CoreChem:9.2 Gas Laws

From ChemEd Collaborative

9.2 GAS LAWS

Avogadro’s Law

For most solids and liquids it is convenient to obtain the amount of sub- stance (and the number of particles, if we want it) from the mass. In Sec. 2.8 numerous such calculations using molar mass were done. In the case of gases, however, accurate measurement of mass is not so simple. Think about how you would weigh a balloon filled with helium, for example. Because it is buoyed up by the air it displaces, such a balloon would force a balance pan up instead of down, and a negative weight would be obtained. Solids and liquids are also buoyed up, but they have much greater densities than gases. For a given mass of a solid or liquid, the volume is much smaller, much less air is displaced, and the buoyancy effect is negligible. The mass of a gas can be obtained by weighing a truly empty container (one in which there is a perfect vacuum), and then filling and reweighing the container. But this is a time-consuming, inconvenient, and sometimes dangerous procedure. (Such a container might implode—explode inward—due to the difference between atmospheric pressure outside and zero pressure within.)

A more convenient way of obtaining the amount of substance in a gaseous sample is suggested by the data on molar volumes in Table 9.1. Remember that a molar quantity (a quantity divided by the amount of substance) refers to the same number of particles.

TABLE 9.1 Molar Volumes of Several Gases at 0°C and 1 atm Pressure.

Substance	Formula	Molar Volume/liter mol^–1
Hydrogen	H₂(g)	22.43
Neon	Ne(g)	22.44
Oxygen	O₂(g)	22.39
Nitrogen	N₂(g)	22.40
Carbon dioxide	CO₂(g)	22.26
Ammonia	NH₂(g)	22.09

The data in Table 9.1, then, indicate that for a variety of gases, 6.022 × 10²³ molecules occupy almost exactly the same volume. If the temperature is 0°C and the pressure is 1.00 atm (101.3 kPa), this volume is 22.4 liters (22.4 dm³). (Since the liter, now defined as exactly 1 dm³, is commonly used as a unit of volume in conjunction with the atmosphere as a unit of pressure, we shall use it that way in this chapter.) That equal volumes of gases at the same temperature and pressure contain equal numbers of molecules was first suggested in 1811 by the Italian chemist Amadeo Avogadro (1776 to 1856). Consequently it is called Avogadro’s law or Avogadro’s hypothesis.

Avogadro’s law has two important messages. First, it says that molar volumes of all gases are the same at a given temperature and pressure. Therefore, even if we do not know what gas we are dealing with, we can still find the amount of substance. Second, we expect that if a particular volume corresponds to a certain number of molecules, twice that volume would contain twice as many molecules. In other words, doubling the volume corresponds to doubling the amount of substance, halving the volume corresponds to halving the amount, and so on.

In general, if we multiply the volume by some factor, say x, then we also multiply the amount of substance by that same factor x. Such a relationship is called direct proportionality and may be expressed mathematically as

V $\propto$ n (9.3)

where the symbol $\propto$ means “is proportional to.” Any proportion, such as Eq. (9.3) can be changed to an equivalent equation if one side is multiplied by a proportionality constant, such a k_A in Eq.(9.4):

V = k_An (9.4)

If we know k_A for a gas, we can determine the amount of substance from Eq. (9.4).

The situation is complicated by the fact that the volume of a gas depends on pressure and temperature, as well as on the amount of substance. That is, k_A will vary as temperature and pressure change. Therefore we need quantitative information about the effects of pressure and temperature on the volume of a gas before we can explore the relationship between amount of substance and volume.

Boyle’s Law

You are probably already familiar with the fact that when you squeeze a gas, it will take up less space. In formal terms, increasing the pressure on a gas will decrease its volume. Studies in 1662 by the English scientist Rob.ert Boyle (1627 to 1691) gave results such as those in Table 9.2. Careful, study of such data reveals that if we double the pressure, we halve the volume;

TABLE 9.2 Variation in the Volume of 0.0446 mol H₂(g) with Pressure at 0°C.

Trial	Pressure/kPa	Pressure/atm	Volume/liter
1	152.0	1.50	0.666
2	126.7	1.25	0.800
3	101.3	1.00	1.00
4	76.0	0.750	1.333
5	50.7	0.500	2.00
6	25.3	0.250	4.00
7	10.1	0.100	10.00

if we triple the pressure, the volume is reduced to one-third; and so on. In general, if we multiply the pressure by some factor x, then we divide the volume by the same factor x. Such a relationship, in which the increase in one quantity produces a proportional decrease in another, is called inverse proportionality.

The results of Boyle’s experiments with gases are summarized in Boyle’s law—for a given amount of gas at constant temperature, the volume is proportional to the pressure. In mathematical terms

$V\propto \frac{\text{1}}{P}\text{ (9}\text{.5)}$

The reciprocal of P indicates the inverse nature of the proportionality. Using the proportionality constant k_A to convert relationship (9.5) to an equation, we have

$V=k_{\text{B}}\text{ }\times \text{ }\frac{\text{1}}{P}\text{=}\frac{k_{\text{B}}}{P}\text{ (9}\text{.6}a\text{)}$

Multiplying both sides of Eq.(9.6a) by P, we have

PV = k_B (9.6b)

where k_A represents a constant value for any given temperature and amount (or mass) of gas.

EXAMPLE 9.4 Using the data printed in color in Table 9.2, confirm that Boyle’s law is obeyed.

Solution Since the data apply to the same amount of gas at the same temperature, PV should be constant [Eq. (9.6b)] if Boyle’s law holds.

P₁V₁ = 1.50 × 0.666 liter = 0.999 atm liter

P₄V₄ = 0.750 atm × 1.333 liter = 1.000 atm liter

P₆V₆ = 0.250 atm × 4.00 liter = 1.00 atm liter

The first product differs from the last two in the fourth significant digit. Since some data are reported only to three significant figures, PV is constant within the limits of the measurements.

If the units atmosphere liter, in which PV was expressed in Example 9.4, are changed to SI base units, an interesting result arises:

$\begin{align} & \text{1 atm }\times \text{ 1 liter}=\text{101}\text{.3 kPa }\times \text{ 1 dm}^{\text{3}}=\text{101}\text{.3 }\times \text{ 10}^{\text{3}}\text{ Pa }\times \text{ 1 dm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 m}}{\text{10 dm}} \right)^{\text{3}} \\ & \text{ }=\text{101}\text{.3 }\times \text{ 10}^{\text{3}}\text{ }\frac{\text{N}}{\text{m}^{\text{2}}}\text{ }\times \text{ 1 dm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 m}^{\text{3}}}{\text{10}^{\text{3}}\text{ dm}^{\text{3}}}=\text{101}\text{.3 N m}=\text{101}\text{.3 kg m s}^{-\text{2}}\text{ m}=\text{101}\text{.3 J} \\ \end{align}$

In other words, PV has the same units (joules) as an energy. While this does not guarantee that PV is an energy (recall the warning in Chap. 1 against relying on cancellation of units unless you know that a relationship between quantities exists), it does suggest that we should explore the possibility. This will be done in Sec. 9.4. The above argument also shows that the product of the units kilopascals times cubic decimeters is the unit joules. In subsequent discussions you will find it handy to remember that

$\text{1 kPa }\times \text{ 1 dm}^{\text{3}}=\text{1 J and 1 kPa}=\frac{\text{1 J}}{\text{1 dm}^{\text{3}}}\text{ and 1 dm}^{\text{3}}=\frac{\text{1 J}}{\text{1 kPa}}$

Boyle’s law enables us to calculate the pressure or volume of a gas under one set of conditions, provided we know the pressure and volume under a previous set of circumstances.

EXAMPLE 9.5 The volume of a gas is 0.657 liter under a pressure of 729.8 mmHg. What volume would the gas occupy at atmospheric pressure (760 mmHg)? Assume constant temperature and amount of gas.

Solution Two methods of solution will be given.

a) Since PV must be constant,

P₁V₁ = k_B = P₂V₂

Initial conditions: P₁ = 729.8 mmHgV₁ = 0657 liter

Final conditions: P₂ = 760 mmHgV₂ = ?

Solving for V₂, we have

$V_{\text{2}}=\frac{P_{\text{1}}V_{\text{1}}}{P_{\text{2}}}=\frac{\text{729}\text{.8 mmHg }\times \text{ 0}\text{.657 liter}}{\text{760 mmHg}}=\text{0}\text{.631 liter}$

b) Note that in method a the original volume was multiplied by a ratio of pressures (P₁/P₂):

V₂ = 0.657 liter × ratio of pressures

Rather than solving algebraically, we can use common sense to decide which of the two possible ratios

$\frac{\text{729}\text{.8 mmHg}}{\text{760 mmHg}}\text{ or }\frac{\text{760 mmHg}}{\text{729}\text{.8 mmHg}}$

should be used. The units cancel in either case, and so units are no help. However, if you reread the problem, you will see that we are asked to find the new volume (V₂) produced by an increase in pressure. Therefore there must be a decrease in volume, and we multiply the original volume by a ratio which is less than 1:

$V_{\text{2}}=\text{0}\text{.657 liter }\times \frac{\text{729}\text{.8 mmHg}}{\text{760 mmHg}}=\text{0}\text{.631 liter}$

It is reassuring that both common sense and algebra produce the same answer.

Charles’ Law

Having explored the effect of pressure on the volume of a gas, we now turn to the effect of temperature. Again you are probably familiar with the fact that increasing the temperature of a gas will cause the gas to expand. This effect was first studied quantitatively in 1787 by Jacques Charles (1746 to 1823) of France. Typical data from such an experiment are given in Table 9.3. You can see that for 0.0466 mol H₂(g) at constant pressure, a 50°C rise in temperature produces a 0.18-liter increase in volume, whether the temperature increases from 0.0 to 50.0° or from 100.0 to 150.0°C.

TABLE 9.3 Variation in the Volume of H₂(g) with Temperature.

When the experimental data of Table 9.3 are graphed, we obtain Fig. 9.4. Notice that the four points corresponding to 0.0446 mol H₂(g) lie on a straight line, as do the points for 0.100 mol H₂(g). If the lines are extrapolated (extended beyond the experimental points) to very low temperatures, we find that both of them intersect the horizontal axis at –273°C. The behavior of H₂(g) (and of many other gases) at normal temperatures suggests that if we cool a gas sufficiently, its volume will become zero at –273°C.

Of course a real substance would condense to a liquid and freeze to a solid as it was cooled. When the pressure is 1.00 atm (101.3 kPa), H₂(g) liquefies at –253°C and freezes at –259°C, and so all experiments involving would have to be performed above –253°C. If we could find a gas that did not condense, however, it would still be impossible to cool it below –273°C, because at that temperature its volume would be zero. Going to a lower temperature would correspond to a negative volume—something that is very hard to conceive of. Hence –273°C is referred to as the absolute zero of temperature—it is impossible to go any lower.

In Fig. 9.4b the zero of the temperature axis has been shifted to absolute zero. The temperature scale used in this graph is called absolute or thermodynamic temperature. It is measured in SI units called kelvins (abbreviated K), in honor of the English physicist William Thomson, Lord Kelvin (1824 to 1907).

Figure 9.4 Charles' law. (a) Volume plotted against Celsius temperature for two samples of H(g) at 1.00 atm (101.3 kPa); (b) volume plotted against thermodynamic temperature for the same two samples. Note how much simpler graph (b) is than graph (a).

The temperature interval 1 K corresponds to a change of 1°C, but zero on the thermodynamic scaleis (0 K) is –273.15°C. The freezing point of water at 1.00 atm (101.3 kPa) pressure is thus 273.15 K. The relationship between the Celsius temperature scale and the thermodynamic temperature scale is discussed further in Appendix 1.

By shifting to the absolute temperature scale, we have simplified the graph of gas volume versus temperature. Figure 9.4b shows that the volume of a gas is directly proportional to its thermodynamic temperature, provided that the amount of gas and the pressure remain constant. This is known as Charles’law, and can be expressed mathematically as where T represents the absolute temperature (usually measured in kelvins).

V $\propto$ T (9.7)

As in the case of previous gas laws, we can introduce a proportionality constant, in this case, k_C:

$V=k_{\text{C}}T\text{ or }\frac{V}{T}=k_{\text{C}}\text{ (9}\text{.8)}$

EXAMPLE 9.6 A sample of H₂(g) occupies a volume of 69.37 cm³ at a pressure of exactly 1 atm when immersed in a mixture of ice and water. When the gas (at the same pressure) is immersed in boiling benzene, its volume expands to 89.71 cm³. What is the boiling point of benzene?

Solution As in the case of Boyle’s law (Example 9.5), two methods of solution are possible.

a) Algebraically, we have, from Eq. (9.8),

$\frac{V_{\text{1}}}{T_{\text{1}}}=k_{\text{C}}=\frac{V_{\text{2}}}{T_{\text{2}}}$

and substituting into the equation

$T_{\text{2}}=\frac{V_{\text{2}}T_{\text{1}}}{V_{\text{1}}}=\frac{\text{89}\text{.71 cm}^{\text{3}}\text{ }\times \text{ 273}\text{.15 K}}{\text{69}\text{.37 cm}^{\text{3}}}=\text{353}\text{.2 K}$

yields the desired result. (The ice-water mixture must be at 273.15 K, the freezing point of water.)

b) By common sense we argue that since the gas expanded, its temperature must have increased. Thus

$T_{\text{2}}=\text{273}\text{.15 K }\times \text{ ration greater than 1}=\text{273}\text{.15 K }\times \text{ }\frac{\text{89}\text{.71 cm}^{\text{3}}}{\text{69}\text{.37 cm}^{\text{3}}}=\text{353}\text{.2 K}$

Note: In this example we have used the expansion of a gas instead of the expansion of liquid mercury to measure temperature.

Because both temperature and pressure affect the volume of a gas, it is convenient to specify a reference temperature and pressure at which all volumes may be compared. Standard temperature and pressure (or STP) is chosen to be 273.15 K (0°C) and 1 atm (101.325 kPa)

Gay-Lussac’s Law

A third gas law may be derived as a corollary to Boyle’s and Charles’ laws. Suppose we double the thermodynamic temperature of a sample of gas. According to Charles’ law, the volume should double. Now, how much pressure would be required at the higher temperature to return the gas to its original volume? According to Boyle’s law, we would have to double the pressure to halve the volume. Thus, if the volume of gas is to remain the same, doubling the temperature will require doubling the pressure. This law was first stated by the Frenchman Joseph Gay-Lussac (1778 to 1850). According to Gay-Lussac’s law, for a given amount of gas held at constant volume, the pressure is proportional to the absolute temperature. Mathematically,

$P\propto T\text{ or }P=k_{\text{G}}T\text{ or }\frac{P}{T}=k_{\text{G}}$

where k_G is the appropriate proportionality constant.

Gay-Lussac’s law tells us that it may be dangerous to heat a gas in a closed container. The increased pressure might cause the container to explode.

EXAMPLE 9.7 A container is designed to hold a pressure of 2.5 atm. The volume of the container is 20.0 cm³, and it is filled with air at room temperature (20°C) and normal atmospheric pressure. Would it be safe to throw the container into a fire where temperatures of 600°C would be reached?

Solution Using the common-sense method, we realize that the pressure will increase at the higher temperature, and so

$P_{\text{2}}=\text{1}\text{.0 atm }\times \frac{\text{(273}\text{.15 + 600) K}}{\text{(273}\text{.15 + 20) K}}=\text{3}\text{.0 atm}$

This would exceed the safe strength of the container. Note that the volume of the container was not needed to solve the problem.

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CoreChem:9.2 Gas Laws

From ChemEd Collaborative

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