7.4: Applying the Ideal Gas Law

Learning Objectives

• To use the ideal gas law to describe the behavior of a gas.

The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables (P, V, T, and n). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters (P, V, T, and n) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample.

Example \(\PageIndex{1}\)

The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft³), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon?

Given: volume, temperature, and pressure

Asked for: amount of gas

Strategy:

1. Solve the ideal gas law for the unknown quantity, in this case n.
2. Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed.

Solution:

A We are given values for P, T, and V and asked to calculate n. If we solve the ideal gas law (Equation \(\ref{10.4.4}\)) for \(n\), we obtain

\[\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm \nonumber\]

B P and T are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:

\[T=273+30=303{\rm K}\nonumber\]

Substituting these values into the expression we derived for n, we obtain

\[\begin{align*} n &=\dfrac{PV}{RT} \\[4pt] &=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K} \\[4pt] &=1.23\times10^3\;mol \end{align*}\]

Exercise \(\PageIndex{1}\)

Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO₂. What is the pressure of the gas at 25°C?

Answer

1.5 atm

In Example \(\PageIndex{1}\), we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example \(\PageIndex{5}\).

General Gas Equation

When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is:

\[\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f) \\ P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array}\]

Both equations can be rearranged to give:

\[R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f}\]

The two equations are equal to each other since each is equal to the same constant \(R\). Therefore, we have:

\[\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\label{10.4.8}\]

The equation is called the general gas equation. The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties.

Example \(\PageIndex{2}\)

Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example \(\PageIndex{1}\)?

Given: temperature, pressure, amount, and volume in August; temperature in January

Asked for: volume in January

Strategy:

1. Use the results from Example \(\PageIndex{1}\) for August as the initial conditions and then calculate the change in volume due to the change in temperature from 30°C to −10°C. Begin by constructing a table showing the initial and final conditions.
2. Simplify the general gas equation by eliminating the quantities that are held constant between the initial and final conditions, in this case \(P\) and \(n\).
3. Solve for the unknown parameter.

Solution:

A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions:

Initial (August)	Final (January)
\(T_i=30\, °C = 303\, K\)	\(T_f=​−10\,°C = 263\, K\)
\(P_i= 0.980 \, atm\)	\(P_f= 0.980\, atm\)
\(n_i=1.23 × 10^3\, mol\)	\(n_f= 1.23 × 10^3\, mol\)
\(V_i=31150\, L\)	\(V_f=?\)

B Both \(n\) and \(P\) are the same in both cases​ (\(n_i=n_f,P_i=P_f\)). Therefore, Equation \ref{10.4.8} can be simplified to:

\[\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f} \nonumber\]

This is the relationship first noted by Charles.

C ​Solving the equation for \(V_f\), we get:

\[\begin{align*} V_f &=V_i\times\dfrac{T_f}{T_i} \\[4pt] &=\rm31150\;L\times\dfrac{263\;K}{303\;K} \\[4pt] &=2.70\times10^4\;L \end{align*}\]

It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change.

Exercise \(\PageIndex{2}\)

At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon?

Answer

0.52 L

Example \(\PageIndex{1}\) illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle (PV = constant) and the relationship between volume and amount observed by Avogadro (V/n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example \(\PageIndex{1}\) can be applied in any such case, as we demonstrate in Example \(\PageIndex{2}\) (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion).

Example \(\PageIndex{3}\)

Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise \(\PageIndex{1}\) (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)?

Given: initial volume, amount, temperature, and pressure; final temperature

Asked for: final pressure

Strategy:

Follow the strategy outlined in Example \(\PageIndex{2}\).

Solution:

Prepare a table to determine which parameters change and which are held constant:

Initial	Final
\(V_i=0.406\;\rm L\)	\(V_f=0.406\;\rm L\)
\(n_i=0.025\;\rm mol\)	\(n_f=0.025\;\rm mol\)
\(T_i=\rm25\;^\circ C=298\;K\)	\(T_i=\rm750\;^\circ C=1023\;K\)
\(P_i=1.5\;\rm atm\)	\(P_f=?\)

Both \(V\) and \(n\) are the same in both cases​ (\(V_i=V_f,n_i=n_f\)). Therefore, Equation can be simplified to:

\[ \dfrac{P_i}{T_i} = \dfrac{P_f}{T_f}\]

By solving the equation for \(P_f\), we get:

\[\begin{align*} P_f &=P_i\times\dfrac{T_f}{T_i} \\[4pt] &=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K} \\[4pt] &=5.1\;atm \end{align*}\]

This pressure is more than enough to rupture a thin sheet metal container and cause an explosion!

Exercise \(\PageIndex{3}\)

Suppose that a fire extinguisher, filled with CO₂ to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher?

Answer

23.4 atm

In Examples \(\PageIndex{1}\) and \(\PageIndex{2}\), two of the four parameters (P, V, T, and n) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P, V, and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions.

Example \(\PageIndex{4}\)

We saw in Example \(\PageIndex{1}\) that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 10³ mol of H₂ gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude?

Given: initial pressure, temperature, amount, and volume; final pressure and temperature

Asked for: final volume

Strategy:

Follow the strategy outlined in Example \(\PageIndex{3}\).

Solution:

Begin by setting up a table of the two sets of conditions:

Initial	Final
\(P_i=745\;\rm mmHg=0.980\;atm\)	\(P_f=312\;\rm mmHg=0.411\;atm\)
\(T_i=\rm30\;^\circ C=303\;K\)	\(T_f=\rm750-30\;^\circ C=243\;K\)
\(n_i=\rm1.2\times10^3\;mol\)	\(n_i=\rm1.2\times10^3\;mol\)
\(V_i=\rm31150\;L\)	\(V_f=?\)

By eliminating the constant property (\(n\)) of the gas, Equation \(\ref{10.4.8}\) is simplified to:

\[\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}\]

By solving the equation for \(V_f\), we get:

\[\begin{align*} V_f &=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i} \\[4pt] &=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K} \\[4pt] &=5.96\times10^4\;L \end{align*}\]

Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/P, the variable that changes the most will have the greatest effect on V. In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation.

We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft:

Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems.

Exercise \(\PageIndex{4}\)

A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.)

Answer

4.07 × 10³

Using the Ideal Gas Law to Calculate Gas Densities

The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain

\[\dfrac{n}{V}=\dfrac{P}{RT}\label{10.4.9}\]

The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass (\(m\), in grams) divided by its molar mass (\(M\), in grams per mole):

\[n=\dfrac{m}{M}\label{10.4.10}\]

Substituting this expression for \(n\) into Equation \(\ref{10.4.9}\) gives

\[\dfrac{m}{MV}=\dfrac{P}{RT}\label{10.4.11}\]

Because \(m/V\) is the density \(d\) of a substance, we can replace \(m/V\) by \(d\) and rearrange to give

\[\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\label{10.4.12}\]

The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL).

Example \(\PageIndex{5}\)

Calculate the density of butane at 25°C and a pressure of 750 mmHg.

Given: compound, temperature, and pressure

Asked for: density

Strategy:

1. Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.
2. Substitute these values into Equation \(\ref{10.4.12}\) to obtain the density.

Solution:

A The molar mass of butane (C₄H₁₀) is

\[M=(4)(12.011) + (10)(1.0079) = 58.123 \rm g/mol \nonumber\]

Using 0.08206 (L•atm)/(K•mol) for R means that we need to convert the temperature from degrees Celsius to kelvins (T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres:

\[P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm \nonumber\]

B Substituting these values into Equation \(\ref{10.4.12}\) gives

\[\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L \nonumber\]

Exercise \(\PageIndex{5}\): Density of Radon

Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics.

Answer

radon, 9.23 g/L; N₂, 1.17 g/L

Using the Ideal Gas Law to Calculate the Molar Mass of a Gas

Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, m, to its amount in moles, n:

\[ℳ=\mathrm{\dfrac{grams\: of\: substance}{moles\: of\: substance}}=\dfrac{m}{n}\]

The ideal gas equation can be rearranged to isolate n:

\[n=\dfrac{PV}{RT}\]

and then combined with the molar mass equation to yield:

\[ℳ=\dfrac{mRT}{PV}\]

This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.

Example \(\PageIndex{6}\): Determining the Molar Mass of a Volatile Liquid

The approximate molar mass of a volatile liquid can be determined by:

1. Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole
2. Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
3. Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (Figure \(\PageIndex{1}\))

Figure \(\PageIndex{1}\): When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At \(t_{l⟶g}\), the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)

Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm³ at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?

Solution

Since

\[ℳ=\dfrac{m}{n} \]

and

\[n=\dfrac{PV}{RT} \]

substituting and rearranging gives

\[ℳ=\dfrac{mRT }{PV}\]

then

\[ℳ=\dfrac{mRT}{PV}=\mathrm{\dfrac{(0.494\: g)×0.08206\: L⋅atm/mol\: K×372.8\: K}{0.976\: atm×0.129\: L}=120\:g/mol} \]

Summary

General gas equation: \(\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\)

Density of a gas: \(\rho=\dfrac{MP}{RT}\)

All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity (P, V, T, or n) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P, V, T, and n) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured.