THE PROPERTIES OF GASES AND THE KINETIC-MOLECULAR THEORY: So far, we have focused on the structure of individual atoms and molecules paying close attention to the relationship between their physical make-up (electronic configuration and 3-D shape...) and the resulting properties (ionization energy, nature of bonds, net dipoles...). Now begin to look at what happens when large numbers of atoms or molecules come together to form solids, liquids, and gases. In the gaseous state, the components (atoms or molecules) are separated by large amounts of empty space. For this reason, gases are the easiest state to study and understand
Try this, list some properties of ALL gases (see how many you can get): some possible answers: ALL gases... 1) completely fill their container. 2) have low density. 3) are highly compressible. 4) exert pressure on their container. 5) have particles with KE propotional to the absolute temp. 6) form homogeneous mixtures with other gases. Here are some INCORRECT answers that are commonly given: some incorrect answers: these are true of some, but not all gases 1) odorless 2) colorless Try this, list some properties that are important when choosing a gas for an airbag: some possible answers: 1) non-toxic 2) non-flammable 3) environmentally safe 4) inexpensive
We can understand these properties if we imagine what is happening in the gaseous phase on the atomic level (See the figure below):
The particles that make up a gas can move freely within their container. Statistically speaking, this leads to the complete filling of the container over time. Since most containers are much larger than the particles that fill them, a container filled with a gas will have mostly empty space with a resulting low density and a high compressibility. Because the particles are not exerting strong forces on each other and because the volume is mostly empty space, all gases form homogeneous mixtures with each other given sufficient mixing. This information is summarized in the kinetic-molecular theory of gases:
1. Gases consist of large numbers of particles in constant motion. 2. The volume occupied by the particles in a gas is negligible compared to the volume of the container. 3. Gas particles do not exert attractive or repulsive forces on each other. 4. Collisions between gas particles is elastic. 5. The KE of the average gas particle is proportional to the temperature (in Kelvin).
So why are the particles able to move around freely? What makes them different than those of a solid or a liquid? All molecules and atoms exert attractive forces on other neighboring molecules and atoms. These forces are called Intermolecular Forces (we'll study them next chapter). Substances that are gases at room temperature have weak intermolecular forces holding the particles together. Weak enough, in fact for the KE of the particles to overcome these attractive forces. The particles end up bouncing off each other sort of like billiard balls, except they fill the container (imagine playing 3-D pool)!
Since the KE of the particles is proportional to the absolute temperature, it is possible to lower the KE enough (by lowering the temp.) so that the forces are now able to hold the particles together (and voila, you have liquid nitrogen or liquid oxygen).
Try this: In your own words, summarize the above discussion of the properties of gases.
As shown in the figure above, when a gas particle collides with the walls of its container, the particle exerts a pressure. This also means that the atmosphere around us is exerting pressure on us. We don't notice it because we are used to the sensation just like fish in the ocean that don't realize the pressure of all the water on them. This pressure in fact, is holding us together and preventing us from popping like a alien ejected from its ship in deep space. (YUCK!)
The pressure of the atmosphere can be determined using a barometer. Typically a barometer is constructed using mercury (Hg) because of its high desity. This is the origin of mmHg as units of pressure. It is helpful to think of the height of the column of mercury as perfectly balancing the height of the column of air. Both columns have the same diameter and the column of air extend to the top of the atmosphere. There are several other common units for pressure that you should be aware of such as: atmosphere, bar, pascal and torr. You do not have to memorize the conversion factors to change between these units, but be sure that if you are given the conversion factors on a test that you are able to use them.
The pressure of the atmosphere can be determined using a barometer. Typically a barometer is constructed using mercury (Hg) because of its high desity. This is the origin of mmHg as units of pressure. It is helpful to think of the height of the column of mercury as perfectly balancing the height of the column of air. Both columns have the same diameter and the column of air extend to the top of the atmosphere.
There are several other common units for pressure that you should be aware of such as: atmosphere, bar, pascal and torr. You do not have to memorize the conversion factors to change between these units, but be sure that if you are given the conversion factors on a test that you are able to use them.
Try this: What is the pressure in atmospheres inside the cabin of a jet plane if the cabin barometer reads 688 mmHg? 1 atm = 760 mmHg answer: 0.905 atm
THE GAS LAWS AND THE IDEAL GAS EQUATION: Hopefully you can now see how this description of gases on the atomic level can explain many of the properties of gases, such as compressibility (the change in volume as the pressure is changed). As scientists it is important to be able to develop general relationships based on experimental results so we can make predictions about new situations. Several hundred years ago scientists develope equations realting the volume of a gas to its pressure, its temperature and to the number of moles. These laws were based on experimental results and are summarized as follows:
The individual gas laws can be combined in the form of the ideal gas equation: PV=nRT where "constant" in the individual laws has been replaced by the gas constant, R = 0.08206 (L atm) / (mol K)
Try this: How many moles of air are in the lungs of an average adult with a lung capacity of 3.8 L ? Assume that the person is at 1.0 atm pressure and has normal body temperature of 37oC. hint: convert T from oC to K start with PV=nRT and solve for n: n = PV / RT answer: n = (1.0 atm) (3.8 L) / (0.08206 Latm/molK) (310 K) n = 0.15 mol That's 90,000,000,000,000,000,000,000 molecules in each breath !!!
One mole of an ideal gas at standard temperature and pressure, STP (0 oC, 1 atm) occupies a volume of 22.4 L
Try this: Prove the above statement by calculating the volume using the ideal gas equation.
Because the value for R doesn't change, we can use R =(PV) / (nT) as a way to predict the effect of making changes on a sample of gas. If we have two sets of conditions (P1, V1, n1, T1 and P2, V2, n2, T2), we can solve for any unknowns. For example, a weather balloon filled with helium has a volume of 1.0 x 104 L at 1.00 atm and 30 oC. It ascends to an altitude at which the ressure is only 0.60 atm and the temperature is -20 oC. What is the volume of the ballon then? Begin by making a table of the varialbles:
P1 = 1.00 atm P2 = 0.60 atm V1 = 1.0 x 104 L V2 = ??? T1 = 30 oC = 303 K T2 = -20 oC = 253 K n1 , n2 = unknown, but it doesn't change so we can cancel it out.
Try this: In a typical automobile engine, the mixture of gasoline and air in a cylinder is compressed from 1.0 atm to 9.5 atm. If the uncompressed volume of the cylinder is 410 mL, what is the volume when the mixture is fully compressed? hint: use (P1)(V1) = (P2)(V2) since T and n don't change answer: the new volume = 43 mL
APPLICATION OF THE IDEAL GAS EQUATION: Here are some important uses of the ideal gas law.
1. Calculation of density and molar mass: Try this: A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/L at 36 oC and 2.88 atm. Calculate the molar mass of the compound and determine its molecular formula. answer to molar mass: M = dRT/P (7.71 g/L)(0.821 Latm/molK)(309K) / (2.88 atm) = 67.9 g/mol hint for determining the formula: this is mainly by trial and error. Start with ClO and determine its molar mass. If it isn't correct try Cl2O and ClO2 .... answer to formula: ClO2 2. Stoichiometry calculations for reactions involving the production or consumption of gases: Try this: A key step in the production of sulfuric acid is the oxidation of sulfur dioxide: 2 SO2(g) + O2(g) 2 SO3(g) Suppose that 6.37 L of O2 (at 420 oC and 1.22 atm) reacts completely with an excess of SO2. The SO3 produced is cooled to 80.0 oC at P = 0.970 atm. Calculate the volume of SO3 produced. hint: try breaking it down in the following steps: 1) Use PV=nRT to determine how many moles of O2 you use 2) Use the balanced reaction to determine how many moles of SO3 you will then make 3) Use PV=nRT to determine the volume of the SO3 answer: 8.16 L of SO3
1. Calculation of density and molar mass:
Try this: A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/L at 36 oC and 2.88 atm. Calculate the molar mass of the compound and determine its molecular formula. answer to molar mass: M = dRT/P (7.71 g/L)(0.821 Latm/molK)(309K) / (2.88 atm) = 67.9 g/mol hint for determining the formula: this is mainly by trial and error. Start with ClO and determine its molar mass. If it isn't correct try Cl2O and ClO2 .... answer to formula: ClO2
2. Stoichiometry calculations for reactions involving the production or consumption of gases: Try this: A key step in the production of sulfuric acid is the oxidation of sulfur dioxide:
Suppose that 6.37 L of O2 (at 420 oC and 1.22 atm) reacts completely with an excess of SO2. The SO3 produced is cooled to 80.0 oC at P = 0.970 atm. Calculate the volume of SO3 produced.
hint: try breaking it down in the following steps: 1) Use PV=nRT to determine how many moles of O2 you use 2) Use the balanced reaction to determine how many moles of SO3 you will then make 3) Use PV=nRT to determine the volume of the SO3 answer: 8.16 L of SO3
REAL GASES: OK, now don't get mad, but I wasn't being completely honest with you. It turns out that REAL gases have some deviation from the IDEAL behavior described by the ideal gas equation and the Kinetic-Molecular theory isn't completely true. If you look at figure 10.22 in your text, you'll see a graph showing this deviation. For an ideal gas, the plot of PV/nRT vs P should always equal 1. Real gases show non-ideal behavior, especially at high pressure. Let's see why...
One of the most important improvements on the ideal qas equation was proposed by Johannes van der Waals in 1873. The van der Waals equation attempts to correct for two assumptions of the Kinetic-Molecular theory that are not always true:
1. Gas particles do occupy volume and though this volume is very small, it is not always negligible, especially at high pressure when gas particles are squeezed together. This actual volume of the container is smaller than the ideal volume since some of the space is occupied by particles and two particles can't occupy the same space. Larger particles will have a larger correction to the V term: 2. Collisions between gas particles are not completely elastic, especially at low temperatures when they are moving more slowly. This "stickiness" results in a correction to the P term. The actual pressure is lower than the ideal pressure because some of the particles form temporary associations so there are fewer idividual collisions as shown in the drawing:
1. Gas particles do occupy volume and though this volume is very small, it is not always negligible, especially at high pressure when gas particles are squeezed together. This actual volume of the container is smaller than the ideal volume since some of the space is occupied by particles and two particles can't occupy the same space. Larger particles will have a larger correction to the V term:
2. Collisions between gas particles are not completely elastic, especially at low temperatures when they are moving more slowly. This "stickiness" results in a correction to the P term. The actual pressure is lower than the ideal pressure because some of the particles form temporary associations so there are fewer idividual collisions as shown in the drawing:
The van der Waals equation is:
Try this: A sample consists of 8.00 kg of gaseous nitrogen and fills a 100-L flask at 300 oC. What pressure is predicted by the ideal gas equation? What pressure is predicted by the van der Waals equation? (for N2: a=1.390 atm L2/mol2, b = 0.03913 L/mol) hint: convert kg to mole and oC to K answer, ideal gas law: P = nRT/V = (286 mol)(0.0821 atmL/Kmol)(573 K) / (100 L) = 135 atm answer, van der Waals equation: P = 140 atm
OTHER TOPICS: 1) Dalton's law of partial pressure states that "the total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone". All this means is that when you measure the pressure of a mixture of gases, it doesn't matter what the gases are, only how many moles of particles there are. Even gaseous compounds with very different molecules will have virtually identical physical properties. Each type of gas will contribute a portion to the total pressure based on how many moles of that gas are present. This makes sense in terms of the Kinetic-Molecular theory since the particles are negligibly small and don't stick together. In terms of real gases, we would anticipate that Dalton's law may not be exactly correct since it really does matter what the gases are (for size and "stickiness" reasons).
You need to be able to do the following:
a. Given a mixture of gases, determine the partial pressures and the total pressure. This is done by applying the ideal gas law to each of the gases, to give each partial pressure. The total pressure is then just, Pt = P1 + P2 + P3 + P4 + ... b. Determine the mole fraction of a gas by relating it to the partial pressure of that gas: Mole fraction, X = n1/nt = P1/Pt Try this: A 1.00-L sample of dry air at 25 oC and 786 mmHg contains 0.925 g N2, plus other gases including oxygen, argon and carbondioxide. What is the partial pressure of N2 in the air sample? What is the mole fraction of N2 in the mixture? answer, partial pressure: P = (0.0330 mol)(0.0821 Latm/molK)(298 K) / (1.00 L) P = 0.808 atm = 614 mmHg answer, mole fraction: X = 614 mmHg / 786 mmHg = 0.781 (not there are no units for X)
a. Given a mixture of gases, determine the partial pressures and the total pressure. This is done by applying the ideal gas law to each of the gases, to give each partial pressure. The total pressure is then just, Pt = P1 + P2 + P3 + P4 + ...
b. Determine the mole fraction of a gas by relating it to the partial pressure of that gas:
Try this: A 1.00-L sample of dry air at 25 oC and 786 mmHg contains 0.925 g N2, plus other gases including oxygen, argon and carbondioxide. What is the partial pressure of N2 in the air sample? What is the mole fraction of N2 in the mixture? answer, partial pressure: P = (0.0330 mol)(0.0821 Latm/molK)(298 K) / (1.00 L) P = 0.808 atm = 614 mmHg answer, mole fraction: X = 614 mmHg / 786 mmHg = 0.781 (not there are no units for X)
2) Effusion and diffusion have application to the determination of unknown gases. The lighter (lower molar mass, M) gases travel at greater average speeds than larger gas particles. This makes sense since we have seen that the kinetic energy (mu2/2) is roughly constant at a given temperature regardless of the actual gas. Make sure you can define effusion and diffusion and use the Graham's law if it is provided on a test.
Try this: Calculate the ratio of effusion rates of CO2 and SO2, from the same container and at the same temperature and pressure. answer: [rate of effusion of CO2) / (rate of effusion of SO2) = [(M, SO2) / (M, CO2)]^1/2 ratio = 1.21 Because CO2 is smaller, it effuses 1.21 times faster than SO2.