Lesson 5.7 Effusion & Diffusion
Suggested Reading
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Zumdahl Chapter 5 Sections 5.6 & 5.7
Essential Questions
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What is effusion and diffusion in gases?
Learning Objectives
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Perform calculations involving root mean square velocity.
Define and differentiate effusion and diffusion.
Determine the molar mass of a gas from rates of effusion.
Introduction
In this lesson you will be learning about Graham's law of effusion. Graham's
Law shows the relationship between the molar mass of a gas and the rate at
which it will effuse. Effusion is the process of gas molecules escaping
through tiny holes in their container. Diffusion, such as perfume diffusing
through a room, can also be considered with Graham's Law.
Watch this You Tube video:
https://www.youtube.com/watch?v=0uBK7VxT00E
Let us first consider why gases effuse. Containers can have small holes or
pores in them. Although these openings are microscopic, they are larger than
the gas molecules. Randomly, the gas molecules move around the inside of
the container until they impact something. This can be another molecule or
the side of the container. A gas can also, instead of hitting the side of the
container, pass through one of those openings by chance. This is effusion: a
random movement of a gas molecule through the container's wall. A common
example of this is a balloon filled with helium: first it is buoyant and floats in
the air, but in a few days it hangs toward the ground or floats a few inches
above the ground (if at all). The Helium has escaped through the small holes
in the balloon.
With Graham's Law, you can find the effusion rates for two gases or the molar
mass of a gas. The gas with the lesser molar mass has a greater rate of
effusion. Graham's law is derived from the KMT description of molecular
speed, so before we can discuss Graham's law we must first learn something
about molecular speed.
Molecular Speeds
Recall from KMT that gas molecules are in constant motion (assumption 2).
Now we will look at the speed of molecules and at some conclusions of KMT
regarding molecular speeds. According to KMT, the speeds of molecules in a
gas vary over a range of values. The British physicist James Maxwell showed
theoretically, and it has since been demonstrated experimentally, that
molecular speeds are distributed as show in the figure. This distribution of
speeds depends on the temperature. At any temperature, molecular speeds
vary widely, but most are close to the average speed. As the temperature
increases, the average speed increases.
The root mean square velocity, urms, is a type of average molecular speed,
given by the equation below (See page 204 of your textbook for the
derivation of this equation).
This equation shows that there is a direct relationship between the speed of a
molecule and the temperature and an inverse relationship between the speed
of a gas and its molar mass. Thus, when comparing two gases, the gas with
the higher molar mass with have the lower rms speed. Values of rms speed
calculated from this formula show that molecular speeds are astonishingly
high. For example, the rms speed of H2 molecules at 20∘C is 1.90 x 103 m/s
(over 4000 mi/hr!).
When using this equation, be careful to use consistent units. If SI units are
used then R (= 8.31 J / K ∙ mol), T (K), M (kg/mol) and the rms speed will
be in m/s. Recall that the 1 J = 1 kg ∙ m2/s2. Note that in these units, H2,
whose molecular weight is 2.02 amu, has a molar mass of 2.02 x 10-3
kg/mol.
Example: Calculating the rms Speed of a Gas Molecules
Calculate the rms speed of O2 molecules in a tank at 21∘C and 15.7 atm.
Solution:
The rms molecular speed is independent of pressure but does depend on
the absolute temperature, which is (21 + 273) K = 294 K. To calculate u, it
is best to use SI units throughout. In these units, the molar mass of O2 is
32.0 x 10-3 kg/mol, and R = 8.31 J / K ∙ mol. Hence,
Effusion
Recall that effusion is the process in which a gas flows through a small
hope in a container. Thomas Graham was the first to study effusion and he
discovered in 1946 that the rate of effusion of a gas is inversely
proportional to the square root of the molecular weight of the gas at
constant temperature and pressure. Stated another way, the relative rates
of effusion of two gases at the same temperature and pressure are given
by the inverse ratio of the square roots of the molar masses of the gases.
This is called Graham's law of effusion and is expressed mathematically as
This equation is derived from the equation for rms speed, and the
derivation can be found on page 207 of your textbook. Thus, the KMT
agrees with the experimental results for effusion.
Diffusion
When you bake cookies, you can usually smell the wonderful aroma
throughout the house. The spread of an odor is easily explained with KMT. All
molecules are in constant, random motion, and eventually a cluster of
molecules of a particular substance will spread out to occupy a larger and
larger space. Gaseous diffusion is the process whereby a gas spreads out
through another gas to occupy the space uniformly.
When you think about the KMT calculations of rms you might wonder why it
takes a gas minutes rather than seconds to diffuse. This was, in fact, one of
the first major criticisms of KMT. The answer is simply that a molecule never
travels very far in one direction before it collides with another molecule and
moves off in another direction. This results in a zig-zagging path that requires
the molecule to travel much further than the straight-line distance.
Although the rate of diffusion certainly depends in part on the rms speed, the
effect of molecular collisions makes the theoretical picture somewhat
complicated. However, Graham's law of effusion can be used to approximate
rates of diffusion. The following video discusses Graham's law of effusion and
diffusion as though they were synonymous. Although this is common, this in
not strictly correct. Graham's law describes rates of effusion, and is used
approximate rates of diffusion.
Watch this You Tube video:
https://www.youtube.com/watch?v=FCVjyS4bQwg
HOMEWORK: Refer to the power point notes examples for effusion
examples. Practice exercise 10.15