it is necessary to write and balance the chemical equation. Recall that in combustion reactions, the given substance reacts with O2 to form CO2 and H2O. Here is the balanced equation for the combustion of ethanol:

C2H5OH(l)+3O2(g)2CO2(g)+3H2O(l)\begin{align*}\mathrm{C_2H_5OH}{(l)}+3\text{O}_{2}{(g)} \rightarrow 2\text{CO}_{2}{(g)}+3\mathrm{H_2O}{(l)}\end{align*}

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

Known

• mass of C2H5OH = 25.21 g
• molar mass of C2H5OH = 46.08 g/mol
• P = 728 mmHg
• T = 54°C = 327 K

Unknown

• volume of CO2 = ? L

The number of moles of carbon dioxide gas is first calculated by stoichiometry. Then, the ideal gas law is used to calculate the volume of CO2 produced.

Step 2: Solve.

25.21 g C2H5OH×1 mol C2H5OH46.08 g C2H5OH×2 mol CO21 mol C2H5OH=1.094 mol CO2\begin{align*}\mathrm{25.21 \ g \ C_2H_5OH \times \dfrac{1 \ mol \ C_2H_5OH}{46.08 \ g \ C_2H_5OH} \times \dfrac{2 \ mol \ CO_2}{1 \ mol \ C_2H_5OH}=1.094 \ mol \ CO_2}\end{align*}

The moles of ethanol (n) is now substituted into PV=nRT to solve for the volume.

V=nRTP=1.094 mol×62.36 LmmHg/Kmol×327 K728 mmHg=30.6 L\begin{align*}\mathrm{V=\dfrac{nRT}{P}=\dfrac{1.094 \ mol \times 62.36 \ L \cdot mmHg/K \cdot mol \times 327 \ K}{728 \ mmHg}=30.6 \ L}\end{align*}

The mass of ethanol is slightly more than one half mole, meaning that the mole ratio results in slightly more than one mole of carbon dioxide being produced. Because of the elevated temperature and reduced pressure compared to STP, the resulting volume is larger than 22.4 L.

Practice Problem
1. Solid copper(II) oxide decomposes to produce copper and oxygen gas. 168 L of oxygen gas is collected at 38°C and 91.4 kPa. Find the mass of copper(II) oxide that decomposed.

## Real Gases and Ideal Gases

An ideal gas is one that follows the gas laws at all conditions of temperature and pressure. To do so, the gas would need to completely abide by the kinetic-molecular theory. The gas particles would need to occupy zero volume, and they would need to exhibit no attractive forces whatsoever toward each other. Since neither of those conditions can be true, there is no such thing as a truly ideal gas. A real gas is a gas that does not behave according to the assumptions of the kinetic-molecular theory. Fortunately, at the conditions of temperature and pressure that are normally encountered in a laboratory, real gases tend to behave very much like ideal gases.

Under what conditions do gases behave least ideally? When a gas is put under high pressure, its molecules are forced closer together, and the empty space between the particles is diminished. A decrease in the empty space means the assumption that the volume of the particles themselves is negligible is less valid. When a gas is cooled, the decrease in the kinetic energy of the particles causes them to slow down. If the particles are moving at slower speeds, the attractive forces between them are more prominent. Put another way, continued cooling of the gas will eventually turn it into a liquid, and a liquid is certainly not an ideal gas anymore (Figure below). In summary, a real gas deviates most from an ideal gas at low temperatures and high pressures. Gases are most ideal at high temperature and low pressure.

Nitrogen gas that has been cooled to 77 K has turned to a liquid and must be stored in a vacuum insulated container to prevent it from rapidly vaporizing.

Shown below (Figure below) is a graph of PV/RT plotted against pressure for 1 mol of a gas at three different temperatures: 200 K, 500 K, and 1000 K. An ideal gas would have a value of 1 for that ratio at all temperatures and pressures, and the graph would simply be a horizontal line, which is clearly not the case. As the pressure begins to rise, the attractive forces become more significant, causing the volume of the gas to be less than expected and the value of PV/RT drops under 1. At very high pressures, the volume occupied by the particles themselves starts to become significant, and the value of PV/RT rises above 1. Notice that the magnitude of the deviation from ideality is greatest for the gas at 200 K and smallest at 1000 K.

This graph shows how real gases deviate from ideal gases at high pressures and at low temperatures.

The ideality of a gas also depends on the strength and type of intermolecular attractive forces that exist between the particles. Gases whose attractive forces are weak, are more ideal than those with strong attractive forces. At the same temperature and pressure, neon is more ideal than water vapor because neon’s atoms are only attracted to one another by weak dispersion forces, while water vapor’s molecules are attracted to each other by relatively stronger hydrogen bonds. Helium is a more ideal gas than neon because it has even weaker dispersion forces, due to its smaller number of electrons per particle.

## Lesson Summary

• The ideal gas law is derived from the combined gas law and allows the number of moles of a gas to be evaluated in a single equation along with pressure, volume, and temperature. The value of the ideal gas constant depends upon the unit of pressure that is used in the problem.
• The ideal gas law can be used to calculate pressure, volume, temperature, or number of moles of a gas when three of the variables are known and the gas is not undergoing a change in conditions.
• The ideal gas law can also be used to calculate the density of a gas and to perform stoichiometry problems with chemical reactions that involve one or more gases.
• Real gases deviate most from ideal behavior at low temperature and high pressure. Strong intermolecular attractive forces between the gas particles results in larger deviations from ideality at a given temperature and pressure.

## Lesson Review Questions

### Reviewing Concepts

1. Explain why it is impossible for a truly ideal gas to exist.
2. Under what conditions of temperature and pressure are gases most ideal? Explain.

### Problems

1. What is the volume of 1.50 moles of an ideal gas at 94.5 kPa and 320. K?
2. At what Celsius temperature does 10.5 g of methane gas (CH4) under a pressure of 1.77 atm occupy 35.6 L? Assume the methane is ideal.
3. 22.0 grams of helium occupies 8.80 L at a temperature of 67.0°C. Calculate the pressure in kPa.
4. A 3.40 L sample of nitrogen gas (N2) is at 920. mmHg pressure and a temperature of 400 K. Calculate the mass of the sample.
5. A student obtains a canister filled with one of the noble gases, but the label is illegible. She determines that the mass of the gas is 382.7 grams. The canister’s volume is 35.00 L and the pressure and temperature are 320. kPa and 22°C. Find the molar mass. Which noble gas is it?
6. What is the density of xenon gas under the following conditions?
1. at STP
2. at 43.2 mmHg and −89°C
7. At what Celsius temperature does F2 gas have a density of 1.15 g/L when the pressure is 1.07 atm?
8. Which gas below would you expect to be the most ideal? The least ideal? Explain your answers.
1. H2 at 200 K
2. H2 at 400 K
3. NH3 at 200 K
4. NH3 at 400 K
9. The reaction of carbon with sulfur dioxide produces carbon disulfide and carbon monoxide. 5C(s)+2SO2(g)CS2(l)+4CO(g)\begin{align*}\text{5C}{(s)}+2\text{SO}_2{(g)} \rightarrow \text{CS}_2{(l)}+4\text{CO}{(g)}\end{align*}
1. What volume of SO2 at 138 kPa and 224°C is required to react completely with 325 g of carbon?
2. A certain reaction produces 42.2 L of CO at 812 mmHg and 500 K. What mass of CS2 was also produced in the reaction?
3. 21.98 g of carbon is reacted with 19.40 L of SO2 at 0.893 atm and 315°C. Determine the limiting reactant. What mass of CS2 will be produced?

## Points to Consider

Dalton’s law is a gas law that deals with mixtures of gases. In a gas mixture, all of the particles are still relatively far apart, so the different gases in the mixture are assumed to interact with one another only during brief collisions.

• How do you determine the pressure of one particular gas when that gas is part of a mixture of gases?
• In the laboratory, gases are often collected by bubbling the gas into a water-filled container. What adjustment needs to be made to determine the amount of a gas when it is collected in this way?

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