Practice problems for these concepts can be found at:
Pressure
When we use the word pressure, we may be referring to the pressure of a gas inside a container or to atmospheric pressure, the pressure due to the weight of the atmosphere above us. These two different types of pressure are measured in slightly different ways. Atmospheric pressure is measured using a barometer (Figure 8.1).
An evacuated hollow tube sealed at one end is filled with mercury, and then the open end is immersed in a pool of mercury. Gravity will tend to pull the liquid level inside thetube down, while the weight of the atmospheric gases on the surface of the mercury pool will tend to force the liquid up into the tube. These two opposing forces will quickly balance each other, and the column of mercury inside the tube will stabilize. The height of the column of mercury above the surface of the mercury pool is called the atmospheric pressure. At sea level the column averages 760 mm high. This pressure is also called 1 atmosphere (atm). Commonly, the unit torr is used for pressure, where 1 torr = 1 mm Hg, so that atmospheric pressure at sea level equals 760 torr. The SI unit of pressure is the pascal (Pa), so that 1 atm = 760 mm Hg = 760 torr = 101,325 Pa (101.325 kPa). In the United States pounds per square inch (psi) is sometimes used, so that 1 atm = 14.69 psi.
To measure the gas pressure inside a container, a manometer (Figure 8.2) is used. As in the barometer, the pressure of the gas is balanced against a column of mercury.
Ideal Gas Equation
The ideal gas equation has the mathematical form of PV = nRT, where:
P = pressure of the gas in atm, torr, mm Hg, Pa, etc.
V = volume of the gas in L, mL, etc.
n = number of moles of gas
R = ideal gas constant: 0.0821 L·atm/K·mol
This is the value for R if the volume is expressed in liters, the pressure in atmospheres, and the temperature in kelvin (naturally). You could calculate another ideal gas constant based on different units of pressure and volume, but the simplest thing to do is to use the 0.0821 and convert the given volume to liters and the pressure to atm. And remember that you must express the temperature in kelvin.
Let's see how we might use the ideal gas equation. Suppose you want to know what volume 20.0 g of hydrogen gas would occupy at 27°C and 0.950 atm. You have the pressure in atm, you can get the temperature in kelvin (27°C + 273 = 300.K), but you will need to convert the grams of hydrogen gas to moles of hydrogen gas before you can use the ideal gas equation. Also, remember that hydrogen gas is diatomic, H2.
First you'll convert the 20.0 g to moles:
(20.0 g/l) × (1 mol H2/2.016 g) = 9.921 mol H2
(We're not worried about significant figures at this point, since this is an intermediate calculation.)
Now you can solve the ideal gas equation for the unknown quantity, the volume.
Finally, plug in the numerical values for the different known quantities:
V = (9.921 mol)(0.0821 L atm/K mol) (300.K)/0.950 atm
Is the answer reasonable? You have almost 10 mol of gas. It would occupy about 224 L at STP (10 mol × 22.4 L/mol) by Avogadro's relationship. The pressure is slightly less than standard pressure of 1 atm, which would tend to increase the volume (Boyle's law), and temperature is greater than standard temperature of 0°C, which would also increase the volume (Charles's law). So you might expect a volume greater than 224 L, and that is exactly what you found.
Remember, the final thing you do when working any type of chemistry problem is answer the question: Is the answer reasonable?
Dalton's Law of Partial Pressures
Dalton's law says that in a mixture of gases (A + B + C …) the total pressure is simply the sum of the partial pressures (the pressures associated with each individual gas). Mathematically, Dalton's law looks like this:
PTotal = PA + PB + PC + …
Commonly Dalton's law is used in calculations involving the collection of a gas over water, as in the displacement of water by oxygen gas. In this situation there is a gas mixture: O2 and water vapor, H2O(g). The total pressure in this case is usually atmospheric pressure, and the partial pressure of the water vapor is determined by looking up the vapor pressure of water at the temperature of the water in a reference book. Simple subtraction generates the partial pressure of the oxygen.
If you know how many moles of each gas are in the mixture and the total pressure, you can calculate the partial pressure of each gas by multiplying the total pressure by the mole fraction of each gas:
where XA = mole fraction of gas A. The mole fraction of gas A would be equal to the moles of gas A divided by the total moles of gas in the mixture.
Graham's Law of Diffusion and Effusion
Graham's law defines the relationship of the speed of gas diffusion (mixing of gases due to their kinetic energy) or effusion (movement of a gas through a tiny opening) and the gases' molecular mass. The lighter the gas, the faster is its rate of effusion. Normally this is set up as the comparison of the effusion rates of two gases, and the specific mathematical relationship is:
where r1 and r2 are the rates of effusion/diffusion of gases 1 and 2 respectively, and M2 and M1 are the molecular masses of gases 2 and 1 respectively. Note that this is an inverse relationship.
For example, suppose you wanted to calculate the ratio of effusion rates for hydrogen and nitrogen gases. Remember that both are diatomic, so the molecular mass of H2 is 2.016 g/mol and the molecular mass of N2 would be 28.02 g/mol. Substituting into the Graham's law equation:
Hydrogen gas would effuse through a pinhole 3.728 times as fast as nitrogen gas. The answer is reasonable, since the lower the molecular mass, the faster the gas is moving. Sometimes we measure the effusion rates of a known gas and an unknown gas, and use Graham's law to calculate the molecular mass of the unknown gas.
Gas Stoichiometry
The gas law relationships can be used in reaction stoichiometry problems. For example, suppose you have a mixture of KClO3 and NaCl, and you want to determine how many grams of KClO3 are present. You take the mixture and heat it. The KCIO3 decomposes according to the equation:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
The oxygen gas that is formed is collected by displacement of water. It occupies a volume of 542 mL at 27°C. The atmospheric pressure is 755.0 torr. The vapor pressure of water at 27°C is 26.7 torr.
First, you need to determine the pressure of just the oxygen gas. It was collected over water, so the total pressure of 755.0 torr is the sum of the partial pressures of the oxygen and the water vapor:
The partial pressure of water vapor at 27°C is 26.7 torr, so the partial pressure of the oxygen can be calculated by:
At this point you have 542 mL of oxygen gas at 728.3 torr and 300. K (27°C + 273). From this data you can use the ideal gas equation to calculate the number of moles of oxygen gas produced:
You will need to convert the pressure from torr to atm:
(728.3 torr) × (1 atm/760.0 torr) = 0.9583 atm
and express the volume in liters: 542 mL = 0.542 L
Now you can substitute the quantities into the ideal gas equation:
(0.9583 atm)(0.542 L)/(0.0821 L · atm/K · mol)(300. K) = n
Now you can use the reaction stoichiometry to convert from moles O2 to moles KClO3 and then to grams KClO3:
Non-Ideal Gases
We have been considering ideal gases, that is, gases that obey the postulates of the Kinetic Molecular Theory. But remember—a couple of those postulates were on shaky ground. The volume of the gas molecules was negligible, and there were no attractive forces between the gas particles. Many times approximations are fine and the ideal gas equation works well. But it would be nice to have a more accurate model for doing extremely precise work or when a gas exhibits a relatively large attractive force. In 1873, Johannes van der Waals introduced a modification of the ideal gas equation that attempted to take into account the volume and attractive forces of real gases by introducing two constants—a and b—into the ideal gas equation. Van der Waals realized that the actual volume of the gas is less than the ideal gas because gas molecules have a finite volume. He also realized that the more moles of gas present, the greater the real volume. He compensated for the volume of the gas particles mathematically with:
corrected volume = V – nb
where n is the number of moles of gas and b is a different constant for each gas. The larger the gas particles, the more volume they occupy and the larger the b value.
The attraction of the gas particles for each other tends to lessen the pressure of the gas, because the attraction slightly reduces the force of gas particle collisions with the container walls. The amount of attraction depends on the concentration of gas particles and the magnitude of the particles' intermolecular force. The greater the intermolecular forces of the gas, the higher the attraction is, and the less the real pressure. Van der Waals compensated for the attractive force with:
corrected pressure = P + an2/V2
where a is a constant for individual gases. The greater the attractive force between the molecules, the larger the value of a. The n2/V2 term corrects for the concentration. Substituting these corrections into the ideal gas equation gives van der Waals equation:
(P + an2/V2)(V – nb) = nRT
The larger, more concentrated, and stronger the intermolecular forces of the gas, the more deviation from the ideal gas equation one can expect and the more useful the van der Waals equation becomes.
Practice problems for these concepts can be found at:
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From 5 Steps to a 5 AP Chemistry. Copyright © 2010 by The McGraw-Hill Companies. All Rights Reserved.
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