Graham’s law of diffusion explains the rate of diffusion. The movement of molecules (gases) from their higher pressure to their lower pressure is called diffusion. For example, a drop of ink is added to the water glass, and a drop diffuses into the whole glass from higher pressure to lower pressure. Effusion (diffusion from a very tiny hole) is the movement of gas particles through a small hole. For example, when the tire gets punctured, the air inside the tire effuse through a tiny hole. Graham’s law of diffusion is also valid for effusion.

## Graham’s Law of Diffusion

According to this law, the rate of diffusion is inversely proportional to the under root of the molar mass of gas provided the pressure, temperature, and area remain constant.

Rate of diffusion (r) ∝ 1 / √molar mass

- The more the molar mass of the gas, the lesser the rate of diffusion.
- The higher the gas, the lesser the rate, or vice versa.

Among hydrogen and oxygen gas, which is more diffuse?

The molar mass of hydrogen and oxygen gas is 2 g and 32 g respectively. So H_{2} is five times more diffuse than O_{2} (provided P, T, and the area remains constant). The rate of diffusion depends upon the pressure, temperature, area, and molar mass.

r ∝ P A / √T M

Where P is pressure, A is the area, T is temperature, and M is the molar mass of the gas. But it is not Graham’s law because as we know that according to this law, temperature, pressure, and the area remains constant so, r ∝ 1 / √M

### Case 1

- P = Constant
- A = Constant
- T =Constant

Let us have two gases, gas 1 and gas 2. The rate of diffusion of both gases is given,

r_{1} ∝ 1 / √M_{1} ———- (1)

r_{2} ∝ 1 / √M_{2} ———- (2)

Combine eq (1) and (2)

**r _{1} / r_{2} = √ M_{2} / M_{1}** ———- (3)

Where,

- r
_{1}= rate of diffusion of gas 1 - r
_{2}= rate of diffusion of gas 2 - M
_{1}= molar mass of gas 1 - M
_{2}= molar mass of gas 2

Graham’s law is also explained by the density of the gas. As we know,

density of gas = d = molar mass / 2

Let us two gases having density d_{1} and d_{2} so,

d_{1} = M_{1} / 2

M_{1} = 2 d_{1}

d_{2} = M_{2} / 2

M_{2} = 2 d_{2}

Putting the values of M_{1} and M_{2} in eq (3)

r_{1} / r_{2} = √ 2d_{2} / 2d_{1}

**r _{1} / r_{2} = √ d_{2} / d_{1}**

So Graham’s law of diffusion also states that

“The rate of diffusion is inversely proportional to the square root of the density of gas”

#### Find the ratio of the rate of diffusion of helium and methane gas under similar pressure, temperature, and area.

r_{1} / r_{2} = √ M_{2} / M_{1}

r_{He} / r_{CH4} = √ M_{CH4} / M_{He}

The molar mass of He = 4

The molar mass of CH_{4} = 16

r_{He} / r_{CH4} = √ 16 / 4

r_{He} / r_{CH4} = 2 / 1

r_{He} : r_{CH4} = 2 : 1

#### Find the ratio of the rate of diffusion of He, CH_{4}, and SO_{2}.

r ∝ 1 / √M

r = k / √M

The molar mass of He, CH_{4}, and SO_{2} is 4, 16, and 64 respectively.

r_{He} : r_{CH4} : r_{SO2}= He : CH_{4} : SO_{2}

r_{He} : r_{CH4} : r_{SO2}= k / √M_{He} : k / √M_{CH4} : k / √M_{SO2}

r_{He} : r_{CH4} : r_{SO2}= k / √4 : k / √16 : k / √64

r_{He} : r_{CH4} : r_{SO2}= 1 / 2 : 1 / 4 : 1 / 8

Multiply by 8

r_{He} : r_{CH4} : r_{SO2}= 4 : 2 : 1

- He is lighter gas so the rate of diffusion is high.
- SO
_{2}is heavier gas so the diffusion rate is low.

#### The rate of diffusion of an unknown gas is 1.73 times smaller than the rate of diffusion of O_{2}. Find the molar mass of an unknown gas.

r_{1} / r_{2} = √ M_{2} / M_{1}

r_{x} / r_{O2} = √ M_{O2} / M_{x}

Let the rate of diffusion of O_{2} is 1.73

1 / 1.73 = √ 32 / M_{x}

1 / √3 = √ 32 / M_{x}

squaring on both sides

1 / 3 = 32 / M_{x}

M_{x} = 96 g

The mass of an unknown gas is 96 g.

##### Meaning of Rate of Diffusion

###### 1) In terms of the volume of gas:

“Volume of gas diffuse of effuse per unit time”

r = volume of gas diffused or effused / time taken

r_{1} / r_{2} = (v_{1}/t_{1}) / (v_{2}/t_{2}) = √ M_{2} / M_{1}

###### 2) In terms of distance travels:

r = distance traveled by the gas/time taken

r_{1} / r_{2} = (d_{1}/t_{1}) / (d_{2}/t_{2}) = √ M_{2} / M_{1}

###### 3) In terms of moles:

r = no of moles effuse or diffuse / time taken

r_{1} / r_{2} = (n_{1}/t_{1}) / (n_{2}/t_{2}) = √ M_{2} / M_{1}

#### 20 ml of He diffused in 5 sec. Find the volume of O_{2} diffused through the same hole in 10 sec.

r_{1} / r_{2} = (v_{1}/t_{1}) / (v_{2}/t_{2}) = √ M_{2} / M_{1}

r_{He} / r_{O2} = (v_{He}/t_{He}) / (v_{O2}/t_{O2}) = √ M_{O2} / M_{He}

r_{He} / r_{O2} = (20/5) / (v_{O2}/10) = √ 32 / 4

r_{He} / r_{O2} = 4×10 / v_{O2} = √8

v_{O2} = 40 /√8

v_{O2} = 14.14 ml

#### 24 g of H_{2} diffused in 5 sec. Find the mass of O_{2} diffused in 10 sec through the same hole.

No of moles of H_{2} = m/M = 24/2 = 12 g

No of moles of O_{2} = m/M = x/32

r_{1} / r_{2} = (n_{1}/t_{1}) / (n_{2}/t_{2}) = √ M_{2} / M_{1}

(12/5) / (x/32 × 10) = √ 32 / 2

(12/5) / (x/32 × 10) = 4

(12/5) / (x/320) = 4

12/5 × 320/x = 4

3840 / 5x = 4

x = 192 g

#### “x” ml of H_{2} effused through a hole in 5 sec. Find the time taken by the same volume of O_{2} to effuse through same hole.

r_{1} / r_{2} = (v_{1}/t_{1}) / (v_{2}/t_{2}) = √ M_{2} / M_{1}

r_{H2} / r_{O2} = (v_{H2}/t) / (v_{O2}/t_{2}) = √ M_{O2} / M_{H2}

(x/5) / (x/t) = √ 32 / 2

(x/5) × (t/x) = 4

t / 5 = 4

t = 20 sec

#### Let us a cylinder, we effuse NH_{3} from one end and effuse HCl from another end of the cylinder. When both gases combine dense white fumes (NH_{4}Cl) is formed. Where the dense white fumes is formed?

NH_{3} + HCl **→** NH_{4}Cl (dense white fumes)

As we know that, lighter the gas more the rate of diffusion, or vice versa.

r ∝ 1 / √M

The molar mass of NH_{3} and HCl is 17 g and 35.5 g so the rate of diffusion of NH_{3} is more so the dense white fumes formed near origin of HCl gas.

### Case 2

- P = not constant
- A = Constant
- T = Constant

As we know that

r ∝ P A / √T M (If A and T is constant)

r ∝ P / √ M

Graham’s law of diffusion when pressure is not constant is:

**r _{1} / r_{2} = P_{1}/P_{2} √ M_{2} / M_{1}**

#### 20 ml of He diffused through a hole in 10 sec at a pressure of 2 atm. Find the volume of CH_{4} diffused in same time at a pressure of 1 atm.

r_{1} / r_{2} = (v_{1}/t_{1}) / (v_{2}/t_{2}) = P_{1}/P_{2}** **√M_{2} / M_{1}

r_{He} / r_{CH4} = (v_{He}/t) / (v_{CH4}/t_{2}) = P_{1}/P_{2}** **√M_{CH4} / M_{O2}

(20/10) / (v_{CH4}/10) = 2/**1 **√16 / 4

20 / v_{CH4} = 2** **√4

20 / v_{CH4} = 2** **× 2

V = 20 / 4

V = 5 ml

#### Let us a cylinder containing helium gas and methane gas having number of moles 2 and 1 respectively. Find the ratio of the rate if diffusion intially.

r_{1} / r_{2} = P_{1}/P_{2} √ M_{2} / M_{1}

According to Dalton’s law, the more the number of moles, the more will be pressure of the gas.

P_{1} / P_{2} = n_{1} / n_{2}

r_{He} / r_{CH4} = n_{He}/n_{CH4} √ M_{CH4} / M_{He}

r_{He} / r_{CH4} = 2/1 √16 / 4

r_{He} / r_{CH4} = 2 × 2 / 1

r_{He} : r_{CH4} = 4 : 1