### Gas Laws (Ideal and Non-Ideal)

Before we begin a discussion of the laws that govern the behavior of gases, we need to define several measures of pressure. Pressure is usually given in atmospheres--1 atm being the standard pressure at sea level. Another common measure is millimeters of mercury, or mm Hg; 760 mm Hg is equivalent to one atmosphere. Another designation for mm Hg is the "torr," named in honor of Evangelista Torricelli, the Italian scientist who invented the barometer. The SI unit, the pascal (Pa) after Blaise Pascal, a French mathematician famous for his triangle, is still less common than the previous two. 101.325 x 103 Pa make up one atmosphere; because the pascal is such a tiny unit, the kilopascal (kPa) is more convenient, with 101.325 kPa making up 1 atm. Finally, the bar is a unit equal to 100,000 Pa or 100 kPa; therefore, 1 atm = 1.01325 bar. So:

1 atm = 760 mm Hg = 760 torr = 101.325 x 103 Pa = 101.325 kPa = 1.01325 bar.

Obviously, there is a profusion of units for pressure! However, in the gas laws that follow, the atm is the most common, followed by the mm Hg or torr.

Now, let's talk about gases. Our knowledge of the behavior of gases is a conglomeration of several partial theories. The first was postulated by Robert Boyle in 1661, stating that the volume of a gas at a constant temperature is inversely proportional to the pressure exerted on the gas. So, if pressure is doubled, volume is halved; if pressure is cut to a third, volume expands by three times. Mathematically, "proportional" means "equal multiplied by a constant," as shown below:

P = CB(1/V), where CB is a proportionality constant that depends on which units are used

We can rewrite the above equation to make it:

PV=CB (pressure times volume is some constant)

The above equation is called Boyle's Law. Or, if we have several pressure/volume pairs, we can say:

P1V1 = CB; P2V2 = CB; P3V3 = CB; P4V4 = CB....

Since the constant is the same in all cases, the pressure/volume pairs become equal, as long as the same units are used throughout:

P1V1 = P2V2 = P3V3 = P4V4 ....

Let's say we have a 5.00 L balloon of gas at 800 mm Hg pressure; what would the pressure be if the volume was 14.0 L? Using the above equation, we know that:

(800 mm Hg) * (5.00 L) = (X mm Hg) * (14.0 L)

Solving for X, we know that the new pressure is about 286 mm Hg.

Another partial theory is Charles' Law, which states that volume is directly proportional to temperature. Using a proportionality constant, we can write that:

V = CCT ,    or CC = V/T

T must be the temperature in kelvins; using degrees Fahrenheit or Celsius will not work. Using the same logic as above, namely that the proportionality constant is the same for each volume/temperature pair as long as the same units are used, we can also say that:

V1/T1 = V2/T2 = V3/T3 ....

As an example of Charles' Law, assume we have a 2.00 L expandable box of gas at 298 kelvins. What would be the volume of the box be if the gas were heated in an oven to 500 kelvins? Using Charles' law:

2 L / 298 K = X L / 500 K

Solving for X, the new volume is 3.36 L.

The final partial theory was developed by both Joseph Gay-Lussac, a French chemist, and Amedeo Avogadro, a lawyer and scientist now famous for his constant. Gay-Lussac, through experimentation with reacting gases, found they always combine in simple, whole-number ratios: for instance, the combustion of two units of hydrogen gas (say 1 L) with one unit (1 L) of oxygen gas gives one unit of water (1 L). This observation called Gay-Lussac's law of combining volumes, led Avogadro to investigate further the behavior of gases. He speculated that a volume of any gas is proportional only to the number of molecules present, as shown below with a proportionality constant:

V = CAn

where n is the number of moles of the gas. There is little practice necessary for this law, since it is so simple.

Now, we have three different laws that govern the behavior of gases:

PV=CB
V = CCT
V = CAn

We can combine these three laws into the following Ideal Gas Law:

PV=nT * CACBCC

We can also combine all these proportionality constants into one value, called the universal gas constant (symbol: R), giving an equation of:

PV=nRT

where P is pressure in atmospheres, V is volume in liters, n is the number of moles of gas, R is the universal gas constant (0.0821 L-atm/K-mol), and T is the temperature in kelvins. This is one of the most useful and powerful equations in chemistry, because it relates so many different properties of a gas. For example, how many moles of hydrogen gas are present in a 6.4 L box, with a pressure of 2.1 atm and a temperature of 342 K? Using the Ideal Gas Law, we can write:

(2.1 atm)(6.4 L) = (X mol)(0.0821 L-atm/K-mol)(342 K)

Solving for X, we get 0.479 moles of hydrogen. Note that the gas could easily have been nitrogen, oxygen, xenon, or any other gas--the Ideal Gas Law works with any gas you want!

Let's go back to vapor pressure for a moment. We wondered how to know whether a sample of water would evaporate completely or establish an equilibrium vapor pressure. There were three factors involved: the volume of the container, the number of moles of water, and the vapor pressure. In the last page, we had no way of equating pressure, volume, and number of moles; with the Ideal Gas Law, however, it's easy.

Let's assume we have a 2-mole (36-gram) sample of water in a 5000 liter room, at room temperature. The way we see whether or not the water will fully evaporate is to calculate its volume at room temperature and its vapor pressure (23.8 mm Hg or 0.03132 atm at 25 °C/298 K):

(0.03132 atm)(X L) = (2 mol)(0.0821 L-atm/K-mol)(298 K)

Solving for X, we find that the water vapor would take up 1562 liters. Since the available volume is much greater than the volume needed, all of the vapor will evaporate. If the room volume were smaller, we could find the amount of water that evaporates by finding how many moles of water would be needed to fill the room at the vapor pressure and 298 K.

The Ideal Gas Law makes some important assumptions. First, it assumes that the gas molecules have no actual volume. At normal conditions, over 99.9% of a gas is empty space, so this a pretty safe assumption. The real volume of the gas would be slightly larger than that predicted by the Ideal Gas Law. Since the volume of the molecules becomes more important at high pressure, the Ideal Gas Law works best at large volumes. Second, it ignores the attractive forces between molecules. These forces, the same ones discussed earlier in this chapter, would reduce the pressure of the gas, because the molecules would hit the walls of the container less often and with less force. This is because they are attracted to the other molecules in the center of the container, making them curve away from the walls. Intermolecular attractions increase with decreasing temperature, since the molecules move slower and are more affected by intermolecular attraction. Therefore, the Ideal Gas Law breaks down at high pressure and low temperature.

To compensate for these assumptions, Johannes van der Waals developed a Real Gas Law that uses several factors to account for molecular volume and intermolecular attraction. It is because of his accomplishments that the class of intermolecular forces discussed earlier was named in his honor. The van der Waals equation is as follows:

[P + a(n/V)2][V-bn] = nRT

P, n, R, and T are the same as in the Ideal Gas Law; a is a correction for intermolecular forces (in atm-L2/mol2) and b accounts for molecular volume (in L/mol). Each of these factors must be determined by experiment. Some common values are:

Gas a b
He 0.034 0.0237
Ar 1.34 0.0322
H2 0.244 0.0266
N2 1.39 0.0391
O2 1.36 0.0318
Cl2 6.49 0.0562
CO2 3.59 0.0427
H2O 2.25 0.0428

Let's use the Ideal Gas Law equation to find the pressure in a 50-L tank of nitrogen gas at 75 kelvins with 50 moles of gas. Then, we'll use the van der Waals equation to find the real value:

(X atm)(50 L) = (50 mol)(0.0821 L-atm/K-mol)(75 K)

Using the above setup, we get a pressure of 6.1575 atmospheres. Next, let's use the van der Waals equation:

[X atm + (1.39 atm-L2/mol2)(50 mol/50 L)2][50 L - 0.0391 L/mol * 50 mol] = 50 mol * (0.0821 L-atm/K-mol) * 75 K

Solving this mess of an equation yields a pressure of 5.0181 atmospheres, in keeping with the theory that pressure decreases due to intermolecular attraction.

Under the vast majority of circumstances, simply use the Ideal Gas Law due to its simplicity; moreover, you will often not know the a and b values for the gas in question. Only when a problem specifically calls for the van der Waals equation should you go to the trouble of using it.

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