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Graham’s Law of Diffusion
Graham’s Law of Diffusion/Effusion:
Gas molecules are in a state of constant motion. So, they intermix with
each other to form a homogenous mixture. Diffusion is a process by
virtue of which two or more gases intermix with each other,
independent of gravitation to form a homogenous mixture.
Example: If hydrogen sulphide (H2S) is released at one corner of room,
its smell can be detected at another corner after some time.
Again the term effusion which is a process in which a gas under
pressure escape out a fine hole e.g. leaking out of a gas from cylinder
or air from punctured tyre etc. however the process of effusion is
always followed by the process of effusion.
Statement of Graham’s Law of Diffusion:
It states that, “Under similar conditions of temperature
and pressure, the rates of diffusion of gases are
inversely proportional to the square root of their
densities.”
Mathematically it may be represented as
r ∝ 1
d
……………………………..(i)
Where ,
r = rate of diffusion
d= density of the gas
Let r1 and r2 be the rates of diffusion of two gases of densities d1 and d2
respectively,
Then according to Graham’s law of diffusion,
r1=K
1
d1
…………(ii)
r2=K
1
d2
…………(iii)
Combining (ii) and (iii)
𝑟1
𝑟2
=
d2
d1
……….(iv)
Equation (iv) is the mathematical expression of Graham’s Law of diffusion. We
know that molecular weight is the twice the vapour density
Molecular weight = 2 × vapour density.
therefore, M = 2d
or, d= M/2…………..(v)
Using relation (v) the Graham’s law equation (iv), becomes
𝑟1
𝑟2
=
M2/2
M1/2
=
M2
M1
………………….(iv)
Equation (vi) is the another form of Graham’s law of diffusion.
Again, rate of Diffusion is defined as the
volume of gas diffused per unit time.
i.e. rate of diffusion =
volume of the gas diffused
time of diffusion
r=
V
t
Let t1 and t2 be the time taken to diffuse V ml of two different gases,
then
r1
r2
=
V/t1
V/t2
=
M2
M1
…………………….(vii)
From above equation(vii) we can say that the time taken for diffusion of
equal volumes of gases is directly proportional to square roots of their
molecular weights or densities.
Importance of Graham’s law of diffusion:
1. It helps in the separation of gases having different densities.
2. It helps in the separation of isotopes of certain elements
3. It helps to determine the density or molecular mass of an unknown
gas by comparing its rate of diffusion with a known gas.
CO2 is heavier than O2 and N2 gases present in the
air but it does not form the lower layer of the
atmosphere. Why?
Gases possess the property of diffusion which is independent of the
force of gravitation. Due to diffusion, the gases mix into each other and
remain almost uniformly distributed in the atmosphere.

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Graham's law of diffusion

  • 1. Graham’s Law of Diffusion
  • 2. Graham’s Law of Diffusion/Effusion: Gas molecules are in a state of constant motion. So, they intermix with each other to form a homogenous mixture. Diffusion is a process by virtue of which two or more gases intermix with each other, independent of gravitation to form a homogenous mixture. Example: If hydrogen sulphide (H2S) is released at one corner of room, its smell can be detected at another corner after some time. Again the term effusion which is a process in which a gas under pressure escape out a fine hole e.g. leaking out of a gas from cylinder or air from punctured tyre etc. however the process of effusion is always followed by the process of effusion.
  • 3. Statement of Graham’s Law of Diffusion: It states that, “Under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square root of their densities.”
  • 4. Mathematically it may be represented as r ∝ 1 d ……………………………..(i) Where , r = rate of diffusion d= density of the gas Let r1 and r2 be the rates of diffusion of two gases of densities d1 and d2 respectively,
  • 5. Then according to Graham’s law of diffusion, r1=K 1 d1 …………(ii) r2=K 1 d2 …………(iii)
  • 6. Combining (ii) and (iii) 𝑟1 𝑟2 = d2 d1 ……….(iv) Equation (iv) is the mathematical expression of Graham’s Law of diffusion. We know that molecular weight is the twice the vapour density
  • 7. Molecular weight = 2 × vapour density. therefore, M = 2d or, d= M/2…………..(v) Using relation (v) the Graham’s law equation (iv), becomes 𝑟1 𝑟2 = M2/2 M1/2 = M2 M1 ………………….(iv) Equation (vi) is the another form of Graham’s law of diffusion.
  • 8. Again, rate of Diffusion is defined as the volume of gas diffused per unit time. i.e. rate of diffusion = volume of the gas diffused time of diffusion r= V t Let t1 and t2 be the time taken to diffuse V ml of two different gases, then r1 r2 = V/t1 V/t2 = M2 M1 …………………….(vii)
  • 9. From above equation(vii) we can say that the time taken for diffusion of equal volumes of gases is directly proportional to square roots of their molecular weights or densities.
  • 10. Importance of Graham’s law of diffusion: 1. It helps in the separation of gases having different densities. 2. It helps in the separation of isotopes of certain elements 3. It helps to determine the density or molecular mass of an unknown gas by comparing its rate of diffusion with a known gas.
  • 11. CO2 is heavier than O2 and N2 gases present in the air but it does not form the lower layer of the atmosphere. Why? Gases possess the property of diffusion which is independent of the force of gravitation. Due to diffusion, the gases mix into each other and remain almost uniformly distributed in the atmosphere.