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Calculation of Mass of a Molecule and an Atom

Hemant More — Thu, 04 Jun 2020 12:05:03 +0000

Science > Chemistry > Molecule and Molecular Mass > Calculation of Mass of a Molecule and an Atom

In this article, we shall study the calculation of the mass of molecule and an atom.

Schematic Diagram for Mole Calculations:

Where, m = Given mass, M = Molar mass

v = Given volume, V = Molar volume = 22.4 dm³

n = Number of moles = m/M

Number of atoms = Number of molecules × Atomicity

Conversions:

From	To	Factor
kg	g	× 10³
mg	g	× 10^-3
μg	g	× 10^-6
metric ton	kg	× 10³
metric ton	g	× 10⁶
cm³	dm³	× 10^-3
m³	dm³	× 10³
litre	dm³	× 1

To Calculate Mass of Given Moles:

Calculate the mass of the following.

2.5 moles of water:

Molecular mass of water (H₂O) = 1 × 2 + 16 × 1 = 2 + 16 = 18 g

Number of moles of water = 2.5 = n

Mass of water = n × Molar mass

Mass of water = 2.5 × 18 = 45 g

1.2 moles of carbon dioxide

Molecular mass of carbon dioxide (CO₂) = 12 × 1 + 16 × 2 = 12 + 32 = 44 g

Number of moles of carbon dioxide = 1.2

Mass of carbon dioxide = n × Molar mass

Mass of carbon dioxide = 1.2 × 44 = 52.8 g

0.25 moles of sulphuric acid

Molecular mass of sulphuric acid (H₂SO₄) = 1 x 2 + 32 x 1 + 16 x 4 = 2 + 32 + 64 = 98 g

Number of moles of sulphuric acid = 0.25 = n

Mass of sulphuric acid = n × Molar mass

Mass of sulphuric acid = 0.25 × 98 = 24.5 g

0.1 moles of ammonia

Molecular mass of ammonia (NH₃) = 14 x 1 + 1 x 3 = 14 + 3 = 17 g

Number of moles of ammonia = 0.1 = n

Mass of ammonia = n × Molar mass

Mass of ammonia = 0.1 × 17 = 1.7 g

3.5 moles of methane

Molecular mass of methane (CH₄) = 12 x 1 + 1 x 4 = 12 + 4 = 16 g

Number of moles of methane = 3.5 = n

Mass of methane = n × Molar mass

Mass of methane = 3.5 × 16 = 56 g

2.4 moles of sulphur dioxide

Molecular mass of sulphur dioxide (SO₂) = 32 x 1 + 16 x 2 = 32 + 32 = 64 g

Number of moles of sulphur dioxide = 2.4 = n

Mass of sulphur dioxide = n × Molar mass

Mass of sulphur dioxide = 2.4 × 64 = 153.6 g

0.6 moles of bromine

Molecular mass of bromine (Br₂) = 40 x 2 = 80 g

Number of moles of bromine = 0.6 = n

Mass of bromine = n × Molar mass

Mass of bromine = 0.6 × 80 = 48 g

To Calculate Mass of a Molecule and Atom:

Calculate the following

mass of one oxygen atom and oxygen molecule in kg.

Molecular mass of oxygen (O₂) = 16 x 2 = 32 g

1 mole of oxygen is 32 g = 32 x 10^-3 kg

1 mole of a substance contains 6.022 x 10²³ molecules

Mass of each molecule of oxygen = (32 x 10^-3)/(6.022 x 10²³)

= 5.314 x 10^-26kg

Atomicity of oxygen (O₂) molecule is 2

Mass of each atom of oxygen = 5.314 x 10^-26/2 = 2.657 x 10^-26 kg

mass of one calcium atom in kg.

Molecular mass of calcium (Ca) = 40 g

1 mole of calcium is 40 g = 40 x 10^-3 kg

1 mole of a substance contains 6.022 x 10²³ molecules

Mass of each molecule of calcium = (40 x 10^-3)/(6.022 x 10²³)

= 6.642 x 10^-26kg

Atomicity of calcium (Ca) molecule is 1

Mass of each atom of calcium = 6.642 x 10^-26/1 = 6.642 x 10^-26 kg

mass of one nitrogen atom and nitrogen molecule in kg.

Molecular mass of nitrogen (N₂) = 14 x 2 = 28 g

1 mole of nitrogen is 28 g = 28 x 10^-3 kg

1 mole of a substance contains 6.022 x 10²³ molecules

Mass of each molecule of nitrogen = (28 x 10^-3)/(6.022 x 10²³)

= 4.650 x 10^-26kg

Atomicity of nitrogen (N₂) molecule is 2

Mass of each atom of nitrogen = 4.650 x 10^-26/2 = 2.325 x 10^-26 kg

mass of one sulphur dioxide molecule in grams.

Molecular mass of sulphur dioxide (SO₂) = 32 x 1 + 16 x 2 = 64 g

1 mole of sulphur dioxide is 64 g

1 mole of a substance contains 6.022 x 10²³ molecules

Mass of each molecule of sulphur dioxide = (64)/(6.022 x 10²³)

= 1.602 x 10^-22kg

mass of 100 molecules of water.

Molecular mass of water (H₂O) = 1 x 2 + 16 x 1 = 18 g

1 mole of water is 18 g = 18 x 10^-3 kg

1 mole of a substance contains 6.022 x 10²³ molecules

Mass of each molecule of water = (18 x 10^-3)/(6.022 x 10²³) = 2.989 x 10^-26kg

Mass of 100 molecules of water = 2.989 x 10^-26x 100 = 2.989 x 10^-24kg

Calculation of Volume at STP

Calculate the volume of following at STP.

8.5 x 10^-4 kg of ammonia

Molecular mass of ammonia (NH₃) = 14 x 1 + 1 x 3 = 14 + 3 = 17 g

= 17 x 10^-3 kg

Number of moles of ammonia = given mass/ molecular mass

= (8.5 x 10^-4)/(17 x 10^-3) = 0.05

1 mol of a gas at STP occupies 22.4 dm³ by volume

Volume of ammonia = number of moles x 22.4

Volume of 8.5 x 10^-4kg of ammonia at STP = 0.05 x 22.4 = 1.12 dm³

3.5 x 10^-3 kg of nitrogen

Molecular mass of nitrogen (N₂) = 14 x 2 = 28 g = 28 x 10^-3 kg

Number of moles of nitrogen = given mass/ molecular mass

= (3.5 x 10^-3)/(28 x 10^-3) = 0.125

1 mol of a gas at STP occupies 22.4 dm³ by volume

Volume of nitrogen= number of moles x 22.4

Volume of 3.5 x 10^-3kg of nitrogen at STP = 0.125 x 22.4 = 2.8 dm³

14 g of nitrogen

Molecular mass of nitrogen (N₂) = 14 x 2 = 28 g = 28 x 10^-3 kg

Number of moles of nitrogen = given mass/ molecular mass

= (14 x 10^-3)/(28 x 10^-3) = 0.5

1 mol of a gas at STP occupies 22.4 dm³ by volume

Volume of nitrogen= number of moles x 22.4

Volume of 3.5 x 10^-3kg of nitrogen at STP = 0.5 x 22.4 = 11.2 dm³

6.023 x 10²² molecules of ammonia

Number of moles of ammonia = Given molecules/ Avogadro’s number

Number of moles of ammonia = (6.023 x 10²²)/(6.023 x 10²³) = 0.1

1 mol of a gas at STP occupies 22.4 dm³ by volume

Volume of ammonia = number of moles x 22.4

Volume of 6.023 x 10²² molecules of ammonia at STP = 0.1 x 22.4 = 2.24 dm³

2.008 x 10²³ molecules of SO₂ at STP.

Number of moles of SO₂ = Given molecules/ Avogadro’s number

Number of moles of SO₂ = (2.008 x 10²²)/(6.023 x 10²³) = 0.3334

1 mol of a gas at STP occupies 22.4 dm³ by volume

Volume of SO₂ = number of moles x 22.4

Volume of 2.008 x 10²³ molecules of SO₂ at STP = 0.3334 x 22.4 = 7.469 dm³

0.2 mole of sulphur dioxide.

1 mol of a gas at STP occupies 22.4 dm³ by volume

Volume of sulphur dioxide = number of moles x 22.4

Volume of 0.2 moles of sulphur dioxide at STP = 0.2 x 22.4 = 4.48 dm³

Science > Chemistry > Molecule and Molecular Mass > Calculation of Mass of a Molecule and an Atom

The post Calculation of Mass of a Molecule and an Atom appeared first on The Fact Factor.

Calculation of Number of Moles, Atoms, and Molecules

Hemant More — Thu, 04 Jun 2020 11:40:44 +0000

Science > Chemistry > Molecule and Molecular Mass > Calculation of Number of Moles, Atoms, and Molecules

In this article, we shall study the calculation of number of moles, number of atoms, and molecules in a given moles.

Schematic Diagram for Mole Calculations:

Where, m = Given mass, M = Molar mass

v = Given volume, V = Molar volume = 22.4 dm³

n = Number of moles = m/M

Number of atoms = Number of molecules × Atomicity

Conversions:

From	To	Factor
kg	g	× 10³
mg	g	× 10^-3
μg	g	× 10^-6
metric ton	kg	× 10³
metric ton	g	× 10⁶
cm³	dm³	× 10^-3
m³	dm³	× 10³
litre	dm³	× 1

To Calculate Number of Moles:

Calculate the number of moles of following.

7.85 g of Fe (at. mass 56)

Given mass of Fe = 7.85 g

Fe is a monoatomic molecule.

Hence molecular mass of Fe = Atomic mass of Fe = 56 g

Number of moles of Fe = Given mass of Fe / Molecular mass of Fe

Number of moles of Fe = 7.85 g / 56 g = 0.1402

Ans: Number of moles of Fe = 0.1402

7.9 mg of Ca (at. mass 40)

Given mass of Ca =7.9 mg = 7.9 × 10^-3 g

Ca is a monoatomic molecule.

Hence molecular mass of Ca = Atomic mass of Ca = 40 g

Number of moles of Ca = Given mass of Ca / Molecular mass of Ca

Number of moles of Ca = 7.9 × 10^-3 g / 40 g = 1.975 × 10^-4

Ans: Number of moles of Ca = 1.975 × 10^-4

1.46 metric tons of Al (at. mass 27)

Given mass of Al = 1.46 metric tons = 1.46 × 10³ kg

= 1.46 × 10³ × 10³g = 1.46 × 10⁶g

Al is a monoatomic molecule.

Hence molecular mass of Al = Atomic mass of Al = 27 g

Number of moles of Al = Given mass of Al / Molecular mass of Al

Number of moles of Al = 1.46 × 10⁶g / 27 g = 5.41 × 10⁴

Ans: Number of moles of Al = 5.41 × 10⁴

65.5 mg of C (at. mass 12)

Given mass of C = 65.5 mg = 65.5 × 10^-3 g

Al is a monoatomic molecule.

Hence molecular mass of C = Atomic mass of C = 12 g

Number of moles of C = Given mass of C/ Molecular mass of C

Number of moles of C = 65.5 × 10^-3 g / 12 g = 5.46 × 10^-3

Ans: Number of moles of C = 5.46 × 10^-3

To Calculate Number of Molecules in a Given mole:

Calculate the number of moles and number of molecules of following.

0.032 mg of methane

Molecular mass of methane (CH₄) = 12 × 1 + 1 × 4 = 12 + 4 = 16 g

Given mass of CH₄ = 0.032 mg = 0.032 × 10^-3 g = 3.2 × 10^-5 g

Number of moles of CH₄ = n = Given mass of CH₄ / Molecular mass of CH₄

n = 3.2 × 10^-5 g / 16 g = 2 × 10^-6

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 2 × 10^-6 × 6.022 × 10²³

Number of molecules = 1.2044 × 10¹⁸

Ans: Number of moles of C = 5.46 × 10^-3 and

Number of molecules = 1.2044 × 10¹⁸

6.4 × 10^-2 kg of sulphur dioxide

Molecular mass of methane (SO₂) = 32 × 1 + 16 × 2 = 32 + 32 = 64 g

Given mass of SO₂ = 6.4 × 10^-2 kg = 6.4 × 10^-2× 10³ g = 64 g

Number of moles of SO₂ = n = Given mass of SO₂ / Molecular mass of SO₂

n = 64 g / 64 g = 1

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 1 × 6.022 × 10²³

Number of molecules = 6.022 × 10²³

Ans: Number of moles of SO₂ = 1 and Number of molecules =6.022 × 10²³

0.065 mg of water

Molecular mass of water (H₂O) = 1 × 2 + 16 × 1 = 2 + 16 = 18 g

Given mass of H₂O = 0.065 mg = 0.065 × 10^-3g = 6.5 × 10^-5 g

Number of moles of H₂O = n = Given mass of H₂O / Molecular mass of H₂O

n = 6.5 × 10^-5 g / 18 g = 3.611 × 10^-6

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 3.611 × 10^-6 × 6.022 × 10²³

Number of molecules = 2.174 × 10¹⁸

Ans: Number of moles of H₂O = 3.611 × 10^-6 and Number of molecules = 2.174 × 10¹⁸

500 mg of carbon dioxide

Molecular mass of carbondioxide (CO₂) = 12 × 1 + 16 × 2 = 12 + 32 = 44 g

Given mass of CO₂ = 500 mg = 500 × 10^-3 g = 0.5 g

Number of moles of CO₂ = n = Given mass of CO₂ / Molecular mass of CO₂

n = 0.5 g / 44 g = 1.136 × 10^-2

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 1.136 × 10^-2 × 6.022 × 10²³

Number of molecules = 6.843 × 10²¹

Ans: Number of moles of CO₂ = 1.136 × 10^-2 and Number of molecules = 6.843 × 10²¹

Calculation of Number of Atoms in a Given Mole:

Calculate the number of moles, number of molecules and number of atoms of following.

1.1. × 10^-4 kg of carbon dioxide

Molecular mass of CO₂ =12 × 1 + 16 × 2 = 12 + 32 = 44 g

Given mass of CO₂ = 1.1. × 10^-4kg = 1.1. × 10^-4× 10³ g = 0.11 g

Number of moles of CO₂ = Given mass of CO₂ / Molecular mass of CO₂

Number of moles of CO₂ = 0.11 g / 44 g = 2.5 × 10^-3

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 2.5 × 10^-3 × 6.022 × 10²³

Number of molecules of CO₂= 1.505 × 10²¹

Atomicity of CO₂ is 3

Hence Number of atoms in CO₂ = Number of molecules of CO₂ × Atomicity of CO₂

Hence Number of atoms of CO₂ = 1.505 × 10²¹× 3 = 4.515 × 10²¹

Ans: Number of moles = 2.5 × 10^-3 , Number of molecules = 1.505 × 10²¹and Number of atoms =4.515 × 10²¹

4.25 × 10^-2 kg of ammonia

Molecular mass of NH₃ =14× 1 + 1 × 3 = 14 + 3 = 17 g

Given mass of NH₃ = 4.25. × 10^-2kg = 4.25. × 10^-2× 10³ g = 42.5 g

Number of moles of NH₃ = Given mass of NH₃ / Molecular mass of NH₃

Number of moles of NH₃ = 42.5 g / 17 g = 2.5

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 2.5 × 6.022 × 10²³

Number of molecules of NH₃= 1.505 × 10²⁴

Atomicity of NH₃ is 4

Hence Number of atoms of NH₃ = Number of molecules of NH₃ × Atomicity of NH₃

Hence Number of atoms in CO₂ = 1.505 × 10²⁴× 4 = 6.020 × 10²⁴

Ans: Number of moles = 2.5, Number of molecules = 1.505 × 10²⁴and Number of atoms = 6.020 × 10²⁴

0.4 g of helium gas.

Molecular mass of He = 4 g

Given mass of He = 0.4 g

Number of moles of He = Given mass of He / Molecular mass of He

Number of moles of He = 0.4 g / 4 g = 0.1

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 0.1 × 6.022 × 10²³

Number of molecules of He= 6.022 × 10²²

Atomicity of He is 1

Hence Number of atoms of He = Number of molecules of He × Atomicity of He

Hence Number of atoms in He = 6.022 × 10²²× 1 = 6.022 × 10²²

Number of moles = 0.1, Number of molecules = 6.022 × 10²²and Number of atoms = 6.022 × 10²²

5.6 cm³ of ammonia at STP.

Given volume of ammonia = 5.6 cm³ = 5.6 × 10^-3 dm³

1 mole of ammonia at STP occupies 22.4 dm³ by volume.

Number of moles of NH₃ = Given volume of NH₃ / 22.4 dm³

Number of moles of NH₃ = 5.6 × 10^-3 dm³ / 22.4 dm³ = 2.5 × 10^-4

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 2.5 × 10^-4 × 6.022 × 10²³

Number of molecules of NH₃= 1.5055 × 10²⁰

Atomicity of NH₃ is 4

Hence Number of atoms of NH₃ = Number of molecules of NH₃ × Atomicity of NH₃

Hence Number of atoms of NH₃ = 1.5055 × 10²⁰× 4 = 6.022 × 10²¹

Ans: Number of moles = 2.5 × 10^-4, Number of molecules = 1.5055 × 10²⁰and Number of atoms =6.022 × 10²¹

5.6 dm³ of ammonia at STP.

Given volume of ammonia = 5.6 dm³

1 mole of ammonia at STP occupies 22.4 dm³ by volume.

Number of moles of NH₃ = Given volume of NH₃ / 22.4 dm³

Number of moles of NH₃ = 5.6 dm³ / 22.4 dm³ = 0.25

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 0.25 × 6.022 × 10²³

Number of molecules of NH₃= 1.5055 × 10²³

Atomicity of NH₃ is 4

Hence Number of atoms of NH₃ = Number of molecules of NH₃ × Atomicity of NH₃

Hence Number of atoms of NH₃ = 1.5055 × 10²³× 4 = 6.022 × 10²³

Ans: Number of moles = 0.25, Number of molecules = 1.5055 × 10²³and Number of atoms =6.022 × 10²³

7.6 dm³ of hydrogen at STP.

Given volume of hydrogen = 7.6 dm³

1 mole of hydrogen at STP occupies 22.4 dm³ by volume.

Number of moles of H₂ = Given volume of H₂ / 22.4 dm³

Number of moles of H₂ = 7.6 dm³ / 22.4 dm³ =0.34

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 0.34 × 6.022 × 10²³

Number of molecules of H₂ = 2.047 × 10²³

Atomicity of H₂ is 2

Hence Number of atoms of H₂ = Number of molecules of H₂ × Atomicity of H₂

Hence Number of atoms of H₂ = 2.047 × 10²³× 2 = 4.094 × 10²³

Ans: Number of moles = 0.34, Number of molecules = 2.047 × 10²³and Number of atoms = 4.094 × 10²³

250.9 dm³ of hydrogen at STP.

Given volume of hydrogen = 250.9 dm³

1 mole of hydrogen at STP occupies 22.4 dm³ by volume.

Number of moles of H₂ = Given volume of H₂ / 22.4 dm³

Number of moles of H₂ = 250.9 dm³ / 22.4 dm³ = 11.2

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 11.2 × 6.022 × 10²³

Number of molecules of H₂ = 6.745 × 10²⁴

Atomicity of H₂ is 2

Hence Number of atoms of H₂ = Number of molecules of H₂ × Atomicity of H₂

Hence Number of atoms of H₂ = 6.745 × 10²⁴× 2 = 1.349 × 10²⁵

Ans: Number of moles = 11.2, Number of molecules = 6.745 × 10²⁴and Number of atoms = 1.349 × 10²⁵

Science > Chemistry > Molecule and Molecular Mass > Calculation of Number of Moles, Atoms, and Molecules

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Mole Concept

Hemant More — Wed, 03 Jun 2020 17:01:09 +0000

Science > Chemistry > Molecule and Molecular Mass > Mole Concept

In this article, we are going to study very important concept of chemistry known as the mole concept.

Berzelius Hypothesis:

According to Dalton’s atomic theory atoms of different elements combine with each other in a simple whole number ratio, whereas Gay Lussac’s law of combining volumes, gases combine with each other in a simple whole ratio by their volumes.

A Swedish chemist Berzelius correlated the two laws and put forward is the concept called Berzelius hypothesis. It states that equal volumes of all gases under similar conditions of temperature and pressure contain an equal number of atoms.

An Italian chemist Avogadro in 1811, modified the theory by differentiating between the ultimate particle of an element that takes part in reaction (atom) and the ultimate particle that has independent existence (molecule). He argued that the smallest particle of a substance which has independent existence is not an atom but molecule.

Avogadro’s Law:

Equal volumes of all gasses under the same conditions of temperature & pressure contain an equal number of molecules.

Explanation:

Let two gasses gas A and Gas B be taken in two containers having equal volume (V). Let the temperature of both the gasses be the same (T).Let the pressure of the two gases be same (P). By Avogadro’s law under such conditions of equal pressure, equal volume & equal temperature, the number of molecules of gas A in the container should be equal to the number of molecules of gas B in the container.

Validation of Avogadro’s Law:

Consider following chemical reactions

H₂ + Cl₂ → 2HCl

1vol 1vol 2 vol

Thus the simple ratio of volumes is 1 : 1 : 2

Suppose 1 volume of hydrogen gas contains ‘n’ molecules. By Avogadro’s law at same temperature and pressure, there should be ‘n’ molecules of chlorine and ‘2n’ molecules of hydrogen chloride are involve.

Thus ‘n’ molecules of H₂ + ‘n’ molecules of Cl₂ → ‘2n’ molecules of HCl

Dividing this equation by n, we have

1 molecule of H₂ + 1 molecule of Cl₂ → 2 molecules of HCl

Hydrogen and chlorine are diatomic molecules.

2 atoms of H₂ + 2 atoms of Cl₂ → 2 molecules of HCl

Dividing this equation by 2, we have

1 atom of H₂ + 1 atom of Cl₂ → 1 molecule of HCl

Now, number atoms of hydrogen, chlorine and hydrogen chloride are the whole numbers the Avogadro’s law is validated.

Importance of Avogadro’s Law:

It differentiates between atoms and molecules of gasses.
It modified Dalton’s atomic theory.
It explains Gay Lussac’s law of combining volume.
It helps in determination of the atomic weight of elements.
It established that at N.T.P.one gram mole of any gas occupies 22.4 dm³ by volume. one mole of a gas contains 6.023 × 10²³ molecules of gas.
It gives the relation between the vapour density & the molecular mass. The relation is

Molecular weight = 2 × vapour density.

Concept of Vapour Density:

The vapour density of a gas is defined as the ratio of the weight of a certain volume of a gas to the weight of the same volume of hydrogen at the same temperature & pressure.

Relation Between Molecular Mass and vapour Density:

The vapour density of a gas is defined as the ratio of the weight of a certain volume of a gas to the weight of the same volume of hydrogen at the same temperature & pressure. By definition of vapour density

This is the relation between the molecular weight and vapour density

Gram Molecular Volume (Molar volume) (GMV):

The volume occupied by one mole of a gas at STP (or NTP) is called as gram molar volume. It is 22.4 dm3 at NTP.

Proof:

Mathematically, Avogadro’s law is stated as “At constant temperature (T) and pressure (P) the volume (V) of a gas is directly proportional to the number of molecules (n).

The molecular mass of gas corresponds to one mole of a gas hence we can say that one mole of a gas occupies 22.4 dm³ at STP.

Atomicity:

The number of atoms present in a molecule of a substance is called the atomicity of that substance.

Examples: The number in bracket indicates atomicity of that compound. Helium He (1), Dioxygen O₂ (2), Ozone O₃ (3), 4 Phosphorous P₄ (4), Sulphur S₈ (8), Carbon dioxide CO₂ (3), Ammonia NH₃ (4).

Atomicity indicates that how many atoms are present in the molecule of the element or the compound.

Atomicity of Element using Avogadro’s Law:

The number of molecules present in a molecule of a substance is called the atomicity of substance.

Let us consider the formation of ammonia. Experimentally it is found that 1 volume of nitrogen reacts with 3 volumes of hydrogen to form 2 volumes of ammonia. Thus,

N₂ + 3H₂ → 2NH₃

1 vol 3 vol 2 vol

1/2 vol 3/2 vol 1vol

According to Avogadro’s law “equal volumes of all gases under the identical condition of temperature and pressure contain the same number of molecules”. If 1 volume of ammonia contains n molecules 1/2 volume of nitrogen would contain n /2 molecules and 3/2 volume of hydrogen would contain 3n/2 molecules.

∴ n/2 molecules of Nitrogen + 3n/2 molecules of Hydrogen → n molecules of Ammonia

∴ 1/2 molecule of nitrogen + 3/2 molecule of hydrogen → 1molecule of ammonia.

As a molecule can be divided into the constituent atoms but the atoms cannot be divided further nitrogen molecule has to be diatomic, hydrogen diatomic and ammonia tetra-atomic.

Thus atomicity of Nitrogen =2, atomicity of hydrogen = 2 and atomicity of ammonia = 4.

Mole Concept:

The term mole was introduced by Wilhelm Ostwald. In Latin mole means heap or pile. He assumed substance as heap or pile of elementary entities like atoms, molecules, ions, etc. and called it as a mole.

When we are studying matter we require the actual number of molecules involved in the reaction. The atoms and molecules are discrete particles present in large number in the matter. To denote this number of atoms or molecules the term mole is used.

The amount of substance that contains as many elementary entities, like atoms, molecules, ions, photons as there are atoms in 0.012 kg of carbon 12 is called mole of a substance.

Example: 1 mole of hydrogen means 2 g of hydrogen. 1 mole of oxygen means 32 g of oxygen.

A mole is a fundamental unit of the amount of substance in the SI system of units. It is a collection of 6.022 × 1023 particles. The number 6.022 × 1023 is also called Avogadro’s number.
1 mole also corresponds to the atomic or molecular mass of an element expressed in gram. Thus 1 mole of gas contains 6.022 × 1023 molecules of gas. A 1-mole atom of gas contains 6.022 × 10²³ atoms of gas. 1-mole ions contain 6.022 × 10²³ ions etc.

Mole and Avogadro ’s Number:

The number of particles such as atoms, molecules, ions, in one mole of a substance is called Avogadro’s number or Avogadro’s constant.

It is denoted by ‘N’. Its value at STP is 6.022 × 10²³ per mole. Thus 1 mole of a gas contains 6.022 × 10²³ molecules of gas. 1-mole atom of a gas contains 6.022 × 10²³ atoms of gas. 1-mole ions contain 6.022 × 10²³ ions etc.

The Significance of Avogadro’s Number:

Avogadro’s number is equal to the number of molecules in one gram mole or one gram molecular mass of any compound. Thus gram molecular mass of any substance is equal to the mass in grams of Avogadro’s number of 6.022 × 10²³ molecules.
Avogadro’s number is equal to the number of atoms in one gram mole or one gram atomic mass of an element. Thus gram atomic mass of any element is equal to the weight in grams of Avogadro’s number of 6.022 × 10²³ atoms.
Avogadro’s number is equal to the number of molecules in 22.4 dm³ of any gas at NTP.
The actual mass of an element or compound can be found using this number.

Concept of one gram atom:

One gram atom of an element is the amount of the element which is equal to the mass of 1 mole (or 6.023 × 10²³ atoms) of the element or atomic mass of the element in grams.

Thus one gram atom of sodium is equal to one mole of sodium corresponds to 23 grams of sodium, because the atomic mass of sodium is 23 grams.

Science > Chemistry > Molecule and Molecular Mass > Mole Concept

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Molecular Mass by Victor Meyer’s Method

Hemant More — Wed, 03 Jun 2020 15:05:55 +0000

Science > Chemistry > Molecule and Molecular Mass > Molecular Mass by Victor Meyer’s Method

In the last few articles, we have studied the molar volume method, Regnault’s method, Graham’s diffusion law method, and ideal gas equation method to determine molecular mass. In this article, we shall study Victor Meyer’s method to determine the molecular mass of a volatile substance.

Molecule:

When two or more atoms are firmly held together by a chemical bond, a molecule is formed. The molecule of an element may consist of one or more atoms of the same kind, while that of the chemical compound consists of different kinds of atoms.

The smallest particle of an element or compound which can exist in a free state and does not take part in a chemical reaction is called molecule.

Molecules are denoted by formula indicating the number of constituent elements in the compound. For example molecular formula for oxygen is O₂. Thus one molecule oxygen consists of two atoms of oxygen

Molecular Mass or Molar Mass:

The molecular mass or molar mass of a substance is defined as the ratio of the mass of one molecule of a substance to 1/12 th of the mass of ⁶C₁₂ isotope taken as 12000 units.

Gram Molecular Mass or Molar Mass:

The molecular mass expressed in grams is called gram molecular mass (GMM)

Method – IV (Molecular Mass by Victor Meyer’s Method):

Principle:

A known mass of volatile substance in Victor Meyer’s tube. As a result, the liquid changes into vapours, the vapours, in turn, displaces an equal volume of air which is collected over water. The volume of air collected is measured at the pressure and temperature of the laboratory. This volume is converted to S.T.P. volume and using gram molar volume concept molecular mass is calculated.

Procedure:

A round bottom flask filled with a liquid whose B.P. is 10 °C more than the volatile liquid acts as an outer glass jacket. A Victor Mayor’s tube with an outer tube acts as inner glass jacket. This outer tube is dipped in a trough fitted with water. Bottom of the Victor Mayor’s tube consists of Hg or asbestos pieces for cushioning.

Outer glass jacket is heated due to which air inside expands and bubbles through the water in the trough. A small. glass tube is known as Hoffman’s bottle. with a stopper is cleaned, washed and dried and weighed. Volatile Liquid is taken in Hoffman’s bottle and weighed. A Eudiometer tube filled with water is placed over the tube connected to Victor Mayor apparatus.

Now Hoffman’s bottle is dropped in the Victor Mayor tube. Due to heat liquid in the bottle vapourises and blows off the stopper and displace air which corresponds to its own volume which is collected in a eudiometer tube by the downward displacement of water.

The water temperature and pressure are also noted. Eudiometer tube is taken into another trough filled with water for equalisation of pressure. The volume of air displaced is noted, which in fact correspond to the volume of the vapours.

Observations:

The weight of the Hoffman’s Bottle = W₁ gm.

The weight of the Hoffman’s bottle + Liquid = W₂ gm.

Weight of the volatile liquid = W₂ – W₁ = W gm.

Pressure = (p – f) mm of Hg.

Temperature = t °C = T K

Volume of vapours = V dm³.

Calculations:

Now the volume of vapours at S.T.P. from the above data is calculated using following formula.

Where the volume of hydrogen at S.T.P. = V₀ dm³

Pressure at S.T.P. = P₀ mm of Hg = 760 m

Absolute temperature at S.T.P.= T K = 273 K

Now, the molecular weight can be calculated by following formula

Merits of Victor Mayer’s Method:

The method is very simple to carry out. Weight.
The sample required for the experiment is very small.

Demerits of Victor Mayer’s Method:

This method is applicable to volatile liquid only.
The method can not be used for the substance which undergoes thermal decomposition.
The liquid in outer jacket is generally water, hence this method is suitable only for the liquids which have boiling point less than 100oC.
Due to manual handling, there is the possibility of personal error.
It is not applicable to volatile substances which are water soluble.

Numerical Problems:

Example – 01:

In the determination of molecular mass by Victor Meyer’s method 0.60 g of volatile substance expelled 123 ml of air measured over water at 20 °C and 757.4 mm pressure. Find the molecular mass of the substance if the aqueous tension at 20 °C is 17.4 mm.

Given: V = 123 ml = 0.123 dm³, t = 20 °C, T = 20 + 273 = 293 K, P = 757.4 mm of Hg , f = 17.4 mm of Hg, P – f = 757.4 – 17.4 = 740 mm of Hg, P_o = 760 mm of Hg, T_o = 273 K

Solution:

Ans: The molecular mass of the substance is 120.43

Method – V (Duma’s Method):

Procedure:

In this method, a known volume of vapour is heated to some higher temperature and its mass is noted. The calculations are the same as Victor Meyer’s method.

Method – VII (Hoffman’s Method):

Procedure:

Some substances are decomposed when they are evaporated at their boiling points. The substance is evaporated at a sufficiently low temperature at reduced pressure. Hoffman’s method is used for calculating the molecular mass of such substances.

A known mass of a substance is vapourised above a mercury column in a barometric tube and the volume of vapour formed is noted, Then the volume is calculated at STP, and calculation is done the same way as in Victor Meyer’s method.

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Molecular Mass Using Ideal Gas Equation

Hemant More — Wed, 03 Jun 2020 14:46:34 +0000

Science > Chemistry > Molecule and Molecular Mass > Molecular Mass Using Ideal Gas Equation

In the last few articles, we have studied the molar volume method, Regnault’s method, and Graham’s diffusion law method to determine molecular mass. In this article, we shall study the ideal gas equation method to determine the molecular mass of a gas.

Molecule:

The smallest particle of an element or compound which can exist in a free state and does not take part in a chemical reaction is called molecule.

Molecular Mass or Molar Mass:

The molecular mass or molar mass of a substance is defined as the ratio of the mass of one molecule of a substance to 1/12 th of the mass of ⁶C₁₂ isotope taken as 12000 units.

Gram Molecular Mass or Molar Mass:

The molecular mass expressed in grams is called gram molecular mass (GMM)

Method – IV (Molecular Mass by Ideal Gas Equation / Law Method):

Principle:

The ideal gas equation is

PV = nRT

Where P = Pressure of the gas

V = Volume of the gas

n = Number of moles of the gas

R = Universal gas constant

T = Absolute temperature of the gas

Using the above relation and knowing the remaining quantities, the molecular mass can be calculated.

Numerical Problems:

Example – 01:

What is the relative molecular mass of the gas if 0.866 g sample is 60.0 ml. The bulb has a pressure of 400 mm at 20 °C. R = 0.0821 lit-atm

Given: w = 0.866 g, V = 60.0 ml = 60 x 10^-3 dm³, P = 400 mm = 400/760 = 0.5263 atm

Solution:

By ideal gas equation. PV = nRT

Ans: The relative molecular mass is 66.00

Example – 02:

0.1348 g of gas was found to occupy a volume of 25.80 ml at 0 OC and 760 mm of Hg pressure. calculate the relative molecular mass of the gas.

Given: w = 0. 1348 g, V = 25.80 ml = 25.80 x 10^-3 dm³, P = 760 mm = 760/760 = 1 atm, T = 0 + 273 = 273 K

Solution:

By ideal gas equation PV = nRT

Ans: The relative molecular mass is 117.1

Example – 03:

3.895 dm³ of a gas at 293 K and 780 mm pressure were found to have a mass of 2.83 g. Calculate the relative molecular mass of the gas.

Given: w = 2.83 g, V = 3.895 dm³, P = 780 mm = 780/760 atm, T = 293 K

Solution:

By ideal gas equation, PV = nRT

Ans: The relative molecular mass is 17.03

Example – 04:

The molecular mass of a gaseous substance is 80. What will be the volume of 1 g of a gas at 0 °C and 720 mm mercury pressure? State whether this gas would diffuse through a porous pot slower or faster than chlorine.

Given: M = 80, w = 1 g, P = 720 mm = 720/760 atm, T = 0 +273 = 273 k

Solution:

By ideal gas equation, PV = nRT

The volume of the gas is 0.2957 dm3

By Graham’s diffusion law “The rate of diffusion of different gases under similar conditions of temperature and pressure are inversely proportional to square root of their densities”. In this case the molecular mass of the gas (80) is greater than the molecular mass of chlorine (71). Hence the gas will diffuse slowly.

Example – 05:

What is the relative molecular mass of the gas if 0.866 g sample is 60.0 ml. The bulb has a pressure of 400 mm at 20 °C. R = 0.0821 lit-atm

Given: w = 0.866 g, V = 60.0 ml = 60 x 10-3 litres, P = 400 mm = 400/760 = 0.5263 atm

Solution:

By ideal gas equation, PV = nRT

Ans: The relative molecular mass is 66.00

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Molecular Mass Using Graham’s Law of Diffusion

Hemant More — Wed, 03 Jun 2020 14:10:20 +0000

Science > Chemistry > Molecule and Molecular Mass > Molecular Mass Using Graham’s Law of Diffusion

In the last few articles, we have studied the molar volume method and Regnault’s method to determine molecular mass. In this article, we shall study Graham’s law diffusion method to determine the molecular mass of a gas.

Molecule:

The smallest particle of an element or compound which can exist in a free state and does not take part in a chemical reaction is called molecule.

Molecular Mass or Molar Mass:

The molecular mass or molar mass of a substance is defined as the ratio of the mass of one molecule of a substance to 1/12 th of the mass of ⁶C₁₂ isotope taken as 12000 units.

Gram Molecular Mass or Molar Mass:

The molecular mass expressed in grams is called gram molecular mass (GMM)

Method – III (Molecular Mass by Graham’s Law Diffusion Method):

Principle:

Graham’s law of diffusion states that “The rate of diffusion of different gases under similar conditions of temperature and pressure are inversely proportional to the square root of their densities”.

But the densities are directly proportional to their molar masses. Hence the law can be restated as “The rate of diffusion of different gases under similar conditions of temperature and pressure are inversely proportional to square root of their molar masses”. Mathematically,

Numerical Problems:

Example – 01:

300 c.c. of oxygen gas takes 50 seconds to effuse through an aperture while the same volume of unknown gas takes 75 seconds. If the relative molecular mass of oxygen is 32, find the relative molecular mass of the unknown gas.

Given: Rate of diffusion of oxygen r_o = 300/50 = 6 c.c./s, Rate of diffusion of gas g = 300/75 = 4 c.c./s, The molecular mass of oxygen M_O = 32,

To Find: Molecular mass of gas M_g =?

Solution:

By Graham’s law of diffusion

Ans: The molecular mass of the gas is 72.

Example – 02:

30 ml of ozone diffuses at the same time as 25 ml of chlorine. If the density of chlorine is 35.4. Find the density and relative molecular mass of ozone.

Given: Rate of diffusion of ozone r_O = 30 / t ml per second, Rate of diffusion of chlorine r_Cl = 25 / t ml per second, Density of chlorine ρ_Cl = 35.4

To Find: Density of ozone ρ_O =?, The Molecular mass of ozone M_O =?

Solution:

By Graham’s law of diffusion

Ans: The molecular mass of the gas is 24.58

Example – 03:

In 75 seconds 400 ml of oxygen diffuses through a glass tube. If 25 seconds are taken by 100 ml of another gas ‘X’ to diffuse through the same tube under similar conditions. calculate the relative molecular mass of the gas ‘X’.

Given: Rate of diffusion of oxygen= r_O = 400 / 75 = 16/3 ml per second, Rate of diffusion of gas ‘X’ = r_X = 100 /25 = 4 ml per second

Solution:

By Graham’s law of diffusion

Ans: The molecular mass of the gas is 56.9

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Molecular Mass by Regnault’s Method

Hemant More — Wed, 03 Jun 2020 13:53:32 +0000

Science > Chemistry > Molecule and Molecular Mass > Molecular Mass by Regnault’s Method

In the last article, we have studied the molar volume method to determine molecular mass. In this article, we shall study the concept of molecular mass (molar mass) and Regnault’s method to determine it.

Molecule:

The smallest particle of an element or compound which can exist in a free state and does not take part in a chemical reaction is called molecule.

Molecular Mass or Molar Mass:

The molecular mass or molar mass of a substance is defined as the ratio of the mass of one molecule of a substance to 1/12 th of the mass of ⁶C₁₂ isotope taken as 12000 units.

Gram Molecular Mass or Molar Mass:

The molecular mass expressed in grams is called gram molecular mass (GMM)

Method – II (Regnault’s Method):

Principle:

In this method vapour densities of gases are determined by direct weighing. The vapour density of a gas is the ratio of the mass of a certain volume of a gas to the mass of the same volume of hydrogen at the same temperature & pressure.

Procedure:

In this method, two hollow glass globes of the same capacity, same mass and same size are taken. They are evacuated and suspended by two sides of a physical balance.

Now one of the glass globes is filled with a gas whose vapour density is to be found. then the mass of the globe is measured. the difference in the filled globe and empty globe gives the mass of the gas in the globe,

Now the globe is evacuated again and filled with hydrogen gas at same temperature and pressure. The mass of globe filled with hydrogen is measured again. The difference in the empty globe and hydrogen filled globe gives the mass of hydrogen.

Vapour density is found by using the formula.

Now, Molecular Mass = 2 x Vapour density

Numerical Problems:

Example – 01:

1 dm³ of hydrogen at S.T.P. has a mass of 0.09 g. If 2 dm3 of gas at S.T.P. has a mass of 2.880 g. Calculate the vapour density and molecular mass of the gas.

Solution:

1 dm³ of hydrogen has a mass of 0.09 g.

2 dm³ of gas has a mass of 2.880 g.

Hence, 1 dm³ of gas has a mass of 1.440 g.

Now, Molecular Mass = 2 x Vapour density = = 2 x 16 = 32

Ans: The molecular mass of the gas is 32.

Example – 02:

The capacity of a glass bulb is 30.5 ml, 0.146 g of gas is filled in a bulb at 22 °C and 755 mm of Hg pressure. 1 dm³ of hydrogen at S.T.P. have a mass of 0.09 g. Calculate the vapour density and molecular mass of the gas.

Given: V = 30.5 ml, t = 22°C, T = 22 + 273 = 295 K, P = 755 mm of Hg , W = 0.146 g, P_o = 760 mm of Hg, T_o = 273 K

Solution:

P_o= 0.02804 litre = 0.02804 dm³

Now, Molecular Mass = 2 x Vapour density = 2 x 57.85 = 115.7

Ans: The molecular mass of the gas is 115.7.

Example – 03:

The capacity of a glass bulb is 127 ml, 0.4524 g of gas is filled in a bulb at 409 K and 758 mm of Hg pressure. 1 ml of hydrogen at S.T.P. has a mass of 0.00009 g. Calculate the vapour density and molecular mass of the gas.

Given: V = 127 ml, T = 409 K, P = 758 mm of Hg , W = 0.4524 g, P_o = 760 mm of Hg, T_o = 273 K

Solution:

Now, Molecular Mass = 2 x Vapour density = 2 x 59.45 = 118.90

Ans: The molecular mass of the gas is 118.90

Example – 04:

In Regnault’s method, 27.32 ml of gas were found to have a mass of 0.1008 g. The experiment was performed at 16.5 °C and 694 mm of mercury. Calculate the molecular mass of the gas.

Given: V = 27.32 ml, t = 16.5 °C, T = 16.5 + 273 = 289.5 K, P = 694 mm of Hg , W = 0.1008 g, P_o = 760 mm of Hg, T_o = 273 K

Solution:

Now, Molecular Mass = 2 x Vapour density = 2 x 47.6 = 95.2

Ans: The molecular mass of the gas is 95.2

Example – 05:

When a glass bulb with a stop cock is evacuated, weighed, and then filled with oxygen, the weight increases by 0.25 g. When the same bulb is evacuated and filled with another unknown gas under the same conditions of temperature and pressure, the mass increases by 0.5525 g. If the molecular mass of oxygen is 32, calculate the molecular mass of the unknown gas. Calculate the volume of the glass bulb assuming that it does not change appreciably with the change in pressure and temperature.

Solution:

The molecular mass of oxygen = 32

Vapour density of oxygen = 32 / 2 = 16

Mass of oxygen = 0.25 g

Now, the vapour density of the gas

Now, Molecular Mass = 2 x Vapour density = 2 x 35.36 = 70.72

Hence, the molecular mass of the gas is 70.72

Ans: The volume of the bulb is 0.1736 dm³.

Science > Chemistry > Molecule and Molecular Mass > Molecular Mass by Regnault’s Method

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Molecular Mass Molar Volume Method

Hemant More — Fri, 29 May 2020 15:21:04 +0000

Science > Chemistry > Molecule and Molecular Mass > Molecular Mass by Molar Volume Method

In this article, we shall study the concept of molecular mass (molar mass) and molar volume method to determine it.

Molecule:

The smallest particle of an element or compound which can exist in a free state and does not take part in a chemical reaction is called molecule.

Molecular Mass or Molar Mass:

The molecular mass or molar mass of a substance is defined as the ratio of the mass of one molecule of a substance to 1/12 th of the mass of ⁶C₁₂ isotope taken as 12000 units.

Gram Molecular Mass or Molar Mass:

The molecular mass expressed in grams is called gram molecular mass (GMM)

Method – I (Molar Volume Method)

Principle:

In this method we find a known volume of a gas at S.T.P. and using the concept that one mole of every gas occupies 22.4 dm³ by volume, we calculate the molar mass of the gas.

Procedure:

Find the volume of the gas at STP, from given data.
Find molecular mass, using the formula

Numerical Problems:

Example – 01:

10 dm³ of gas at 14 °C and 729 mm pressure has a mass of 17.925 g. Calculate the relative molecular mass of the gas.

Given: V = 10 dm³, t = 14 °C, T = 14 + 273 = 287 K, P = 729 mm of Hg , W = 17.925 g

Solution:

Thus the mass of 9.125 dm³ of gas is 17.925 g. Hence the mass of 22.4 dm³ of a gas

Molecular mass = (17.975/9.125) x 22.4 = 44

Ans: The relative molecular mass of the gas is 44 g

Example – 02:

5 litres of a gas at NTP has a mass of 14.4 g. Find the relative molecular mass of the gas.

Given: V_o = 5 litres = 5 dm³, W = 14.4 g

Solution:

The mass of 5 dm³of a gas at NTP is 14.4 g. Hence mass of 22.4 dm³ of a gas

Molecular mass = (14.4/5) x 22.4 = 64.51

Ans: The molecular mass of the gas is 64.51 g.

Example – 03:

240 ml of a dry gas measured at 300 K and 750 mm of mercury has a mass of 0.42336 g. calculate the relative molecular mass of the gas.

Given: V = 240 ml = 0.240 dm³, T = 300 K, P = 750 mm of Hg , W = 0.42336 g

Solution:

The mass of 0.2155 dm³ of gas is 0.42336 g. Hence mass of 22.4 dm³ of a gas

Molecular mass = (0.42336/0.2115) x 22.4 = 44

Ans: The molecular mass of the gas is 44 g.

Example – 04:

1.25 g of a pure carbonate on ignition leave a residue of 0.70 g and evolved is 312 mm at 27 °C and 755 mm of mercury. Calculate the molecular mass of the gas.

Given: Mass of carbonate = 1.25 g, Mass of residue = 0.70 g, Mass of the gas = 1.25 – 0.70 = 0.55 g, V = 312 ml = 0.312 dm³ T = 27 °C = 27 + 273 = 300 K, P = 755 mm of Hg , W = 0.55 g

Solution:

The mass of 0.2821 dm3 of gas is 0.55 g. Hence the mass of 22.4 dm³ of a gas

Molecular mass = (0.55/0.2821) x 22.4 = 43.67

Ans: The molecular mass of the gas is 43.67 g.

Science > Chemistry > Molecule and Molecular Mass > Molecular Mass by Molar Volume Method

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Correction to the Ideal Gas Equation

Hemant More — Tue, 07 Apr 2020 10:34:19 +0000

Science > Physics > Themodynamics > Correction to the Ideal Gas Equation

In this article, we shall study the correction to ideal gas equation.

Ideal Gas Equation:

The equation of state for an ideal gas is given by

PV = n RT

Where, P = Pressure of gas, V = Volume of gas, n = No. of moles of gas

R = Universal gas constant, T = Absolute temperature of gas.

For one mole of gas, PV = RT

Van Der Wall’s Correction to Ideal Gas Equation:

Necessity of Correction of Ideal Gas Equation:

While deriving the ideal gas equation PV = RT, we had assumed that

The volume occupied by the gas molecules themselves is negligible compared with the total volume of the gas, and
The molecules exert no appreciable force on one another.

Both of these assumptions cannot be true at high pressure. When the gas is at high pressure, it has a small volume and therefore volume actually occupied by an individual molecule of a gas cannot be neglected in comparison with the volume of the entire gas.

Also, at high pressure, the molecules come closer, therefore considerable cohesive forces ‘will be acting on them. Thus at high pressure and low temperature, the real gases do not obey the above relation.

While modifying the perfect gas equation, PV = RT, both these factors were considered by van der Waals.

Correction for Volume:

Let us consider a container containing the gas. The gas molecules are like hard elastic spheres. Let us assume that ‘r’ is their radius. Thus the distance between two molecules cannot be less than ‘2r’. Thus the presence of one molecule in the container will reduce the space available for another molecule by amount

Thus the space available for free motion of gas molecules is less than the actual volume of the gas. Therefore, the corrected volume is taken as (V – b).

The correction term ‘b’ is called co-volume and is equal to four times the actual volume occupied by the molecules.

Where N₀ is no. of molecules of gas in the container

Correction for Intermolecular Attraction:

Consider a molecule A in the interior, far from the boundary. It is surrounded by other molecules equally distributed in all directions. Due to the similar forces (cohesive forces) acting upon it, symmetrically, from all sides the net intermolecular force acting on it is zero.

Now, let us consider another gas molecule B near a wall of the container as shown in the figure. The cohesive force on B is due to adjacent molecules and the adhesive force due to the attraction between atoms of the wall with which molecule collides. Due to this inward pull acts on molecule B. Thus there is a small inward pressure p in addition to observed pressure P. Therefore corrected pressure is (P + p). The inward pressure depends upon

The number of molecules striking per unit area of the wall per unit time (p ∝ n) and
The number of attracting molecules per unit volume which attract the colliding molecule. (p ∝ n). Both these factors are proportional to the density of the gas.

Thus, p ∝ n²

If N be the number of molecules present in volume V of the gas, then n = N / V

where ‘a’ is some constant and V is the volume of the gas. Therefore, corrected pressure ( P + p) becomes

Thus van der walls corrected equation for one mole of gas is

Drawbacks of Van der wall’s Correction:

According to Van der walls, a and b are constants for given gas, but it is found that the constants a and b change in temperature.
The shape of theoretical isotherm plotted using Van der walls equation is different from the experimentally drawn isotherm.

Science > Physics > Themodynamics > Correction to the Ideal Gas Equation

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