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Graham’s Law of Diffusion

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  • Last Updated : 29 Jul, 2022
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Graham’s law of diffusion is the relationship between a gas’s rate of diffusion or effusion and its molecular weight. The law of diffusion’s basic tenet is that any gas’s rate of diffusion, at any given temperature and pressure, is inversely proportional to the square root of its density. The mechanism by which a gas can escape from the container is known as effusion, and the ability of a gas to spread and occupy all of the volumes that are available to it is known as diffusion.

Graham’s Law of Diffusion

In the year 1848, a scientist by the name of Thomas Graham made the initial discovery of Graham’s Law. During his experiments with the effusion of gas, he developed an essential theory that the lighter gas molecules will move through the air more quickly than the heavier gas molecules. Graham’s law of effusion is the common name for this law.

This law states that under the constant temperature and pressure of a gas, lighter atoms and molecules will diffuse through the air more quickly than heavier atoms and molecules. 

According to this law, the rate at which a gas molecule effuses is inversely related to the square root of the gas’s density or molecular mass.

Graham’s Law of Diffusion Formula

Graham’s Law of Diffusion Formula can be expressed as

Rate 1/Rate 2 = \frac{√d2}{√d1}

Rate 1/Rate 2 = \frac{√M2}{√M1}

Where,

  • Rate 1 = Rate of effusion of the first gas,
  • Rate 2 = Rate of effusion of the Second gas,
  • d1 = First gas density,
  • d2 = second gas density,
  • M1 = Molar mass of first gas,
  • M2 = Molar mass of second gas.

Rate of Diffusion

Diffusion of a gas is the term used to describe the net movement of the substance from the area of higher concentration to the area of lower concentration. Each gas particle starts crashing into one another. The maximum particle density zone of a gas causes the particles to begin bouncing off of each other and the border container at a faster pace than the particles in the lower particle density regions.

The rate of gas diffusion, according to effusion, is inversely proportional to the square root of the density of a gas molecule. A gas’s density is determined by dividing its mass by its volume to determine the density of a given gas molecule. It is possible to compare two gases if the gas molecule’s volume is kept constant.

Rate of Effusion

When air particles escape or leak through a hole that has a width much smaller than the mean free path of the molecules, this action is known as an effusion of a gas. Due to the small amount of molecule-to-molecule collisions that occur in these regions, all particles and molecules that reach the hole will pass through throughout this procedure.

The process by which material particles from the enclosed space begin to escape over time might be referred to as the rate of effusion of a gas. An illustration can help make this process more understandable. For instance, imagine that when we cut a hole in a balloon, the gas inside would begin to escape into the atmosphere, causing the balloon to deflate from the inside out. The effusion of gas into the atmosphere is what this is known as.

As a result, we may state that a gas’ rate of effusion is inversely related to its density and molar mass.

Rate of Effusion = 1/√(density) = 1/√(Molar mass of the gas molecule)

Derivation of Graham’s Law of Diffusion

According to Kinetic Molecular theory,

PV = mNc2 / 3

If we have 1 mole of gas then N = NA

PV = mNAc2 / 3

Where,

P = Pressure, V = Volume, m = Mass, NA = Avogadro’s number of molecule, c2 = Velocity

∴ PV = Mc2 / 3  …(mNA = M (Molar mass))

∴ c2 = 3PV/M

∴ c2 = 3P/d  …(density (d) = M/V)

Take Square root on both side,

∴ √c2 = √(3P/d)

∴ √c2 = Rate of Diffusion

∴ Rate = 1/√d

Important Points to remember

  1. The gas goes from higher concentrations to lower concentrations during the phenomenon of diffusion, whereas during the process of effusion, the gas moves from lower to higher concentrations.
  2. A gas’s entrance system becomes significantly disorganized due to diffusion. Both solid and liquid gases have slower rates of action.
  3. The length of time a gas particle spends in the diffusion process is directly inversely proportional to its molecular weight.
  4. The molecular masses and vapor density of a gas molecule are calculated using Graham’s law of diffusion.
  5. Additionally, this is employed in the separation of isotopes from the same element as well as the separation of various gases from a mixture of gases.

Problems based on Graham’s Law of Diffusion

Problem 1: Identify the gas’s molar mass by calculating its diffusion coefficient, which is 4.11 times that of ammonia (NH3).

Solution:

Rate of diffusion of gas = 4.11

As a result, we may state that the supplied gas’s ratio of diffusion rates should be 1/4.11

So, Rate1/Rate2 = 1/4.11

Molar mass of the Ammonia gas = 17.03

According Graham’s law of effusion,

Rate 1/Rate 2 = \frac{√M2}{√M1}

∴1/4.11 =  \frac{√M2}{√17.03}

Squaring Both sides, we get

∴1/16.8291 =  \frac{M2}{17.03}

∴ M2 = 0.0591 × 17.03

∴ M2 = 1.006 g/mol

Problem 2: Determine the molar mass of gas whose diffusion rate is two times that of water.

Solution:

Rate of diffusion of gas = 2

As a result, we may state that the supplied gas’s ratio of diffusion rates should be 1/2

So, Rate1/Rate2 = 1/2

Molar mass of the water = 18

According Graham’s law of effusion,

Rate 1/Rate 2 = \frac{√M2}{√M1}

∴1/2 = √\frac{M2}{18}

Squaring Both sides, we get

∴1/4 = \frac{M2}{18}

∴ M2 = 0.25 × 18

∴ M2 = 4.5 g/mol

Problem 3: What are the respective diffusion rates between hydrogen and nitrogen with a 14 molar mass?

Solution:

Molar mass of Hydrogen (M1) = 2, Molar mass of Nitrogen (M2) = 14

Let’s now assume that hydrogen has a slower rate of diffusion than other elements and that hydrogen has a diffusion rate of one.

Rate2 = 1

According Graham’s law of effusion,

Rate 1/Rate 2 = \frac{√M2}{√M1}

∴ Rate1/1 = √(14/2)

Squaring Both sides, we get

∴ (Rate 1)2/1 = 14/2

∴ (Rate1)2 = 7

∴ Rate1 = 2.64

Problem 4: Compare the relative diffusion rates of hard water (molar mass=20.0276) and water (molar mass=18.0152).

Solution:

Molar mass of water (M1) = 18.0152, Molar mass of Hard water (M2) = 20.0276

Given that heavy water has a slower diffusion rate, let’s assume that the rate of diffusion formula for heavy water is one.

Rate2 = 1

According Graham’s law of effusion,

Rate 1/Rate 2 = \frac{√M2}{√M1}

∴ Rate 1/1 = √\frac{20.0276}{18.0152}

Squaring Both sides, we get

∴ (Rate 1)2/1 = √\frac{20.0276}{18.0152}

∴ (Rate1)2 = 1.1117

∴ Rate1 = 1.05

Problem 5: Determine the molar mass of gas whose diffusion rate is 5 times that of the molar mass of the second gas is 7.2.

Solution:

Rate of diffusion of gas = 5

As a result, we may state that the supplied gas’s ratio of diffusion rates should be 1/5

So, Rate1/Rate2 = 1/5

Molar mass of the second gas (M2) = 7.2

According Graham’s law of effusion,

Rate 1/Rate 2 = \frac{√M2}{√M1}

∴ 1/5 = √(7.2/M1)

Squaring Both sides, we get

∴1/25 = 7.2/M1

∴ M1 = 7.2 / 0.04

∴ M1 = 180 g/mol

Problem 6: Find out what a gas’s molar mass is that has a diffusion rate four times that of chlorine.

Solution:

Rate of diffusion of gas = 4

As a result, we may state that the supplied gas’s ratio of diffusion rates should be 1/4

So, Rate1/Rate2 = 1/4

Molar mass of the Cl = 35.453

According Graham’s law of effusion,

Rate 1/Rate 2 = \frac{√M2}{√M1}

∴14=  √\frac{M2}{35.453}

Squaring Both sides, we get

∴1/16 = M2/(35.453)

∴ M2 = 0.0625 × 35.453

∴ M2 = 2.2158 g/mol

Frequently Asked Questions

Question 1: Which gas effuses the quickest?

Answer:

As per Graham’s law of effusion, the rate of a gas’s effusion is inversely related to the square root of the gas’ molecular masses. Thus, the gas with the lowest molecular weight will effuse more quickly than the gas with a higher molecular weight. So, helium is the lightest and quickest gas, according to our definition.

Question 2: List a few examples where Graham’s Law is required.

Answer:

To determine the effusion or prolixity rate of a certain gas, Graham’s law of effusion or prolixity is typically applied. The effusion or prolixity rates of two distinct feasts might be compared as well. hence assisting scientists in estimating the amount of time a specific gas will take to escape from the vessel or girding in which it is located. To create the safety procedures in case of a gas leak, this effusion or prolixity rate assessment is required.

By igniting an incense stick and watching the movement of the bank motes it causes, one can easily see how Graham’s law operates in real life. This is so that the scent of the burning incense stick can quickly spread over the surrounding area thanks to the prolixity process.

Question 3: State Graham’s Law. How is it calculated?

Answer:

According to Graham’s Law, the square root of a gas’s molar mass has an inverse relationship with the rate of diffusion or effusion of that gas. The equation for this law is shown below. 

r is inverse proportional to square root of M

In these equations,

M = molar mass and r = rate of diffusion or effusion.

Question 4: Are there any applications of Graham laws in real life?

Answer:

Graham’s law has a variety of applications in daily life, as we now know. Graham’s legislation is mostly used in divorce proceedings. This law allows us to separate different gases with different densities. Graham’s law’s formula can be used to compare diffusion rates and compute the molar masses of an unknown gas using a known gas. Using this formula, we may even divide the isotopes of a certain gas. The uranium isotopes serve as a classic illustration of this process. We mostly utilize the heavy and light isotopes of uranium that our planet naturally produces.


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