Humboldt State University ® Department of Chemistry

Richard A. Paselk
Chem 109

General Chemistry

Summer 2002
Lecture Notes:: 20 June

© R. Paselk 2002
  
     
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Gas Laws, cont.

Gas Stoichiometry

Last time we determined the stoichiometry in the example below:

Assume you have an engine with a 500 mL cylinder with a 10:1 compression ratio. If 0.025 L of octane (C8H18) vapor and 0.300 L of oxygen (both measured at 765 mmHg and 25 °C) are introduced into the cylinder, what will be the pressure "at the top of the stroke" if ignition gives a temperature of 557 °C?

We can begin by determining the stoichiometry and thus the amounts of reactants and products left after reaction. We will come back tommorow to use the gas law to find pressure.

First we need to write a balanced equation, determine the stoichiometry, and the before and after situation.

The easiest way to do this is to assume no temperature or pressure change during the reaction - we'll take care of that later:

Note the volume change in this reaction: start with 0.325 L end up with 0.409 L
Equation:  C8H18 + O2 Æ CO2 + H2O
Balancing:  2 C8H18 + 25 O2 Æ 16 CO2 + 18 H2O
Stoichiometry (n or V): 2 : 25 : 16 : 18
Before reaction:  0.025 L   0.300 L   0   0

From the stoichiometry can see that oxygen is limiting - some octane will be left over. (vol octane required to react with 0.300 L oxygen = {2/25}{0.300 L} = 0.024 L)
After reaction: 0.001 L   0 L   0.192 L   0.216 L

Now lets find the pressure as requested:

Use Gas Laws to solve: PV = nRT,

putting constants together, PV/nT = R, or P1V1/n1T1 = P2V2/n2T2

Rearranging: P2 = (P1)(V1/V2)(n2/n1)(T2/T1)

    • P1 = 765 mmHg
    • V1 = 0.500 L
    • V2 = 0.050 L
    • "n1" = 0.325 L
    • "n2" = 0.409 L
    • T1 = 25° + 273 ° = 298 K
    • T2 = 557 ° + 273 ° = 830 K

Substituting: P2 = (765 mmHg)(0.500 L/0.050 L)(0.409/0.325)(830 K/298 K)

P2 = 2.68 x 104 mmHg = 35.3 atm

 

Graham's Laws of Effusion and Diffusion

Effusion refers to the passage of a substance through a small orifice.

Graham's Law of Effusion states that the effusion of a gas through a small orifice is inversely proportional to its density.

or, since the density of a gas is proportional to its molecular weight

Equivalently, the relative rates of effusion of two gases at the same pressure and temperature is given by the inverse square roots of their densities.

Example: What is the relative rate of effusion of H2 vs. O2?

RateH2/RateO2 = (32/2)1/2 = 161/2 = 4

Diffusion refers to the passage of one substance through another. An example for gases would be the passage of an aroma, such as a perfume or skunk smell, through still air. Given that gases are mostly empty space this interpenetration is not surprising. What we want to look at now is the rate of this process:

Graham's Law of Diffusion states that "The rate of diffusion of a gas is inversely proportional to the square root of its density."

or, since the density of a gas is proportional to its molecular weight

This turns out to be not quite the case. That is, the ratios of rates of diffusion of different gases will not quite fit prediction. The problem is that, although the average velocities of the molecules follow the inverse proportionality, as in effusion, the molecules are impeded by collisions with the gas they are passing through. Not surprisingly, the description of this more complex process is not quite the simple law originally postulated by Graham. It does still give a useful first order picture however.

 

Kinetic Molecular Theory of Gases

We have been looking at the various properties of gases, now we want to look at a theory to explain those behaviors. A simple model is the kinetic-molecular theory. There are four basic postulates:

  1. A gas is composed of a large number of tiny particles (molecules, or atoms for the inert gases). These particles are so small that the sum of the particles volumes is negligible compared to the volume of their container - most of the container volume is empty space.
  2. The particles of a gas are in rapid, linear motion. They make frequent collisions with each other and the walls of any vessel containing them. All collisions between gas particles and between gas particles and container walls are elastic. (There is no net loss of kinetic energy in collision - energy can be exchanged between particles, but the total stays the same.)
  3. Except when they are colliding, the particles are completely independent of each other. That is, there are no forces of attraction or of repulsion between them.
  4. The particles in a gas have a wide range of velocities: some may be nearly still, while others move at great speed. Thus there is a wide range of kinetic energies in any gas. However, the average kinetic energy for any gas is the same at a given temperature. The average kinetic energy for the particles in a gas is proportional to the absolute temperature of the gas. (KE = 3/2 RT, R is still the Gas constant, but different units.)

Consequences/predictions:

  1. Gases are easy to compress - expected if there is lots of empty space between them.
  2. This explains why gases rapidly fill their containers. We also note that they don't condense out as a liquid or solid if they are left in an insulated container (they don't lose energy as they collide with walls.) Brownian motion is also a consequence of their rapid movement.
  3. Three is a bit more subtle, and we won't worry about it.
  4. From this postulate we expect a distribution of velocities. (Overhead 42, Zumdahl figure 5.20, p 219)

Note that for kinetic energy, KE = 1/2 mV2, so V varies as the square root of the mass (m1/2). Notice also that the energy increases with the square of the velocity. (This is why an accident at 60 mph (88 ft/s) is much worse that one at 30 mph - four times as much energy is involved!)

 

Real Gases

van der Waals Equation: Real gas data, such as the fact that many real gases cool upon expansion, or the molar volumes tabulated in Table 5.2 (p 202 of Zumdahl, they do not all occupy 22.42 L) is not always as PV = nRT would predict (Zumdahl overheads 44 & 45, figures 5.24, & 5.25, pp 222). Thus the question arises: How can we model the behavior of real gases? The best known equation for real gases is the van der Waals equation. This equation attempts to model real gases by including two obvious differences between real and ideal gases:

To account for attraction van der Waals made the simple assumption that in order to attract, the particles must come close, and in fact collide. The pressure will then be reduced by a constant describing the degree of attraction (specific to each kind of gas) multiplied by the probability of collision which is known to be = (n/V)2 (statistically calculated probability of occupying the same place at the same time). Thus:

Pobs = P - a(n/V)2 is the corrected pressure

To account for the actual volume available to a real gas particle we simply have to subtract the volume of the particles in the container, n (vol/mol). If we know the molar volume = b we get:

Vavail = V - bn as the corrected volume

Substituting these corrected values into PV = nRT gives the van der Waals Equation:

This gives a better approximation than PV = nRT for most real gases, though it is still just an approximation. The constant a & b are determined empirically. They are sometimes taken as giving an indication of attraction (a) and particle volume (b).

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