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 H2 is five times more diffuse than O2 (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,
r1 ∝ 1 / √M1 ———- (1)
r2 ∝ 1 / √M2 ———- (2)
Combine eq (1) and (2)
r1 / r2 = √ M2 / M1 ———- (3)
Where,
- r1 = rate of diffusion of gas 1
- r2 = rate of diffusion of gas 2
- M1 = molar mass of gas 1
- M2 = 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 d1 and d2 so,
d1 = M1 / 2
M1 = 2 d1
d2 = M2 / 2
M2 = 2 d2
Putting the values of M1 and M2 in eq (3)
r1 / r2 = √ 2d2 / 2d1
r1 / r2 = √ d2 / d1
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.
r1 / r2 = √ M2 / M1
rHe / rCH4 = √ MCH4 / MHe
The molar mass of He = 4
The molar mass of CH4 = 16
rHe / rCH4 = √ 16 / 4
rHe / rCH4 = 2 / 1
rHe : rCH4 = 2 : 1
Find the ratio of the rate of diffusion of He, CH4, and SO2.
r ∝ 1 / √M
r = k / √M
The molar mass of He, CH4, and SO2 is 4, 16, and 64 respectively.
rHe : rCH4 : rSO2= He : CH4 : SO2
rHe : rCH4 : rSO2= k / √MHe : k / √MCH4 : k / √MSO2
rHe : rCH4 : rSO2= k / √4 : k / √16 : k / √64
rHe : rCH4 : rSO2= 1 / 2 : 1 / 4 : 1 / 8
Multiply by 8
rHe : rCH4 : rSO2= 4 : 2 : 1
- He is lighter gas so the rate of diffusion is high.
- SO2 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 O2. Find the molar mass of an unknown gas.
r1 / r2 = √ M2 / M1
rx / rO2 = √ MO2 / Mx
Let the rate of diffusion of O2 is 1.73
1 / 1.73 = √ 32 / Mx
1 / √3 = √ 32 / Mx
squaring on both sides
1 / 3 = 32 / Mx
Mx = 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
r1 / r2 = (v1/t1) / (v2/t2) = √ M2 / M1
2) In terms of distance travels:
r = distance traveled by the gas/time taken
r1 / r2 = (d1/t1) / (d2/t2) = √ M2 / M1
3) In terms of moles:
r = no of moles effuse or diffuse / time taken
r1 / r2 = (n1/t1) / (n2/t2) = √ M2 / M1
20 ml of He diffused in 5 sec. Find the volume of O2 diffused through the same hole in 10 sec.
r1 / r2 = (v1/t1) / (v2/t2) = √ M2 / M1
rHe / rO2 = (vHe/tHe) / (vO2/tO2) = √ MO2 / MHe
rHe / rO2 = (20/5) / (vO2/10) = √ 32 / 4
rHe / rO2 = 4×10 / vO2 = √8
vO2 = 40 /√8
vO2 = 14.14 ml
24 g of H2 diffused in 5 sec. Find the mass of O2 diffused in 10 sec through the same hole.
No of moles of H2 = m/M = 24/2 = 12 g
No of moles of O2 = m/M = x/32
r1 / r2 = (n1/t1) / (n2/t2) = √ M2 / M1
(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 H2 effused through a hole in 5 sec. Find the time taken by the same volume of O2 to effuse through same hole.
r1 / r2 = (v1/t1) / (v2/t2) = √ M2 / M1
rH2 / rO2 = (vH2/t) / (vO2/t2) = √ MO2 / MH2
(x/5) / (x/t) = √ 32 / 2
(x/5) × (t/x) = 4
t / 5 = 4
t = 20 sec
Let us a cylinder, we effuse NH3 from one end and effuse HCl from another end of the cylinder. When both gases combine dense white fumes (NH4Cl) is formed. Where the dense white fumes is formed?
NH3 + HCl → NH4Cl (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 NH3 and HCl is 17 g and 35.5 g so the rate of diffusion of NH3 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:
r1 / r2 = P1/P2 √ M2 / M1
20 ml of He diffused through a hole in 10 sec at a pressure of 2 atm. Find the volume of CH4 diffused in same time at a pressure of 1 atm.
r1 / r2 = (v1/t1) / (v2/t2) = P1/P2 √M2 / M1
rHe / rCH4 = (vHe/t) / (vCH4/t2) = P1/P2 √MCH4 / MO2
(20/10) / (vCH4/10) = 2/1 √16 / 4
20 / vCH4 = 2 √4
20 / vCH4 = 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.
r1 / r2 = P1/P2 √ M2 / M1
According to Dalton’s law, the more the number of moles, the more will be pressure of the gas.
P1 / P2 = n1 / n2
rHe / rCH4 = nHe/nCH4 √ MCH4 / MHe
rHe / rCH4 = 2/1 √16 / 4
rHe / rCH4 = 2 × 2 / 1
rHe : rCH4 = 4 : 1
