Grahams Law

Introduction

Graham's law or also called Graham's law of diffusion was given or established by Thomas Graham (Scottish chemist) in the year 1848. Graham’s law (for both effusion and diffusion) describes that the rate (amount) of (e) effusion or diffusion of any particular gas (g) is inversely (inv.) proportional ($\mathrm{\propto}$) to the square (sq.) root of its molecular weight (m.w.). Graham’s law is useful for separating (distinction) isotopes by diffusion, which is further very useful in the production of atomic (atom) bombs.

This law is more accurate (suitable) for molecular type diffusion which further involves the motion of one gas (g) at a given time (t) through a small hole. And, if the conditions like temperature (t) and pressure (p) are the same (similar), then the molar mass (M) is proportional ($\mathrm{\propto}$) to the mass density. Examples- spraying perfumes, room fresheners, burning of incense sticks, e.t.c.

State Grahams Law

Graham’s law (for both effusion and diffusion) states (describes) that the rate (amount) of (e) effusion or diffusion of any particular gas (g) is inversely (inv) proportional ($\mathrm{\propto}$) to the square (sq.) root of its molecular weight (m.w.). This law also states that the ratio (division) of the diffusion rate (amount) of two (2) gases is the same (similar) as the ratio (division) of the square (sq.) root (√) of the molar mass or weight (m.w.) of the two gases. Graham’s law is a direct (straight) outcome of the fact that the gaseous (g) molecules present at the same (similar) temperature (t) have the same (equal) average (avg.) kinetic energy (K.E.). The formula representation for this law is − $\mathrm{\frac{Rate\: 1}{Rate\: 2}=\sqrt{\frac{M2}{M1}}}$, where Rate 1 and, Rate 2 are the rate (amount) of diffusions for the first (1st) and second (2nd) gases respectively.

Astrang13, Effusion, CC BY-SA 3.0

Write the mathematical form of Graham’s law of diffusion.

As we know, according to graham’s law, at constant pressure (p) and temperature (t), atoms or molecules with lower molecular mass (m.w) will effuse faster as compared to the molecules or atoms with higher molecular mass. This law also states (describes) that the rate (amount) of (e) effusion of any gas is inversely (inv.) proportional ($\mathrm{\propto}$) to the square (sq.) root (√) of its molar (m) or, molecular mass (m.w), i.e.$\mathrm{r\propto 1/(M)^{1/2}}$. The formula generated here is used to compare the rates of two different gases (g) present at equal pressures (p) and temperature (t). And, the formula involved is − $\mathrm{\frac{Rate\: 1}{Rate\: 2}=\sqrt{\frac{M2}{M1}}}$, where Rate 1 represents the rate (amount) of effusion (e) of the 1st gas, Rate 2 denotes the amount of effusion (e) for the 2nd gas, M1 denotes the (m) molar mass (m.w.) of gas1 and M2 denotes the (m) molar mass (m.w.)of gas2.

Give an example of this law

Some of the real-life examples of Graham’s law are as follows −

• Whenever a room or air fresheners or perfumes are applied in any corner of the room or sprayed, their fragrance (smell) covers the entire room. This is due to the diffusion phenomenon.

• This law is also very useful in calculating the rate (amount) of effusion or diffusion for a particular gas.

• Burning or lightening of incense stick is also a very familiar application of Graham’s law.

• This law is useful in comparing the nature or behaviour of the various or different gases.

• This law is also used for the separation process for separating different (distinct) isotopes of an element.

Rate of Effusion

Effusion is defined (described) as the process or method in which a gas (g) escapes (comes out) from the container (closed surface) with help of a hole (small pinhole). The rate of effusion for any gaseous substance (g) is inversely (inv.) proportional ($\mathrm{\propto}$) to the square (sq.) root (√) of its (m) molar or molecular mass. The rate at which a molecule (single) or a mole (group) of molecules effuses is directly dependent or related to the speed (s) at which it is moving that time. The formula representation for the same is −

$$\mathrm{\frac{rate\: of\: effusion\: A}{rate\: of\: effusion\: B}=\sqrt{\frac{M_B}{M_A}}.}$$

Give some solved numericals

Some solved examples or numerical for this law are −

Q1.If the same (similar) amounts (quantity) of helium (He) and argon (Ar) gas is placed (kept) in a container (porous) and then permitted to come out, then find which of the gas will escape (come out) faster?

Sol:- Let the rate of diffusion (r1) for He gas be x and, the rate of effusion (r2) for the Ar gas be 1. Now, the atomic (molar) weight (m.w.) of He = 4.00 and, the atomic (molar) weight (m.w.) of Ar = 39.95.

Now, according to Graham’s law, $\mathrm{\frac{r1}{r2}=\frac{M2}{M1}}$. Now, substituting the values, we will get −

$$\mathrm{\frac{x}{1}=\frac{39.95}{4.00}\Rightarrow x=3.16.}$$

Therefore, Helium gas will escape faster than argon.

Q2. If 32 ml of H₂ gas diffuses through a small fine hole in 1 minute, then find the volume of CO₂ gas that will diffuse in 1 minute under the same condition?

According to Graham's law −

$$\mathrm{\frac{V1}{V2}= \sqrt{\frac{M_B}{M_A}}}$$

Here, V1= 32 ml, M_(CO2)=44, and M_(H2)= 2.

$$\mathrm{\Rightarrow \frac{32}{V2}= \sqrt{\frac{44}{2}} = 6.82 ml.}$$

Conclusion

Graham’s law of both effusion and diffusion states (describes) that the rate (amount) of (e) effusion or diffusion (d) of any particular gas (g) is inversely (inv) proportional ($\mathrm{\propto}$) to the square (sq.) root (√) of its molecular weight (m.w.). This law also states that the ratio (division) of the diffusion rate (amount) of two (2) gases is the same (similar) as the ratio (division) of the square (sq.) root (√) of the molar mass or weight (m.w.) of the two gases. Graham’s law is a direct (straight) outcome of the fact that the gaseous (g) molecules present at the same (equal) temperature (t) have a similar average (avg.) kinetic energy (K.E.). The formula representation for this law is − $\mathrm{\frac{Rate\: 1}{Rate\: 2}=\sqrt{\frac{M2}{M1}}}$.

FAQs

1. What do you mean by Graham’s law?

Graham’s law (effusion or diffusion) states (describes) that the rate (amount) of effusion or diffusion of any particular gas (g) is inversely (inv) proportional ($\mathrm{\propto}$) to the square (sq.) root (√) of its molecular weight (m.w.).

2. Write the formula for finding Graham’s law.

The formula representation for Graham’s law is − $\mathrm{\frac{Rate\: 1}{Rate\: 2}=\sqrt{\frac{M2}{M1}}}$.

3. Which two parameters are constant in Graham’s law?

Temperature (t) and pressure (p) are constant parameters in graham’s law.

4. Define effusion.

Effusion is described as the process or method in which a gas (g) escapes (comes out) from the container (closed surface) through a hole (small pinhole).

5. What is the colour of fumes generated while experimenting with graham’s law?

Dense white fumes are produced.

6. What is a real life example of Graham's law?

One can easily observe the application of Graham's law in real life by lighting an incense stick and observing the behaviour of the smoke molecules generated by it. This is because the smoke coming out of the burning incense stick spreads in the surroundings quickly with the help of the diffusion process.