Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of...
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Transcript of Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of...
![Page 1: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/1.jpg)
Graham’s LawRate of Diffusion and Effusion
![Page 2: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/2.jpg)
Introduction
When we first open a container of ammonia, it takes time for the odor to travel from the container to all parts of a room.
This shows the motion of gases through other gases.
In this case, ammonia gas, NH3, moves through air.
This is an example of diffusion and effusion.
![Page 3: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/3.jpg)
Introduction
Diffusion is the tendency of a gas to move toward areas of lower density.
Ammonia moving throughout a room.
Effusion is the escape of a gas from a container from a small hole.
Air escaping from a car tire.
![Page 4: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/4.jpg)
Introduction
In 1831, the Scottish physical chemist, Thomas Graham, first showed the relationship between the mass of a gas molecule and its rate of diffusion or effusion.
This is called Graham’s Law.
“The rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass.”
![Page 5: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/5.jpg)
Introduction
The law comes from the relationship between the speed, mass, and kinetic energy of a gas molecule.
At a given temperature, the average kinetic energy of all gas molecules in a mixture is the same value.
If gas A has KEA = ½mAvA2
If gas B has KEB = ½mBvB2
Then KEA = KEB ➙ ½mAvA2 = ½mBvB
2
![Page 6: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/6.jpg)
Introduction
½mAvA2 = ½mBvB
2
mAvA2 = mBvB
2
vA2 mB
vB2 mA
=
vA2 mB
vB2 mA
=
vA mB
vB mA
=
The speed of an individual gas molecule is inversely proportional to its mass.
The ½’s cancel out.
Get all speed and mass terms together.Take the square root of both sides.Simplify.
![Page 7: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/7.jpg)
Introduction
If we extend this to all of the gas,
vA mB
vB mA
=
the speed becomes the rate
rateA mB
rateB mA
=
the mass becomes the molar mass
rateA MB
rateB MA
=
“The rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass.”
➙ ➙
Which leads us back to Graham’s Law:
![Page 8: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/8.jpg)
Application
This is how we apply Graham’s law.
We compare the rates of effusion of different gases.
rateA MB
rateB MA
=
![Page 9: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/9.jpg)
Example 1
Compare the rate of effusion of hydrogen gas to the rate of effusion of oxygen gas at a constant temperature.MH2 = 2.00 g/mol
rateH2
MO2rateO2
MH2
=
MO2 = 32.00 g/mol
32.00 g/mol 2.00 g/mol
= = 16.00= 4.00
Hydrogen gas effuses at a rate 4 times faster than oxygen.
![Page 10: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/10.jpg)
Example 2
A sample of helium, He, effuses through a porous container 6.04 times faster than does unknown gas A. What is the molar mass of the unknown gas?MHe = 4.00 g/mol
rateHe
MArateA
MHe
=
MA = A g/mol
rateHe = 6.04
rateA = 1.00
➙ (rateHe)2 MA
(rateA)2
MHe = ➙ (MHe)(rateHe
)2
(rateA)2
MA =
(4.00 g/mol)(6.04 )2 (1.00)2
MA = ➙ = 146 g/mol
![Page 11: Graham’s Law Rate of Diffusion and Effusion. Introduction When we first open a container of ammonia, it takes time for the odor to travel from the container.](https://reader036.fdocuments.in/reader036/viewer/2022062422/56649ebb5503460f94bc42f3/html5/thumbnails/11.jpg)
Summary
Diffusion is the tendency of a gas to move toward areas of lower density.
Effusion is the escape of a gas from a container from a small hole.
Graham’s Law: the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass.rateA
MB
rateB MA
=