- Pierre Louis Dulong (12 February 1785 – 19 July 1838) was a French physicist and chemist, remembered today largely for the law of Dulong and Petit. He worked on the specific heat capacity and the expansion and refractive indices of gases.
- Alexis Thérèse Petit (2 October 1791, 21 June 21 1820) was a French physicist. Petit is known for his work on the efficiencies of air- and steam-engines. His well-known discussions with the French physicist Sadi Carnot, founder of thermodynamics, may have stimulated Carnot in the development theories of thermodynamic efficiency in heat engines.

**Specific Heat:**

- The amount of heat required to raise the temperature of one mole of an element from 287.5 K to 288.5 K is called specific heat of the element. Its S.I. unit is J/mol/ K.
- The product of atomic mass and specific heat of the element is called atomic heat of the element.

**Dulong Petit ‘s Law:**

- The product of specific heat and the atomic mass of an element in the solid state is approximately equal to 6.4. OR Atomic heat of a solid element is nearly equal to 6.4.

**LImitations of Dulong Petit’s Law:**

- This law is applicable to elements which are in solid state.
- This law Is applicable to the heavier element. It is not applicable to lighter elements having high melting points.
- This law gives approximate atomic mass.

**Steps Involved:**

- Find the equivalent mass of the element by any method mentioned in topic equivalent mass.
- Find approximate atomic mass using relation, Approx. atomic mass × Specific heat = 6.4
- Find the valency of the element using relation, Approx. atomic mass = equivalent mass × valency
- Find the nearest whole number for the calculated valency and use this whole number as valency of that element.
- Use following formula to calculate the corrected atomic mass of the element, Corrected atomic mass = Equivalent mass × valency

**Example – 1:**

- The specific heat of a metal A is found to be 0.03; Its equivalent mass is 69.66. calculate the valency and exact atomic mass of an element.
**Solution:**

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.03 = 213.33

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 213.33 / 69.66 = 3.06

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 69.66 × 3 = 208.98 u

Ans: Hence the valency is 3 and its atomic mass is 208.98 u.

**Example – 2:**

- A solid element of equivalent mass 9 has specific heat 1 J/g/K. calculate its atomic mass.
**Solution:**

Specific heat = 1 J/g/K = 1 / 4.188 = 0.2388

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.2388 = 26.80

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 26.80 / 9 = 2.98

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 9 × 3 = 27 u

Ans: Hence the valency is 3 and its atomic mass is 27 u.

**Example – 3:**

- Equivalent mass of barium is 68.68 and its valency is 2. Find its atomic mass.
**Solution:**

Equivalent mass of barium = 68.68, Valency of barium = 2

Atomic mass = Equivalent mass x valency = 68.68 x 2 = 137.6

Thus atomic mass of barium is 137.6.

**Example – 4:**

- The specific heat of a metal was found to be 0.03 and its equivalent mass is 103.5. Find the atomic mass of the metal.
**Solution:**

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.03 = 213.33

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 213.33 / 103.5 = 2.05

∴ Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 103.5 × 2 = 207 u

Ans: atomic mass is 207 u.

**Example – 5:**

- The specific heat of metal M that forms sulphide MS is 0.032 cal g
^{-1}deg^{-1}. What is the equivalent mass of the metal? **Solution:**

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.032 = 200

Now metal sulphide has formula MS

Hence valency of metal is 2

Now, approx. atomic mass = equivalent mass × valency

∴ Equivalent mass = approx. atomic mass / valency = 200 / 2 = 100

Ans: equivalent mass is 100 u.

**Example – 6:**

- The chloride of a metal was found to contain 26.2 % of chlorine. The specific heat of the metal is 0.033. Calculate the atomic mass and equivalent mass of the metal. Also write the molecular formula of metal chloride.
**Solution:**

Consider 100 g of metal chloride

% of chlorine = 26.2

% of metal = 100 – 26.2 = 73.8

Mass of chlorine = 26.2 g, Mass of metal = 73.8 g

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.033 = 194

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 194 / 100 = 1.94

∴ Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 100 × 2 = 200 u

Valency of metal (M) is 2 and that of chlorine is 1. Hence the formula of metal chloride is MCl_{2}.

Ans: the atomic mass of the metal is 200 and its equivalent mass is 100.

the formula of metal chloride is MCl_{2}.

**Example – 7:**

- 1 g of metal having specific heat 0.060205 combines with oxygen to form 1.08 g of oxide. Find atomic mass and valency of the metal. Also, write the molecular formula of the metal oxide.
**Solution:**

Mass of metal = 1 g, Mass of oxide = 1.08 g

Mass of oxygen = 1.08 g – 1 g = 0.08 g

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.060205 = 103.1

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 103.1 / 100 = 1.03

∴ Valency = 1 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 100 × 1 = 200 u

Valency of metal (M) is 1 and that of oxygen is 2. Hence the formula of metal chloride is M_{2}O.

Ans: the atomic mass of the metal is 100 and its valency is 1,

the formula of metal chloride is M_{2}O.

**Example – 8:**

- A metallic oxide contains 47.06 % of oxygen. The specific heat of metal is 0.25. Calculate the atomic mass of the metal. Write molecular formula of metal oxide.
**Solution:**

Consider 100 g of metal oxide

% of oxygen = 47.06

% of metal = 100 – 47.06 = 52.94

Mass of oxygen = 47.06 g, Mass of metal = 52.94 g

Equivalent mass = (52.94 × 8) / 47.06 = 9

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.25 = 25.6

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 25.6 / 9 = 2.8

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 9 × 3 = 27 u

Valency of metal (M) is 3 and that of oxygen is 2. Hence the formula of metal chloride is M_{2}O_{3}.

Ans: the atomic mass of the metal is 27 and the formula of metal oxide is M_{2}O_{3}.

**Example – 9:**

- 0.54 g of a metal combines with 0.48 g of oxygen to form its oxide. Its specific heat is 0.22 cal per deg. What is the atomic mass of the metal?
**Solution:**

Mass of oxygen = 0.48 g, Mass of metal = 0.54 g

Equivalent mass = (0.54 × 8) / 0.48 = 9

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.22 = 29.09

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 29.09 / 9 = 3.2

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 9 × 3 = 27 u

Ans: the atomic mass of the metal is 27.

**Example – 10:**

- 0.45 g of metal displaced 560 ml of hydrogen at STP from acid. specific heat of metal is 0.214. Calculate the equivalent mass and atomic mass of the metal.
**Solution:**

Mass of metal = 0.45 g

Volume of hydrogen = 560 ml = 0.560 dm³ at STP

Mass of 0.560 dm³ of hydrogen at STP = (Molecular mass of gas x volume of gas in dm³) / 22.4

Mass of hydrogen displaced = (2 x 0.560) / 22.4 = 0.05 g

Equivalent mass of metal = 0.45 / 0.05 = 9

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.214 = 29.9

Now, approx. atomic mass = equivalent mass × valency

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 29.9 / 9 = 3.3

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 9 × 3 = 27 u

Ans: the atomic mass of the metal is 27.

Ans: the atomic mass of the metal is 27 and its equivalent mass is 9.

**Example – 11:**

- 0.2160 g of metal, when treated with an excess of dilute sulphuric acid gave 80.4 cc of moist hydrogen measured at 20 °C and 770 mm of pressure. The specific heat of the metal is 0.0955. Calculate the valency and exact atomic mass of the metal The aqueous tension at 20 °C is 17.5 mm.
**Solution:**

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.0955 = 67.02

Now, approx. atomic mass = equivalent mass × valency

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 67.02 / 32.36 = 2.07

∴ Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 32.36 × 2 = 64.72 u

Ans: the atomic mass of the metal is 64.72 and its valency is 2.

**Example – 12:**

- 0.45 g of metal gave 176.6 ml of hydrogen at 23 °C and 743 mm pressure when treated with dilute sulphuric acid. calculate the equivalent mass of the metal. aqueous tension at 23 °C is 21 mm. If the specific heat of the metal is 0.091, calculate the exact atomic mass of the metal.
**Solution:**

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.091 = 70.33

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 70.33 / 32.32 = 2.1

∴ Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 32.32 × 2 = 64.64 u

Ans: the atomic mass of the metal is 64.64 and its valency is 2.

**Example – 13:**

- A metal M forms a volatile chloride, containing 80% of chlorine. The vapour density of the chloride is 66.75. Calculate the exact atomic mass of the element.
**Solution:**

Molecular mass = 2 x Vapour Density = 2 x 66.75 = 133.5

% of chlorine = 80

% of the element = 100 – 80 = 20

Mass of element = 20 g, Mass of chlorine = 80 g

Equivalent mass of metal = (20 x 35.5) / 80 = 8.875

Let valency of the metal be ‘x’, hence the formula of the chloride is MCl_{x}.

Atomic mass of metal = Equivalent mass x valency = 8.875 x

Molecular mass of chloride = Atomic mass of metal + Atomic mass of chlorine × x

Molecular mass of chloride = 8.875 x + 35.5 × x = 133.5

∴ 8.875 x + 35.5 × x = 133.5

∴ 44.375 x = 133.5

∴ x = 133.5 /44.375 = 3.01

∴ Valency = 3 (Corrected to nearest whole number)

Actual atomic mass = Equivalent mass x valency = 8.875 x 3 = 26.625

Thus, exact atomic mass of the element is 26.635

**Example – 14:**

- The specific heat of a metal A is found to be 0.03. 10 g of metal on evaporation of nitric acid gave 18.9 g of pure dry nitrate. Calculate the equivalent mass and exact atomic mass of the metal.
**Solution:**

Mass of metal = 10 g

By double decomposition method

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.03 = 213.33

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 213.33 / 69.66 = 3.06

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 69.66 × 3 = 208.98 u

Ans: the atomic mass of the metal is 208.98 and its equivalent mass is 69.66.

**Example – 15:**

- 1 g of metallic bromide dissolved in water gave with the excess of silver nitrate, 1.88 g of silver bromide. Calculate the accurate atomic mass of the element, if its specific heat is 0.15 cal (atomic mass of Ag is 108 and that of bromine is 80).
**Solution:**

Molecular mass of silver bromide = 108 + 80 = 188 g

Mass of bromine in 1.88 g of silver bromide = (80/180) x 1.8 = 0.8 g

Mass of metallic bromide = 1g, Mass of bromine = 0.8 g

Mass of metal = 1 – 0.8 = 0.2 g

By double decomposition method

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.15 = 42.67

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 42.67 / 20 = 2.1

∴ Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 20 × 2 =40 u

Ans: the atomic mass of the metal is 40.