Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2013

Lecture Notes 16: 27 February

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Gas Stoichiometry

Last time we left off having figured out the gas volumes in our engine problem, assuming no pressure or temperature chenges in one cylinder (2.5/4 = 0.625):

Note the volume change in this reaction: start with 0.625 L end up with 0.7125 L

Equation:	C8H18	+	O2		CO2	+	H2O
Balancing:	2 C8H18	+	25 O2		16 CO2	+	18 H2O
Stoichiometry (n or V):	2	:	25	:	16	:	18
Before reaction:	0.025 L		0.600 L		0		0
From the stoichiometry can see that octane is limiting - some octane will be left over. (vol octane required to react with 0.600 L oxygen = {2/25}{0.600 L} = 0.048 L); O2 = 0.600 - (25/2)(0.025) = 0.600-0.3125 = 0.288 L, CO2 = (16/2)(0.025) = 0.200 L, H2O = (18/25)(0.025) = 0.225 L
After reaction:	0 L		0.288 L		0.200 L		0.225 L

Now lets find the pressure as requested:

• P₁ = 765 mmHg
• V₁ = 0.625 L
  V₂ = 0.0625 L (1/10 of 2.5L)
• "n₁" = 0.625 L
  "n₂" = 0.7125 L
• T₁ = 25° + 273 ° = 298 K
  T₂ = 557 ° + 273 ° = 830 K
Substituting: P₂ = (765 mmHg)(0.625 L/0.0625 L)(0.7125/0.625)(830 K/298 K) = 2.429 x 10⁴

P₂ = 2.43 x 10⁴ mmHg = 32.0 atm (or 470 psi)

 

Graham's Laws of Effusion and Diffusion

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

Graham's Law of Effusion

This law states that the effusion of a gas through a small orifice is inversely proportional to the square root of its density.



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

(n = mass/MW; from PV = nRT, n/V = const. = (mass/MW)/V; multiplying both sides by MW gives (MW)(const.) = mass/V = density.)



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?

Rate_H2/Rate_O2 = (32/2)^1/2 = 16^1/2 = 4

What is the MW of a molecule that effuses 6.5 times slower than nitrogen?

Rate_N2/Rate_unk = 6.5 = (MW_unk)^1/2/(MW_N2)^1/2 = (MW_unk)^1/2/(28)^1/2

(MW_unk)^1/2 = (6.5)(28)^1/2, square both sides,

MW_unk = (42.2)(28) = 1183g/mol = 1.2 x 10³g/mol

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

This law 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



Unlike in effusion, this turns out to be not quite the case for diffusion. 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 particle's 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.)
   

   

   public domain image via Wikipedia Creative Commons‡

   
   
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 205)



public domain image via Wikipedia Creative Commons‡

Note that for kinetic energy, KE = 1/2 mV², so V varies as the square root of the mass (m^1/2). 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!)

 

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Last modified 27 February 2013