The Ideal Gas Law

10.4 The Ideal Gas Law

Learning Objective

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

In the previous section, you learned how the volume of a gas changes when its pressure, temperature, or amount is changed, as long as the other two variables are held constant. In this section, we describe how these relationships can be combined to give a general expression that describes the behavior of a gas.

Deriving the Ideal Gas Law

Any set of relationships between a single quantity (such as V) and several other variables (P, T, and n) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions derived in Section 10.3 "Relationships among Pressure, Temperature, Volume, and Amount" are as follows:

Boyle’s law

V \propto \frac{1}{P} (at constant n, T)

Charles’s law

V \propto T (at constant n, P)

Avogadro’s law

V \propto n (at constant T, P)

Combining these three expressions gives

Equation 10.9

V \propto \frac{n T}{P}

which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as

Equation 10.10

V = (constant) (\frac{n T}{P})

By convention, the proportionality constant in Equation 10.10 is called the gas constantA proportionality constant that is used in the ideal gas law., which is represented by the letter R. Inserting R into Equation 10.10 gives

Equation 10.11

V = \frac{R n T}{P} = \frac{n R T}{P}

Clearing the fractions by multiplying both sides of Equation 10.11 by P gives

Equation 10.12

PV = nRT

This equation is known as the ideal gas lawA law relating pressure, temperature, volume, and the amount of an ideal gas..

An ideal gasA hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces. is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed.

Note the Pattern

Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures.

Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then

Equation 10.13

R = 0.082057 (L·atm)/(K·mol)

Because the product PV has the units of energy, as described in Section 5.1 "Energy and Work" and Essential Skills 4 (Section 5.6 "Essential Skills 4"), R can also have units of J/(K·mol) or cal/(K·mol):

Equation 10.14

R = 8.3145 J/(K·mol) = 1.9872 cal/(K·mol)

Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and 1 atm pressure, referred to as standard temperature and pressure (STP)The conditions 0°C (273.15 K) and 1 atm pressure for a gas.. We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation 10.11:

Equation 10.15

V = \frac{n R T}{P} = \frac{(1.000 mol) [0.082057 (L \cdot atm)/ (K \cdot mol)] (273.15 K)}{1.000 atm} = 22.41 L

Thus the volume of 1 mol of an ideal gas at 0°C and 1 atm pressure is 22.41 L, approximately equivalent to the volume of three basketballs. The quantity 22.41 L is called the standard molar volumeThe volume of 1 mol of an ideal gas at STP (0°C and 1 atm pressure), which is 22.41 L. of an ideal gas. The molar volumes of several real gases at STP are given in Table 10.3 "Molar Volumes of Selected Gases at Standard Temperature (0°C) and Pressure (1 atm)", which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at STP. The relationships described in Section 10.3 "Relationships among Pressure, Temperature, Volume, and Amount" as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed.

Table 10.3 Molar Volumes of Selected Gases at Standard Temperature (0°C) and Pressure (1 atm)

Gas	Molar Volume (L)
He	22.434
Ar	22.397
H₂	22.433
N₂	22.402
O₂	22.397
CO₂	22.260
NH₃	22.079

If n, R, and T are all constant in Equation 10.11, the equation reduces to

Equation 10.16

V = (constant) (\frac{1}{P}) or V \propto \frac{1}{P}

which is exactly the same as Boyle’s law in Equation 10.6.

Similarly, Charles’s law states that the volume of a fixed quantity of gas is directly proportional to its temperature at constant pressure. If n and P in Equation 10.11 are fixed, then

Equation 10.17

V = \frac{n R T}{P} = (constant) (T) or V \propto T

which is exactly the same as Equation 10.7.

Applying the Ideal Gas Law

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 5

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:

A Solve the ideal gas law for the unknown quantity, in this case n.

B 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 10.12) for n, we obtain

n = \frac{P V}{R T}

B P and T are given in units that are not compatible with the units of the gas constant [R = 0.082057 (L·atm)/(K·mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:

\begin{matrix} P = (745 mmHg) (\frac{1 atm}{760 mmHg}) = 0.980 atm \\ V = 31,150 L (given) \\ T = 30 + 273 = 303 Κ \end{matrix}

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

n = \frac{P V}{R T} = \frac{(0.980 atm) (31,150 L)}{[0.082057 (L \cdot atm) / (K \cdot mol)] (303 K)} = 1.23 \times 10^{3} {mol H}_{2}

Exercise

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 5, 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 6.

Example 6

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 5?

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

Asked for: volume in January

Strategy:

A Use the results from Example 5 for August as the initial conditions and then calculate the change in volume due to the change in temperature from 86°F to 14°F. Begin by constructing a table showing the initial and final conditions.

B Rearrange the ideal gas law to isolate those quantities that differ between the initial and final states on one side of the equation, in this case V and T.

C Equate the ratios of those terms that change for the two sets of conditions. Making sure to use the appropriate units, insert the quantities and 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:

	August (initial)	January (final)
T	30°C = 303 K	−10°C = 263 K
P	0.980 atm	0.980 atm
n	1.23 × 10³ mol H₂	1.23 × 10³ mol H₂
V	31,150 L	?

Thus we are asked to calculate the effect of a change in temperature on the volume of a fixed amount of gas at constant pressure.

B Recall that we can rearrange the ideal gas law to give

V = (\frac{n R}{P}) (T)

Both n and P are the same in both cases, which means that nR/P is a constant. Dividing both sides by T gives

\frac{V}{T} = \frac{n R}{P} = constant

This is the relationship first noted by Charles.

C We see from this expression that under conditions where the amount (n) of gas and the pressure (P) do not change, the ratio V/T also does not change. If we have two sets of conditions for the same amount of gas at the same pressure, we can therefore write

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

where the subscripts 1 and 2 refer to the initial and final conditions, respectively. Solving for V₂ and inserting the given quantities in the appropriate units, we obtain

V_{2} = \frac{V_{1} T_{2}}{T_{1}} = \frac{(31,350 L) (263 K)}{303 K} = 2.70 \times 10^{4} L

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

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 6 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 6 can be applied in any such case, as we demonstrate in Example 7 (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 7

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 5 (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 6.

Solution:

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

	Initial	Final
V	0.406 L	0.406 L
n	0.025 mol	0.025 mol
T	25°C = 298 K	750°C = 1023 K
P	1.5 atm	?

Once again, two parameters are constant while one is varied, and we are asked to calculate the fourth. As before, we begin with the ideal gas law and rearrange it as necessary to get all the constant quantities on one side. In this case, because V and n are constant, we rearrange to obtain

P = (\frac{n R}{V}) (T) = (constant) (T)

Dividing both sides by T, we obtain an equation analogous to the one in Example 6, P/T = nR/V = constant. Thus the ratio of P to T does not change if the amount and volume of a gas are held constant. We can thus write the relationship between any two sets of values of P and T for the same sample of gas at the same volume as

\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}

In this example, P₁ = 1.5 atm, T₁ = 298 K, and T₂ = 1023 K, and we are asked to find P₂. Solving for P₂ and substituting the appropriate values, we obtain

P_{2} = \frac{P_{1} T_{2}}{T_{1}} = \frac{(1.5 atm) (1023 K)}{298 K} = 5.1 atm

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

Exercise

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 Example 10.6 and Example 10.7, 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. If we rearrange the ideal gas law so that P, V, and T, the quantities that change, are on one side and the constant terms (R and n for a given sample of gas) are on the other, we obtain

Equation 10.18

\frac{P V}{T} = n R = constant

Thus the quantity PV/T is constant if the total amount of gas is constant. We can therefore write the relationship between any two sets of parameters for a sample of gas as follows:

Equation 10.19

\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}

This equation can be solved for any of the quantities P₂, V₂, or T₂ if the initial conditions are known, as shown in Example 8.

Example 8

We saw in Example 5 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 6.

Solution:

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

	Initial	Final
P	745 mmHg = 0.980 atm	312 mmHg = 0.411 atm
T	30°C = 303 K	−30°C = 243 K
n	1.23 × 10³ mol H₂	1.23 × 10³ mol H₂
V	31,150 L	?

Thus all the quantities except V₂ are known. Solving Equation 10.19 for V₂ and substituting the appropriate values give

V_{2} = V_{1} (\frac{P_{1} T_{2}}{P_{2} T_{1}}) = (31,150 L) [\frac{(0.980 atm) (243 K)}{(0.411 atm) (303 K)}] = 5.96 \times 10^{4} L

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:

V = \frac{n R T}{P} = \frac{(1.23 \times 10^{3} mol) [0.082057 (L \cdot atm) / (K \cdot mol)] (243 K)}{0.411 atm} = 5.97 \times 10^{4} L

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

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 and Molar Masses

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

Equation 10.20

\frac{n}{V} = \frac{P}{R T}

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):

Equation 10.21

n = \frac{m}{M}

Substituting this expression for n into Equation 10.20 gives

Equation 10.22

\frac{m}{M V} = \frac{P}{R T}

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

Equation 10.23

d = \frac{P M}{R T}

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 9

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

Given: compound, temperature, and pressure

Asked for: density

Strategy:

A Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.

B Substitute these values into Equation 10.23 to obtain the density.

Solution:

A The molar mass of butane (C₄H₁₀) is

(4)(12.011) + (10)(1.0079) = 58.123 g/mol

Using 0.082057 (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:

(750 mmHg) (\frac{1 atm}{760 mmHg}) = 0.987 atm

B Substituting these values into Equation 10.23 gives

d = \frac{P M}{R T} = \frac{(0.987 atm) (58.123 g/ mol)}{[0.082057 (L \cdot atm)/ (K \cdot mol)] (298 K)} = 2.35 g/L

Exercise

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

A common use of Equation 10.23 is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example 10.

Example 10

The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound.

Given: pressure, temperature, mass, and volume

Asked for: molar mass and chemical formula

Strategy:

A Solve Equation 10.23 for the molar mass of the gas and then calculate the density of the gas from the information given.

B Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass.

C Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas.

Solution:

A Solving Equation 10.23 for the molar mass gives

M = \frac{d R T}{P}

Density is the mass of the gas divided by its volume:

d = \frac{m}{V} = \frac{0.289 g}{0.157 L} = 1.84 g/L

B We must convert the other quantities to the appropriate units before inserting them into the equation:

\begin{matrix} T = 18 + 273 = 291 K \\ P = (727 mmHg) (\frac{1 atm}{760 mmHg}) = 0.957 atm \end{matrix}

The molar mass of the unknown gas is thus

M = \frac{d R T}{P} = \frac{(1.84 g/ L) [0.082057 (L \cdot atm) / (K \cdot mol)] (291 K)}{0.957 atm} = 45.9 g/mol

C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations:

\begin{matrix} NO = 14 + 16 = 30 g/mol \\ N_{2} O = (2) (14) + 16 = 44 g/mol \\ {NO}_{2} = 14 + (2) (16) = 46 g/mol \end{matrix}

The most likely choice is NO₂ which is in agreement with the data. The red-brown color of smog also results from the presence of NO₂ gas.

Exercise

You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it.

Answer: 44 g/mol; CO₂

Summary

The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant and has the value 0.08206 (L·atm)/(K·mol), 8.3145 J/(K·mol), or 1.9872 cal/(K·mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume. 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.

Key Takeaway

The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known.

Key Equations

Ideal gas law

Equation 10.12: PV = nRT

Relationship between initial and final conditions

Equation 10.19: $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} n is constant$

Density of a gas

Equation 10.23: $d = \frac{P M}{R T}$

Conceptual Problems

For an ideal gas, is volume directly proportional or inversely proportional to temperature? What is the volume of an ideal gas at absolute zero?
What is meant by STP? If a gas is at STP, what further information is required to completely describe the state of the gas?
For a given amount of a gas, the volume, temperature, and pressure under any one set of conditions are related to the volume, the temperature, and the pressure under any other set of conditions by the equation $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} .$ Derive this equation from the ideal gas law. At constant temperature, this equation reduces to one of the laws discussed in Section 10.3 "Relationships among Pressure, Temperature, Volume, and Amount"; which one? At constant pressure, this equation reduces to one of the laws discussed in Section 10.3 "Relationships among Pressure, Temperature, Volume, and Amount"; which one?
Predict the effect of each change on one variable if the other variables are held constant.
1. If the number of moles of gas increases, what is the effect on the temperature of the gas?
2. If the temperature of a gas decreases, what is the effect on the pressure of the gas?
3. If the volume of a gas increases, what is the effect on the temperature of the gas?
4. If the pressure of a gas increases, what is the effect on the number of moles of the gas?
What would the ideal gas law be if the following were true?
1. volume were proportional to pressure
2. temperature were proportional to amount
3. pressure were inversely proportional to temperature
4. volume were inversely proportional to temperature
5. both pressure and volume were inversely proportional to temperature

Given the following initial and final values, what additional information is needed to solve the problem using the ideal gas law?

Given	Solve for
V₁, T₁, T₂, n₁	n ₂
P₁, P₂, T₂, n₂	n ₁
T₁, T₂	V ₂
P₁, n₁	P ₂

Given the following information and using the ideal gas law, what equation would you use to solve the problem?

Given	Solve for
P₁, P₂, T₁	T ₂
V₁, n₁, n₂	V ₂
T₁, T₂, V₁, V₂, n₂	n ₁

Using the ideal gas law as a starting point, derive the relationship between the density of a gas and its molar mass. Which would you expect to be denser—nitrogen or oxygen? Why does radon gas accumulate in basements and mine shafts?
Use the ideal gas law to derive an equation that relates the remaining variables for a sample of an ideal gas if the following are held constant.
1. amount and volume
2. pressure and amount
3. temperature and volume
4. temperature and amount
5. pressure and temperature
Tennis balls that are made for Denver, Colorado, feel soft and do not bounce well at lower altitudes. Use the ideal gas law to explain this observation. Will a tennis ball designed to be used at sea level be harder or softer and bounce better or worse at higher altitudes?

Answer

1. P/T = constant
2. V/T = constant (Charles’ law)
3. P/n = constant
4. PV = constant (Boyle’s law)
5. V/n = constant (Avogadro’s law)

Numerical Problems

Calculate the number of moles in each sample at STP.
1. 1580 mL of NO₂
2. 847 cm³ of HCl
3. 4.792 L of H₂
4. a 15.0 cm × 6.7 cm × 7.5 cm container of ethane
Calculate the number of moles in each sample at STP.
1. 2200 cm³ of CO₂
2. 1200 cm³ of N₂
3. 3800 mL of SO₂
4. 13.75 L of NH₃
Calculate the mass of each sample at STP.
1. 36 mL of HI
2. 550 L of H₂S
3. 1380 cm³ of CH₄
Calculate the mass of each sample at STP.
1. 3.2 L of N₂O
2. 65 cm³ of Cl₂
3. 3600 mL of HBr
Calculate the volume in liters of each sample at STP.
1. 1.68 g of Kr
2. 2.97 kg of propane (C₃H₈)
3. 0.643 mg of (CH₃)₂O
Calculate the volume in liters of each sample at STP.
1. 3.2 g of Xe
2. 465 mg of CS₂
3. 5.34 kg of acetylene (C₂H₂)
Calculate the volume of each gas at STP.
1. 1.7 L at 28°C and 96.4 kPa
2. 38.0 mL at 17°C and 103.4 torr
3. 650 mL at −15°C and 723 mmHg
Calculate the volume of each gas at STP.
1. 2.30 L at 23°C and 740 mmHg
2. 320 mL at 13°C and 97.2 kPa
3. 100.5 mL at 35°C and 1.4 atm
A 8.60 L tank of nitrogen gas at a pressure of 455 mmHg is connected to an empty tank with a volume of 5.35 L. What is the final pressure in the system after the valve connecting the two tanks is opened? Assume that the temperature is constant.
At constant temperature, what pressure in atmospheres is needed to compress 14.2 L of gas initially at 25.2 atm to a volume of 12.4 L? What pressure is needed to compress 27.8 L of gas to 20.6 L under similar conditions?
One method for preparing hydrogen gas is to pass HCl gas over hot aluminum; the other product of the reaction is AlCl₃. If you wanted to use this reaction to fill a balloon with a volume of 28,500 L at sea level and a temperature of 78°F, what mass of aluminum would you need? What volume of HCl at STP would you need?
An 3.50 g sample of acetylene is burned in excess oxygen according to the following reaction:
2 C₂H₂(g) + 5 O₂(g) → 4 CO₂(g) + 2 H₂O(l)
At STP, what volume of CO₂(g) is produced?
Calculate the density of ethylene (C₂H₄) under each set of conditions.
1. 7.8 g at 0.89 atm and 26°C
2. 6.3 mol at 102.6 kPa and 38°C
3. 9.8 g at 3.1 atm and −45°C
Determine the density of O₂ under each set of conditions.
1. 42 g at 1.1 atm and 25°C
2. 0.87 mol at 820 mmHg and 45°C
3. 16.7 g at 2.4 atm and 67°C
At 140°C, the pressure of a diatomic gas in a 3.0 L flask is 635 kPa. The mass of the gas is 88.7 g. What is the most likely identity of the gas?
What volume must a balloon have to hold 6.20 kg of H₂ for an ascent from sea level to an elevation of 20,320 ft, where the temperature is −37°C and the pressure is 369 mmHg?
What must be the volume of a balloon that can hold 313.0 g of helium gas and ascend from sea level to an elevation of 1.5 km, where the temperature is 10.0°C and the pressure is 635.4 mmHg?
A typical automobile tire is inflated to a pressure of 28.0 lb/in.² Assume that the tire is inflated when the air temperature is 20°C; the car is then driven at high speeds, which increases the temperature of the tire to 43°C. What is the pressure in the tire? If the volume of the tire had increased by 8% at the higher temperature, what would the pressure be?
The average respiratory rate for adult humans is 20 breaths per minute. If each breath has a volume of 310 mL of air at 20°C and 0.997 atm, how many moles of air does a person inhale each day? If the density of air is 1.19 kg/m³, what is the average molecular mass of air?
Kerosene has a self-ignition temperature of 255°C. It is a common accelerant used by arsonists, but its presence is easily detected in fire debris by a variety of methods. If a 1.0 L glass bottle containing a mixture of air and kerosene vapor at an initial pressure of 1 atm and an initial temperature of 23°C is pressurized, at what pressure would the kerosene vapor ignite?

Answers

1. 7.05 × 10⁻² mol
2. 3.78 × 10⁻² mol
3. 0.2138 mol
4. 3.4 × 10⁻² mol
1. 0.21 g HI;
2. 840 g H₂S;
3. 0.988 g CH₄
1. 0.449 L Kr
2. 1510 L C₃H₈
3. 3.13 × 10⁻⁴ L (CH₃)₂O
1. 1.5 L
2. 4.87 mL
3. 650 mL
281 mmHg
20.9 kg Al, 5.20 × 10⁴ L HCl
1. 1.0 g/L
2. 1.1 g/L
3. 4.6 g/L
2174 L