Gases

Though much of chemistry performed in classes involves aqueous (water-containing) liquid solutions, the chemistry and physics of gases is of tremendous importance. For example, one of the most important industrial chemical reactions, the formation of ammonia from nitrogen and hydrogen takes place in gas phase. The air that we breathe is obviously a gas, and its chemical physics is of extreme importance to SCUBA divers who can suffer nitrogen narcosis (the "bends") as a result of too rapid an ascent. Finally, many a driver has opened up the radiator of his or her car in order to add coolant or antifreeze without letting the water vapor inside cool; the resulting burn is a painful reminder of the importance of the gas laws. Without further ado, then,

Gases are an important part of chemistry, and the first step to understanding them is to understand their simplified physics, as exemplified by a group of relationships collectively known as the gas laws.

Though much of this lesson will seem to have little to do with chemistry (and more to do with thermal physics), such knowledge is a critical prerequisite for more "chemically oriented" subjects such as chemical equilibrium and chemical thermodynamics.

From here on, it is assumed that you have some basic algebra skills. If you need a primer on math, we highly recommend http://www.sosmath.com/. Also, if you have forgotten any dimensional analysis, we suggest that you read the lesson on that topic. Some elementary physics wouldn't hurt either.

What, precisely, is a gas? Some people might say, "air is a gas," or, "helium is a gas." Unfortunately, definition by example is not adequate for a rigorous discussion of a topic. Thus we present here a broad definition of a gas:

Most substances can, under the proper conditions, become a gas. (These conditions are typically high temperature and/or low pressure, as will be discussed later.)

Much of our discussion of gases hinges on a precise understanding of several physical concepts. The first three of these that we shall deal with are temperature, work, and energy.

Matter is constantly in motion at the sub-microscopic (atomic or molecular) level. Molecules and atoms shift around, bump into one another, and may, on occasion, stick together or split apart due to a collision (when this happens, a chemical reaction occurs). No matter what a substance is, or what phase (solid, liquid, gas, plasma, gel, liquid crystal, or Bose-Einstein condensate) it is in, there is no escaping from the eternal motion of atoms and/or molecules.

Kinetic energy is a quantity associated with moving objects. The faster an object moves, the greater its kinetic energy. If an object possesses kinetic energy, it is possible to harness that energy to do work.

Most people realize that there is something intrinsically dangerous about a bullet fired from a rifle. The fear associated with the bullet arises from its high kinetic energy: if it strikes someone, that person's body must do a great deal of work to counter the kinetic energy of the bullet. However, this knowledge, though perfectly valid, is not entirely sufficient for us as chemists or physicists. We must find a way to quantify work and energy. To do this, we first define the term velocity:

Velocity is a vector quantity (that is, consists of both magnitude and direction) that describes both the speed of an object and, if moving, its relative direction.

For our purposes, "direction" will consist of two possibilities: moving (in one dimension) toward a defined point or away from a defined point. In more advanced physics, velocity can take any direction in any number of dimensions and is defined through use of the mathematical language of calculus. Velocity can be constant (i.e. driving straight at 10 miles per hour) or can change (accelerating from standing still to 50 miles per hour). Working with the latter situation is not too difficult once some more advanced mathematics (calculus) is applied, but working with the former is exceedingly simple. For any situation, we simply apply the idea of average velocity:

Average velocity can be found by determining the total distance traveled in one direction by an object and dividing it by the time it took the object to travel that distance.

Mathematically, we write it as follows, where

is average velocity,

is change in distance (

is always "change in"), and

is change in time:

Practice Exercise 1:

Methane (chemical formula CH₄), or natural gas, is commonly used to fuel laboratory equipment such as Bunsen burners and is an important reactant in many organic reactions. It also heats homes, runs stoves, and produces electricity in many areas. It is a gas at room temperature and standard atmospheric pressure. Let us suspend disbelief and suppose that a physical chemist has developed a "methane gun" that can shoot individual molecules of methane down a long cylindrical tube. He fires a molecule down the tube at t = 6.0 s and records that it has traveled 200 m at t = 8.2 s. What is the average velocity of the methane molecule during its travel?

Solution:

Here, the change in distance is given to us: = 200 m. The change in time is found by = 8.2 s - 6.0 s = 2.2 s. Then the velocity is given by Equation 1 as follows:

We can now define kinetic energy in more precise mathematical language, where KE is kinetic energy, m is the mass of the moving object, and v is its velocity (not average):

The derivation of Equation 2 requires some math and physics beyond the scope of this lesson.

Note that Equations 1 and 2 only strictly apply to situations where the velocities involved are small relative to the speed of light. At very high velocities, special mathematical modifications (known as Lorentz transformations) have to be applied to these equations in order for them to hold (this is the basis of Einstein's special theory of relativity). However, all velocities that will be considered in this lesson, though high, are not high enough to cause concern. Also note that average velocity can be substituted for velocity to find average kinetic energy; this will be important later on.

In Equation 2, if the unit of mass is the kilogram (kg) and the unit of velocity meters per second (m/s), then the unit of kinetic energy is the joule (J).

Now, out of necessity we must constrict our definition of kinetic energy, for there are many kinds of motion and many kinds of moving objects. For example, a wheel spinning around an axle possesses kinetic energy by virtue of its movement (this form is known as rotational kinetic energy), but can be ignored in the theory that we are slowly but surely building. Thus we restrict ourselves to translational kinetic energy:

Translational kinetic energy is the form of kinetic energy resulting from linear ("in a line") motion.

Properly speaking, Equation 1 operates under the assumption that translational kinetic energy is the form of kinetic energy involved; this just formalizes the assumption. Linear motion includes most forms of motion, such as back-and-forth vibration and movement straight from one point to another. It excludes forms of motion such as rotation and waves, which become very tricky to work with.

We are finally in a position to deal with temperature. Temperature is a macroscopic (can be observed without elaborate instrumentation) that is directly related to motion of atoms and molecules:

Temperature is a macroscopic property corresponding to the average translational kinetic energy of the particles (atoms and/or molecules) making up a substance whose temperature is being measured.

Here, we are taking an average of the kinetic energies at any given instant of all the particles in a system being observed. If this average rises, the temperature rises. The inverse is also true. Thus if we had a device that was created using concepts outside the bounds of this discussion (watch for the lesson on thermochemistry and chemical thermodynamics to go into detail on the theory of thermometry) that could measure temperature, we could indirectly measure the translational kinetic energy of a substance such as a gas. Luckily for us, such a device exists: it is a thermometer. Note that it is not actually possible to measure the average kinetic energy of a substance with a thermometer, because another property (entropy, or the disorder of the atoms or molecules in the substance) is a major player in temperature. However, if we measure the temperature of a substance at one instant, and measure its temperature sometime later, the temperatures let us infer the change in translational kinetic energy of the composing particles.

Finally, we can begin discussing gases. Modern chemists view gases according to something called the Kinetic-Molecular Theory. This theory integrates concepts to which we have already alluded and applies them to gases. There are two major assumptions in this theory. First,

Gases are composed of discrete particles (atoms or molecules), and anything that happens at the microscopic level of these particles is somehow reflected at the macroscopic level.

If we are to accept that matter is composed of atoms and/or molecules, and that their interactions govern the macroscopic interactions of matter (both fundamental axioms in chemistry), then this first assumption is easily justified and understood. Second,

The particles composing gases are constantly in motion, and it is this motion that gives gases their characteristic physical properties.

As earlier stated, we are working under the assumption (experimentally demonstrated to be true) that all matter is composed of moving particles. Thus the first part of this assumption is implied. The second assumption will be built up in the remainder of this lesson.

If you fill a balloon with a gas (such as air from your lungs or helium from a tank), it will expand to a certain volume (perhaps a few liters). If you tie it off at room temperature, and then put it in the freezer for a few hours, the balloon will shrink. If you then take the balloon out of the freezer and allow it to warm to room temperature, it will return to approximately its original volume. Feel free to try this at home! This experiment demonstrates that there is some relationship between gas temperature and the volume of a gas. Specifically, it shows that at higher temperatures, a given gas occupies more volume than it does at lower temperatures.

Let us make the above experiment a bit more scientific. Let us use a thermometer (calibrated in Celsius; scientists rarely use Fahrenheit) to measure room temperature (say 20.00 ºC) and measure the volume of the balloon filled with helium (3.000 L). Let us then turn up the room's thermostat to a toasty 30.00 ºC, and measure the new volume of the balloon (found to be 3.102 L). Let us then put it in a freezer with a known temperature (8.00 ºC) and measure the volume of the balloon (now 2.877 L). Let us finally turn the freezer's thermostat down to a chilly 0.00 ºC and measure the balloon's volume (2.795 L). (These data are from an actual experiment, tweaked a bit to suit our purposes.) We now create a graph of balloon volume versus temperature; this is displayed in Figure 1.

We immediately see that the relationship between temperature and volume of a gas appears to be linear. We can thus easily figure out both the slope and T-intercept of the graph; that is, we can figure out the constant that relates temperature and volume for this sample of gas, and can figure out the temperature at which the gas should shrink to zero volume. Figure 2 shows the same graph as Figure 1, but with the line's equation shown (T=temperature, V=volume) and the line extrapolated to the T-intercept.

Logically, no gas can have a negative volume, so we can create a new temperature system based around the point at which all gases would have zero volume: in Celsius, this point is -273.15 ºC. We assign this temperature a value of zero, and let the increment of temperature be equal in magnitude to the degree Celsius. This temperature scale is known as the Kelvin scale (originally created in 1848 by Lord Kelvin, William Thomson), and 0 K is known as absolute zero. No substance can actually achieve absolute zero, though scientists have created substances with temperatures within a few millionths of a Kelvin of absolute zero. By definition, at absolute zero, molecular motion would halt. (Well, in reality there would still be some motion, due to a phenomenon known as zero-point energy, but that is outside the scope of this lesson.) This is worth emphasizing.

At absolute zero (-273.15 ºC or 0 K), gases should have zero volume and all molecular or atomic motion should halt.

The Kelvin temperature scale is extremely important in chemical physics, and temperatures can be readily converted from Celsius to Kelvin and back. Since 0 K corresponds to -273.15 ºC, to convert from Celsius to Kelvin, simply add 273.15. To convert in the opposite direction, subtract 273.15.

A concise summary of the relationship between temperature and volume of gases (assuming all other factors, such as amount of gas and pressure are held constant) is given by Charles' Law, named for the discoverer (in 1787) of the temperature-volume relationship, French physicist Jacque Charles. A modern statement of his law follows:

Charles' Law states that the volume of a gas (V) is directly proportional to its temperature in Kelvin (T), provided that all other conditions are held constant.

Figure 3 displays the balloon data with temperature in Kelvin, and is thus a graphical depiction of Charles' Law.

A kinetic-molecular explanation of Charles' Law must wait until we consider the topic of pressure.

Practice Exercise 2:

Methane, discussed in Practice Exercise 1, obeys Charles' Law. A balloon is filled with methane at a natural gas outlet in a lab until it has a radius of 10 cm at room temperature (20.00 ºC). It is then placed in an incubator and heated to 50.00 ºC. What will be the volume and balloon radius at 50.00 ºC? Assume that the methane does not combust, the balloon does not burst, and that the balloon is spherical at both temperatures. Hint: The volume V of a sphere of radius r is given by V = (4/3) (pi) r³, where (pi) equals about 3.1415.

Solution:

First we must find the initial volume. The volume at 20.00 ºC is given by the volume equation V_initial = (4/3) (pi) r_initial³ = (4/3) (3.1415) (10 cm)³ = 4189 cm³. Applying Equation 3, we find that 4189 cm³ = k (293.15 K), which implies that k = 14.29 cm³/K, which will be the same after heating as well. Thus after heating, k = 14.29 cm³/K = V_final cm³/ 323.15 K, so V_final = 4618 cm³. This is half of our answer (the final balloon volume). To find the radius, we apply the volume equation 4618 cm³ = (4/3) (pi) r_final³, so r_final³= 1102 cm³, which implies that r_final = 10.33 cm.

Recipes (say, for cooking rice or baking a cake) often have two sets of instructions: one for most people, and one for people at high altitude (i.e. those living in a mountain range). Though ingredients are not typically modified, cooking procedures usually are. Such modifications may include cooking food at higher temperature or boiling food longer. Why does elevation have an impact on cooking? The answer lies in another property of gases (and consequently the atmosphere): pressure. We begin with a definition:

Pressure is a measure of the force applied (i.e. by a gas or a an object due to its weight) per unit area of a material against which the force is applied.

Conceptually, pressure can be difficult to comprehend, but is simplified by an example. Consider a nail. We can choose to hammer it into the wall in two ways: with the pointy end going in or the flat head going in. We always do it the former way because it is easier. Why? The area of the point is much smaller than the area of the head, but the force applied (i.e. the strength with which we strike the hammer) is the same regardless. Thus the same amount of force is spread over two different areas, so more pressure is applied to the wall when the small point is driven in. In effect, we make more efficient use of our swing. It is for this reason that the points of needles used to give injections are exceedingly small; it would be much more difficult (and painful) to use needles the size of nails to give injections.

By definition, anything (such as gas molecules) exerting a force on a surface (such as the interior of a balloon) has an associated pressure. Gases have pressures that can be measured with a variety of devices, including pressure gauges, manometers, or barometers. A simple manometer is shown in Figure 4; it employs the fact that the atmosphere exerts a constant pressure at a given altitude. Click "Measure Pressure" to watch the manometer in action.

There is an astounding array of units for measuring pressure, and all ultimately reduce to force per unit area. The accepted SI unit is the Pascal (Pa), equal to one Newton per square meter (N/m²). The problem is that this unit is actually very small compared to most pressures normally encountered (such as atmospheric pressure); atmospheric pressure at sea level equals about 101325 Pa. To simplify this, the more commonly used unit, albeit not SI, is simply called the atmosphere (atm), and is equal to sea level atmospheric pressure. Thus 1 atm = 101325 Pa = 101325 N/m². Further complicating matters is that the unit of pressure used in the days of old mercury manometers (millimeters mercury, mmHg, equal to the displacement of mercury as shown in Figure 4) is still in use, especially in meteorology and blood pressure. 1 atm = 760 mmHg. The mmHg is also sometimes called the Torricelli, or torr. So 1 mmHg = 1 torr. In summary,

1 atm = 760 torr = 760 mmHg = 101325 Pa = 101325 N/m² = atmospheric pressure at sea level. (Equation 4)

Gases exert a force on the walls of their container. It is this pressure that causes flexible containers (such as balloons) to possess any volume at all. Since an entirely flexible such as a balloon expands to form a sphere (or close to one), we can infer that force over any given area A of the balloon's internal surface is equal to that exerted over any other area A in the balloon. In simple terms, pressure at any infinitesimal point in the balloon (or any other container, for that matter, flexible or not) is equal to that at any other point.

At sea level, the atmosphere exerts 1 atm pressure by virtue of its weight; since the atmosphere has a roughly homogenous composition worldwide, air pressure at sea level does not change much with location. However, air pressure changes rather rapidly with change in altitude. As we climb up into the mountains, pressure markedly drops, sometimes resulting in ears "popping" (equilibrating with the new pressure). This effect is amplified further when we fly in an airplane; at such altitudes the atmospheric pressure is so low that the airplane cabin must be pressurized and oxygenated to prevent a painful and potentially lethal journey.

As may be guessed, it is this change in pressure with change in altitude that necessitates the odd cooking instructions printed on many foods. Unfortunately, we are not in a position yet to talk about why such instructions are necessary (see Vapor Pressure and Boiling Points). First, let us consider the effect of

Pressure on a gas (in a container such as a balloon) can be externally altered with a device such as a vacuum pump. In Figure 5, click "Turn On" to watch the pressure drop in a closed vessel holding a balloon, and watch the accompanying balloon volume change. A similar effect would be observed with a real vacuum pump and desiccator (Figure 6 shows an actual pump and desiccator, but sadly no balloon).

It turns out that, when all other factors such as temperature are held constant, changing the pressure of a gas causes an inverse change in its volume and vice-versa. Another way to state this is to say that the pressure-volume product for a given sample of gas is constant. Early English experimental physicist Robert Boyle is credited with this discovery (in 1662), and we honor him by calling this principle Boyle's Law:

Boyle's Law states that for a given sample of gas at constant temperature, volume (V) is inversely proportional to pressure (P) and vice-versa.

Boyle's Law leads to a characteristic graph (sometimes called a PV graph) that is shown in Figure 7. Note that the axes labels are not shown because they vary with other factors, such as temperature.

Practice Exercise 3:

The balloon from Practice Exercise 2 is allowed to equilibrate (reach an internal pressure of 1 atm to exactly balance the 1 atm of atmospheric pressure exerted on the outside of the balloon) at room temperature, and (as before) has a radius of 10 cm. The balloon is placed in a pressure chamber (like the kind used by deep sea SCUBA divers for slow decompression) and, after equilibration (still at room temperature), the balloon now has a radius of 8.3 cm. Assuming that the balloon remains spherical throughout the compression process, what is the final pressure inside the pressure chamber? Also, does it matter what the balloon's shape is during active compression?

Solution:

First we must find the balloon's initial volume (at 1 atm pressure). We know from the hint in Practice Exercise 2 that the volume of a sphere of radius r is given by V = (4/3) (pi) r³, so the volume of the balloon is given by V_initial = (4/3) (3.1415) (10 cm)³ = 4189 cm³. At 1 atm pressure, then, by Equation 5 we find (1 atm) (4189 cm³) = k' = 4189 atm cm³. This is the same at higher pressure, where the volume is given by V_final = (4/3) (3.1415) (8.3 cm)³ = 2395 cm³. So, by Equation, (P_final) (2395 cm³) = 4189 atm cm³, which implies that P_final = 1.75 atm.

As for the shape of the balloon during compression, it does not matter. Volume as a function of pressure (or vice-versa) is an example of a state function, one that depends only on the initial and final states of a system; we can completely ignore whatever happened between two states, provided that all intermediate states obey the laws governing the initial and final states of the system. Don't worry if this doesn't make sense just yet; this becomes much more important when studying thermodynamics.

If we view a gas as a system of particles flying around almost randomly, then we can easily see that gases get their volumes from particles striking the walls of a container, exerting a slight pressure, and causing the container to expand (in reality, a container is not a prerequisite for volume; the same principles apply, though, and a container allows us to easily visualize what is going on at the molecular level). If we increase the temperature of a gas, we by definition increase the particles' average kinetic energy and thus their average velocity. Each particle will then strike the container with greater force at higher temperature than at lower temperature. Since pressure is directly proportional to force but inversely proportional to area struck, then for pressure to remain constant the "average area" struck by a particle, and by consequence the volume of the container, must increase. The phrase "average area" is somewhat vague, and relies on the idea that a given area a is, statistically, struck by n particles in a given amount of time. Then the "average area" struck by one particle is thus given by a / n.

What if pressure is varied but temperature (and thus average particle velocity) is held constant? The opposite of the above clearly holds true. If pressure goes up while the force exerted by a particle is held constant (due to constant velocity), then the only explanation can be a decrease in the average area struck by a particle (i.e. a decrease in container or gas volume). The converse is logically also true.

Let us consider a container filled with 1 mol gas at a constant temperature and pressure (i.e. volume can vary). The addition of 1 mol of gas (i.e. doubling of gas quantity) results in a two-fold volume change. The addition of 3 mol gas (i.e. quadrupling of gas quantity) results in a four-fold volume change. Such experiments were first performed by Amedeo Avogadro in 1811, and the following principle is named Avogadro's Law in his honor:

The volume of a gas is proportional to the number of moles of gas present, provided that all other variables are held constant.

Mathematically, we write V = nk'', where n is moles of gas and k'' is a constant. (Equation 6)

Practice Exercise 4:

Background -

At the time of writing, there is a lot of talk in the popular news media about Saddam Hussein's alleged nuclear weapons program and the search for instruments needed to drive it. On the day of writing, CNN revealed that a scientist had hidden parts for a gas centrifuge needed to enrich uranium to the weapons-grade form (uranium-235). Most people, though, do not know why a centrifuge (a device that spins samples around an axis very rapidly) is needed to enrich uranium. Let us investigate, and tie the investigation into a gas problem.

Uranium exists naturally in predominantly two isotopes: uranium-235 and uranium-238 (see the lesson on Nuclear Chemistry). Only the former will readily undergo fission and thus is the only usable form for nuclear weapons (and reactors). The question is, how might we separate atoms that weigh almost the same and react chemically in identical fashions?

Uranium can be made to react with fluorine gas (F₂) to form a chemical known as uranium hexafluoride (UF₆). At moderately high temperature, its molecules vaporize and the compound becomes a gas. The kind containing U-235 has a mass of 352 amu (or g/mol), while the kind containing U-238 has a mass of 355 amu (or g/mol). Since the two molecules occupy identical volumes and possess identical intermolecular interactions, the U-238 form has a higher density than the U-235 form.

Spinning a mixture of gaseous UF₆ in a centrifuge causes both types of molecules to be pushed outward (due to centripetal force), but their different densities cause movement at different rates. The first molecules to come off will be the less useful U-238-containing UF₆, leaving behind slightly enriched UF₆ (i.e. has a higher-than-average U-235 fraction). Repetition of this enrichment process can eventually result in so-called weapons-grade uranium, suitable for use in an explosive. It is thus for this reason that the U.S. and IAEA are so interested in gas centrifugation technology in Iraq.

Question -

So, let us apply our new knowledge to an exercise. 10 g of elemental uranium (U) containing the naturally occurring isotope mixture are reacted with sufficient fluorine to completely consume all the solid uranium. Reaction byproducts are removed, yielding pure UF₆. The compound is next heated to 350 K to induce vaporization. All UF₆ is vaporized, and the vessel holding the gas at constant 1 atm pressure (actually, UF₆ is quite corrosive as a gas and would not be capable of containment in a conventional flexible container, but let's suspend disbelief) possesses a volume of .8159 L. Gas-phase centrifugation is performed, and .0571 L (at the same temperature and pressure) fully enriched UF₆ is extracted. Subsequent processing returns the uranium to elemental form. What is the mass of this lump of weapons grade uranium?

Solution:

We recognize that the ratio of the new gas volume to old gas volume is equivalent to the ratio of moles of enriched UF₆ to non-enriched UF₆. Mathematically, if {U} is the number of moles of non-enriched UF₆ initially, then .8159 L / .0571 L = (10 g / 238.0289 g/mol) / {U}, which implies that {U} = .00294 mol. This is equal to the number of moles of U-235 present in the final lump, so the mass equals .00294 mol. x 235 g/mol = .6909 g uranium.

Note that in reality this much enriched uranium would never be recovered from such a small sample, and that 100% enrichment is not needed to "weaponize" a sample of uranium.

The equations we have thus far discovered or proved can be combined into a single, powerful expression dubbed the Ideal Gas Law: Because of its utility and simplicity, THIS MAY BE THE MOST IMPORTANT EQUATION PRESENT IN THIS LESSON!

The Ideal Gas Law states the the pressure-volume product of a gas is proportional to the amount of gas present in the sample and to the temperature of the gas.

Mathematically, we write PV = nRT, where R is a constant known as the universal gas constant (see below). (Equation 7)

This equation is derived in the following manner. Since volume is proportional to both n (moles of gas) and to T (Kelvin temperature), we write V = rnT, where r is a new constant. Since volume is inversely proportional to P (pressure), as stated by Equation 5, we can further write V = RnT / P, where R is a constant that incorporates r. Rearrangement yields Equation 7.

Equation 7 contains a constant R, the universal gas constant. Unlike the previous constants, which can (but do not necessarily) differ from gas to gas and from situation to situation, R is constant for all gases at all temperatures, pressures, volumes, and gas quantities. It takes several forms:

Note that temperature is always in Kelvin and quantity is always in moles. It is important to be able to deduce which value of R should be implemented when solving an ideal gas problem. R should be selected based on the units used in a problem and the units desired in the result. In general, a well selected R should allow cancellation of several units. Now, let us solve an interesting problem:

Practice Exercise 5:

Some time ago, the author heard about a college physics class told to find, as part of a national competition, the number of atoms of helium (He) in the Goodyear Blimp. The winners, if memory serves correctly, got a free trip to some theme park. Let us see if we can do what a group of college students struggled to accomplish! Let us assume the following:

The number of atoms of He is constant (i.e. no leakage).
The blimp has a volume of 202,700 ft³ (from Goodyear Blimp website).
Pressure on the blimp is 1 atm (on the ground), and the pressure inside matches this pressure.
The nominal temperature is 25.0 ºC.

Find the number of atoms of He in the Goodyear blimp.

Solution:

The task of determining the number of atoms of He in the blimp is essentially the problem of finding the number of moles (n) in the blimp, for the molar mass of He is known. The Ideal Gas Law allows us to find n from the information given. First, we rearrange Equation 7 into a form that solves for n:

n = PV / RT

We then convert the temperature to Kelvin by adding 273.15, giving 298.15 K. We also convert the volume to a more sane (read: metric or SI) form, 5,739,824 L. We now are in a position to select an R value based on our units in use. We choose 8.205746 x 10^-2 (L*atm) / (mol*K), and solve the equation:

n = (1 atm)(5,739,824 L) / (8.205746 x 10^-2 L*atm / mol*K)(298.15 K) = 234629 mol He.

Using stoichiometry, we find that there are (234629 mol)(6.022 x 10²³ atoms/mol) = 1.41 x 10²⁹ atoms He.

A previous version of this exercise contained an error in the gas constant. A vigilant reader named Ian discovered this error, and it has since been corrected.

One interesting (or painful) consequence of the Ideal Gas Law is the potential for steam burns upon uncapping the radiator in a car. As the engine in a car runs, the water vapor in the radiator heats up. Since the volume and quantity (number of moles) of the vapor are fixed, the pressure must rise in accordance with the Ideal Gas Law. When the radiator cap is released, the high pressure rapidly equilibrates with the lower atmospheric pressure, resulting in a (predictable) rise in volume to offset the dropping pressure. The gas expands out of the radiator, and - yikes! - can severely burn anyone and anything nearby. By allowing a radiator to cool to the outside temperature, the pressure inside drops, as does the burn potential.

Ideal gases are those for which two conditions hold true. First, the particles in the gas have no volume (i.e. the gas is entirely empty space), and second, there is no interaction among gas particles. These conditions are never entirely met in any actual gas (a real gas), but can be approximated under many conditions.

In general, for most gases at non-extreme temperatures and pressures, the Ideal Gas Law can be used to obtain a reasonably correct solution to problems. If working on, say, homework or exams, unless instructions state otherwise, the Ideal Gas Law can almost certainly be used.

So, under what conditions might the Ideal Gas Law break down? Let us consider a few scenarios and the possible outcomes.

Gas particles are widely spaced, making their individual volumes negligible. For this reason, and the high particle velocity associated with temperature, intermolecular forces are negligible as well. The Ideal Gas Law serves well under these conditions.

At high pressure, the particles of gas are pushed close to one another. The gas becomes decreasingly compressible, for particles cannot be pushed into each other. The particles are still moving fast enough (indicated by high temperature) that intermolecular forces are negligible, however. Thus the volume is slightly higher than expected at the given temperature. The Ideal Gas Law probably does not give good results.

This is similar to the second scenario, but now intermolecular forces act to draw the particles even closer to one another, perhaps offsetting some of the error caused by particle volume. It is unclear whether the Ideal Gas Law will give good results.

Here, particle volume may be negligible, but particles are moving slow enough for interactions among particles to exist. The volume will thus be reduced slightly from what was expected. The Ideal Gas Law does not yield a good approximation of reality.

Now that we have identified several cases in which the Ideal Gas Law may not suffice, how can we calculate gas variables? The answer lies with tweaking the Ideal Gas Law to somewhat account for real gas behavior. One fairly successful (and surprisingly simple) attempt at such a result is the van der Waals equation, named after Johannes Diderik van der Waals, a late-nineteenth and early-twentieth century Dutch physicist; intermolecular forces that he articulated are now also named 'Van der Waals Forces' in his honor:

The van der Waals equation employs two constants specific to each individual gas, known as a and b, in a modified Ideal Gas Law that allows for good analyses of real gases under even extreme conditions. It is given below mathematically.

As is apparent, the van der Waals equation is identical to the Ideal Gas Law, but with scaling variables to adjust the pressure and volume to correspond with reality. The variables a and b are specific for each gas, and are available in many chemistry textbooks and in references such as the CRC Handbook of Chemistry and Physics.

Now that all the major macroscopic properties of gases have been considered, let us return to the molecular details of gases. We will focus on the molecular kinetics (or particle motion), and will justify all that math and physics presented earlier.

Since temperature corresponds to the average translational kinetic energy of the particles in a substance, we can infer that for all gases at a particular temperature, the average translational kinetic energy is constant. This means that for any individual gas, the average velocity (or, rather, speed) of the gas particles is dependent only on the temperature of the gas. Furthermore, the average speed of the particles in any gas is a function of only two variables: the gas temperature and the mass of the individual gas particles (in accordance with Equation 7). After some reasonably complex mathematics, we can arrive at an elegant equation, known as the root-mean-square speed equation, that relates average particle speed to temperature and molar mass (analogous to particle mass):

Note that the units for Equation 10 are somewhat quirky. To find speed in meters per second, molar mass must be in kilograms per mole, and R must be in J / (mol*K).

In reality, this gives only the average speed. However, the true speeds of particles can conceivably range from near zero to near infinite speeds. However, the majority of the particles stay at around the root-mean-square average speed. This is equivalent to saying that an individual particle has the highest probability of having a speed equal to the average speed. A probability curve, showing the probability that a given gas particle will move at a given speed can be constructed (such a curve is known as a Gaussian function), but requires substantially more mathematics than this lesson demands.

What good is calculating the average particle speed of a gas? Let us say that we wish to analyze the rates of diffusion and effusion of a gas. We begin with two definitions:

Diffusion is the process of mixing of two gases (i.e. the opening a bottle of perfume in a room and the subsequent drifting of perfume molecules across the room). Effusion is the process of expansion of a gas from a previously closed container through a small opening in the container (i.e. vapor escaping through a tiny hole in a hose).

We would expect that the average speed of gas particles is related to the rates of diffusion and effusion. However, at least in the case of diffusion, the relationship (though present) is not easily discernable. Even though gas molecules often travel at hundreds of meters per second, it can take minutes for perfume to be detectable a mere ten meters away from where the bottle was opened. This is primarily due to the interference and resultant scattering of particles caused by other (air) particles in the way. As a result, diffusion is a surprisingly difficult subject to consider, and even rudimentary calculations are beyond the scope of this lesson. However, effusion is a far simpler phenomenon to study.

The rate of effusion of a gas is proportional to its average particle (root-mean-square) speed, and the relative rate of effusion of two gases at a given temperature can be found by the ratio of the root-mean-square speed of one to the root-mean-square of the other.

This second observation has an important implication, as the following calculations show (M₁ and M₂ represent molar masses of gases 1 and 2):

We will now do two exercises with the idea of average particle speed and with Equation 11.

Practice Exercise 6:

Rank the following molecules in order of decreasing average particle speed at any fixed temperature at which they are all gaseous:

C₂H₄ (ethylene, a plant hormone that causes fruit to ripen)
Ne (neon, used in making reddish 'neon' signs)
C₇H₆O (benzaldehyde, an organic compound that gives almonds their distinctive smell)
Xe (xenon, used for producing noble gas compounds as well as exotic lighting)
CO₂ (carbon dioxide, a product of cellular respiration and a component of 'synthesis gas')
NO (nitric oxide, a component of photochemical smog as well as an important intercellular signaling molecule)
H₂ (hydrogen, used in early blimps and in production of ammonia)

Solution:

This problem does not require precise solutions using Equation 10, for all we care about is relative average particle speed. We must merely recognize that average speed decreases with increasing molecular weights or molar masses. We can thus rank the compounds in order of increasing molecular weight:

H₂ (~2 amu)
Ne (~20 amu)
C₂H₄ (~28 amu)
NO (~30 amu)
CO₂ (~44 amu)
C₇H₆O (~106 amu)
Xe (~131 amu)

Practice Exercise 7:

Calculate the relative rate of effusion of UF₆(352 g/mol) and UF₆ (355 g/mol), and suggest a possible use for this knowledge.

Solution:

By Equation 11, the solution is given by:

A possible use for this knowledge, suggested by the background information in Practice Exercise 4, is the refining of uranium. Though separation by density may be useful, gas phase centrifugation also is used to force UF₆ gas through pores in a membrane, creating, in essence, effusion. Since the two gases do not effuse at identical rates, isotopic separation can be achieved.

So far, most of our discussion has been focused on unmixed gases (i.e. pure nitrogen gas). However, most gases exist in mixtures (for example, the air is approximately 78.1% nitrogen, 20.9% oxygen, 0.9% argon, and 0.1% "other" [carbon dioxide, et cetera]). Do the gas laws become cumbersome or useless? Hardly.

A gas mixture is nothing more than the sum of its components. Two laws make it possible to "dissect" a gas mixture and analyze the component gases individually.

The partial pressure of a component of an ideal gas mixture is equivalent to the pressure that the component gas would possess if it alone occupied the volume of the gas mixture at the same temperature as the mixture. The partial volume of a component of an ideal gas mixture is equivalent to the volume that the component gas would occupy if it alone possessed the pressure of the gas mixture at the same temperature as the mixture.

The pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases. This is known as Dalton's Law of Partial Pressures.

Mathematically, if P_i is the partial pressure of the i^th gas in a mixture of n ideal gases (with partial pressures P₁, P₂, etc.), where P_m is the pressure of the gas mixture, then:

Another important result of Dalton's Law of Partial Pressures is that the partial pressure of a component gas in an ideal gas mixture is equal to the mole fraction of the component gas times the pressure of the mixture. Mole fraction is merely the number of moles of the component gas divided by the total number of moles of all gases present in the mixture.

Practice Exercise 8:

Using the data presented at the beginning of this section, calculate the combined partial pressure of oxygen and argon at sea level.

Solution:

As stated, oxygen composes 20.9% of the air, while argon composes 0.9%. Together, they account for 20.9% + 0.9% = 21.8% of the air. Assuming that these percentages are given as percentage of moles of gas (not percent mass), a reasonable assumption , then this corresponds to a mole fraction of .218. Since at sea level there is a total pressure of 1 atm, oxygen and argon together contribute a collective partial pressure of .218 atm.

A very similar (albeit less commonly known) law is the Amagat-Leduc Law, which is as follows:

The volume of a mixture of ideal gases is equal to the sum of the partial volumes of the component gases. This is known as the Amagat-Leduc Law of Additive Volumes.

Mathematically, if V_i is the partial volume of the i^th gas in a mixture of n ideal gases (with partial volumes V₁, V₂, etc.), where V_m is the volume of the gas mixture, then:

Similar to the result of Dalton's Law, a result of the Amagat-Leduc Law of Additive Volumes is that the partial volume of a component gas in an ideal gas mixture is equal to the mole fraction of the component gas times the volume of the mixture. Observe that the two laws just given are essentially converses of one another, and can be selectively used to gain information about partial pressures or partial volumes. Since the laws are so similar, we will omit an exercise for the Amagat-Leduc rule.

Under ideal circumstances (as earlier defined), these laws are equivalent (i.e. pressures gleaned from Dalton's Law can be used to determine volumes using the Amagat-Leduc Law, and vice-versa). However, when dealing with real gases, these laws may not be equivalent. In general, when dealing with real gases, the Amagat-Leduc Law is best for high(er) pressures and Dalton's Law is best for low(er) pressures.

One important gas mixture results from the vaporization of a liquid, or boiling. We close with this topic.

Gases can be formed in a couple of different ways. Some elements (such as the noble gases) exist almost exclusively as gases at any non-extreme temperature. Solid carbon dioxide ("dry ice") sublimes, changing from solid to gas without ever going through a liquid phase. By and large, however, gases are formed from liquids under the right conditions.

Every liquid slowly gives up individual particles to the surrounding environment regardless of environmental conditions. This process is known as evaporation. The result is that the environment now contains gaseous liquid. However, the gas can be deposited back onto the surface of the liquid (or a solid), in the reverse of evaporation: condensation.

Evaporation is the spontaneous generation of a gas from a liquid. Condensation is the spontaneous generation of a liquid from a gas. These processes almost always occur in tandem, with one working just slightly faster than the other.

For all substances, at constant pressure, at higher temperatures evaporation is favored, while at lower temperatures condensation is favored. This is why puddles disappear more rapidly when the sun comes out; the water in the puddles is heated.

Also, for all substances, at constant temperature, at lower pressures evaporation is favored, while higher pressure favors condensation. If both temperature and pressure vary, the rate of evaporation or condensation is regulated by an interplay of both variables. This principle is put to practical use in a laboratory device known as a rotary evaporator, or Rotavapor (®). Such a device is shown in Figures 8a through 8c. The captions contain an explanation of its purpose and theory of its operation.

Figure 8a: Büchi Rotavapor assembly.
A rotary evaporator such as this one at Gene Tools, LLC, removes solvents that are vaporized at temperatures much lower than the solute (dissolved compounds), leaving behind solvent-free solid compounds. The water bath at right heats up the flask contents, while a vacuum pump (far left, behind the exhaust tubing) lowers the pressure. Both processes promote evaporation. Click to open a larger version in another window.

Figure 8b: Büchi Rotavapor condenser.
Though a solvent may vaporize at room temperature (or higher) when placed under vacuum, the vacuum is not enough to overcome condensation favored by ice packed in the condenser. The vaporized solvent condenses on glass inside the assembly and drips into a storage flask, also on ice to inhibit evaporation. Click to open a larger version in another window.

Figure 8c: Büchi Rotavapor rotary flask attachment.
The contents of a round-bottom flask (solvent plus solute, in this case an experimental fluorescent dye in dichloromethane) are constantly spun by the rotary flask attachment. Such spinning helps alleviate the possibility of violent boiling, a very undesirable process in the industrial chemical world. Click to open a larger version in another window.

The caption for Figure 8c describes how boiling is often undesirable, and suggests a possible way to avoid it. Everyone who has ever cooked has an intuitive understanding of what boiling is: water gets really hot, starts evaporating quickly, and then hits the "boiling point," when it starts bubbling uncontrollably. The observant cook might even notice that water reaches the "boiling point" slightly more quickly when a cover is put on; this is the main reason for using lids on pots and saucepans. But what, physically, makes boiling different from evaporation? What defines the "boiling point?" Is the boiling point fixed at one temperature for a liquid, and if not, how can it be changed? We now turn to these questions.

The vapor pressure of a liquid is defined as the pressure exerted by evaporated liquid once dynamic equilibrium is achieved. Vapor pressure varies with temperature.

Since evaporation typically results in a mix of gases previously present above the liquid, and the newly evaporated vapor, the vapor pressure can also be the partial pressure of the liquid vapor once dynamic equilibrium is achieved.

The term dynamic equilibrium is one that we have not yet encountered, and indeed an entire future lesson is devoted to chemical equilibrium. For our current purposes, we can say that dynamic equilibrium is the condition where no net gas exchange is occurring; that is, no condensation or evaporation is visibly observed. Vapor pressures are determined experimentally by heating liquids to various temperatures, and measuring the final pressure change that results from newly evaporated liquid.

Our definition of vapor pressure is important, because we use it to pin down two previously poorly defined concepts:

Evaporation will occur when the partial pressure of a liquid's vapor is less than the defined vapor pressure of the liquid at the temperature in question. Boiling of a liquid occurs when the atmospheric (or total) pressure above the liquid is less than the defined vapor pressure of the liquid in question.

It becomes clear, then, that applying a vacuum (such as that in the rotary evaporator) is responsible for accelerating evaporation and even boiling. The macroscopic evidence for boiling (i.e. bubbles) results from spontaneous vaporization of liquid inside the liquid, while evaporation occurs only at the surface (technically, the liquid-gas interface). The bubbling, however, requires seed points within the liquid, where vaporization can begin. Movement of the liquid (as in the rotary evaporator), or the use of extremely clean glassware and pure chemicals, can impede bubbling by eliminating seed points. In chemistry, boiling is to be avoided because of potential splash damage to instruments, loss of chemicals, and injuries to workers.

As stated, the vapor pressure is a function of temperature. It is for this reason that boiling is dependent on temperature: as the temperature rises, the vapor pressure of liquids rises, increasing the likelihood that atmospheric pressure is less than the vapor pressure. Finally, we can also answer a question posed in the Pressure and Volume section: why does elevation affect cooking times and temperatures? At higher elevations, there is less "atmosphere" above us than at sea level, so the pressure logically drops, simultaneously lowering the minimum temperature for boiling. Since temperature does not usually rise much after boiling commences (due to reasons left for another lesson), the final cooking temperature is lower at higher elevations than at sea level. Thus to ensure thorough (or safe) cooking, foods must be boiled longer!

Okay, let's be honest. few people think that gases are the most fun things in the world to study. But, the first quantitative portions of chemistry involved gases. In a way, precise measurements of gases brought chemistry out of the ages of alchemy and trial-and-error pharmacology and into the modern era. Some of us will go on to work in fields such as aerospace, petrochemicals, and catalysis research, where this lesson, however detailed, is just the stepping stone into the rich disciplines of gas-phase physical chemistry, fluid dynamics, and more. But even for the majority of us who (like the author) end up more concerned with liquid phase organic reactions, macromolecular interactions, or the preparation of new crystal lattices than with the intricacies of gas compression, the study of gases is a fascinating and ancient study, with something to be gained by everyone!

Author: C. Shultz

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