The increase in size of a body on heating is called thermal expansion.

Expansion of Solids

In solids, molecules are held by strong intermolecular forces between the molecules. So, expansion in solids is small. When it is heated, its atoms vibrate about their mean position and their amplitude increases. So, equilibrium distance between them increases, which increases its size.

1. Linear Expansion

If a substance in the form of rod or wire is heated, its length increases. Such type of expansion is called linear expansion. It does not mean that there is no expansion in area and volume. However, the expansion of area and volume in insignificant as compared to the expansion in length.

Let a metal rod has length l₁ at temperature [latex]\theta_1[/latex]. When it is heated up to the temperature [latex]\theta_2[/latex], its length increases & becomes l₂. Thus, the change in length [latex](\Delta l) = l_2-l_1[/latex] & change in temperature [latex](\Delta \theta) = \theta_2 – \theta_1[/latex]. Thus, the change in length [latex](\Delta l)[/latex] is directly proportional to
(i) The original length, i.e.
[latex]\Delta l \propto[/latex]  l₁
The rise in temperature, i.e.
[latex]\Delta l \propto \Delta \theta = (\theta_2 – \theta_1)[/latex]
Combining both, we get,
[latex]\Delta l \propto l_1 (\theta_2 – \theta_1)[/latex]
[latex]\Delta l = \alpha l_1(\theta_2 – \theta_1)[/latex] ……….. (1)
Where, [latex]\alpha[/latex] is a proportionality constant, called coefficient of linear expansion.
From equation (1), we get,
[latex]\alpha = \frac{\Delta l}{l_1 (\theta_2 – \theta_1)}[/latex] ………………………………….. (2)

      = [latex]\frac{change\ in\ length}{original\ length \times change\ in\ temperature}[/latex]
If l₁ = 1 unit, [latex]\theta_2 – \theta_1[/latex] = 1 unit, then, [latex]\alpha = \Delta l[/latex].
Thus, the coefficient of linear expansion may be defined as the change in length to the per unit original length per unit change in temperature. Its value is different for different materials.
From eq^n. (1), we get,
l₂-l₁ = [latex]\alpha l_1 \Delta \theta[/latex]
Or, l₂ = l₁ + l₁[latex]\alpha \Delta \theta[/latex]
[latex]l_2 = l_1 (1+\alpha \Delta \theta)[/latex] …………………………… (3)
Unit of [latex]\alpha[/latex]
The unit of [latex]\alpha[/latex] is ^oC^-1 or K^-1.

2.  Superficial Expansion

The increase in area with rise in temperature is called superficial expansion.

Let us consider, a metal sheet has an area A₁ at temperature [latex]\theta_1[/latex]. When it is heated up to [latex]\theta_2[/latex], its area increases & becomes A₂. Thus, the change in area, [latex]\Delta A[/latex] = (A₂-A₁) & change in temperature [latex]\Delta \theta[/latex] = [latex](\theta_2 – \theta_1)[/latex]. Experimentally, the change in area is:

i.    Directly proportional to the original area. i.e.

[latex]\Delta A \propto A_1[/latex] ……………….. (1)

ii.   Directly proportional to the change in temperature. i.e.

[latex]\Delta A[/latex] ∝ (θ₂-θ₁).  ……………. (2)

Combining both, we get,

[latex]\Delta A \propto A_1(\theta_2 – \theta_1)[/latex]

Or, [latex]\Delta A = \beta A_1(\theta_2 – \theta_1)[/latex] ………………………… (3)

Where, β is the proportionality constant, called coefficient of superficial expansion.

From eq^n. (3), we get,

[latex]\beta = \frac{\Delta A}{A_1(\theta_2 – \theta_1)}[/latex]

= [latex]\frac{\text{Change in Area}}{\text{original area}\times \text{change in temperature}}[/latex]

If A₁ = 1 unit, [latex]\theta_2 – \theta_1[/latex] = 1 unit, then β = [latex]\Delta A[/latex]

Hence, the coefficient of superficial expansion may be defined as the change in area per unit original area per unit change in temperature.

From eq^n. (3), we get,

A₂-A₁ = β A₁ [latex]\Delta \theta[/latex]

Or, A₂ = A₁ (1+β[latex]\Delta \theta[/latex]) 

Unit of β

Its unit is ^oC^-1 or K^-1.

3. Cubical Expansion:

The expansion in volume with rise in temperature is called cubical expansion.

Let us consider, an object has volume [latex]V_1\ at\ θ_1[/latex] temperature. When it is heated upto temperature [latex]θ_2[/latex], its volume increases and becomes [latex]V_2[/latex]. Thus, the change in volume ∆V = [latex](V_2-V_1)[/latex] and change in temperature ∆θ = [latex](θ_2-θ_1)[/latex]. Thus, the change in volume is directly proportional to:

i. The original volume. i.e.

[latex]\Delta V \propto V_1[/latex]

ii. The rise in temperature. i.e.

[latex]\Delta V \propto (\theta_2 – \theta_1)[/latex]

Combining both, we get,

[latex]\Delta V \propto V_1(\theta_2 – \theta_1)[/latex]

[latex]\Delta V = \gamma V_1(\theta_2 – \theta_1) \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots (1)[/latex]

Where, γ is the proportionality constant called coefficient of cubical expansion.

From eqn. (1), we get,

[latex]\gamma = \dfrac{\Delta V}{V_1(\theta_2-\theta_1)} = \dfrac{change\ in\ volume}{original\ volume\ \times change\ in\ temperature}[/latex]

If [latex]V_1[/latex] = 1 unit & [latex](θ_2-θ_1)[/latex] = 1 unit, then, γ = ∆V.

Thus, cubical expansion may be defined as the change in volume per unit original volume per unit change in temperature.

From eqn. (1), we get,

[latex]V_2-V_1 = \gamma V_1 \Delta\theta[/latex]

[latex]V_2 = V_1(1+\gamma\Delta\theta)[/latex]

Unit of γ:

Its unit is [latex]^oC^{-1}\ or\ K^{-1}[/latex]

Relation between coefficients of expansion:

Relation between α and β:

Let us consider, a square plate of side [latex]l_1[/latex] at temperature [latex]θ_1[/latex] & [latex]l_2[/latex] at [latex]θ_2[/latex] as shown in the given figure. Then,

[latex]l_2 = l_1(1+\alpha\Delta\theta)[/latex]

Where, α is the coefficient of linear expansion.

Area of the plate at [latex]θ_1 (A_1) = l_1^2[/latex]

Area of the plate at [latex]θ_2 (A_2) = l_2^2[/latex]

[latex]= l_1^2(1+\alpha\Delta\theta)^2[/latex]

[latex]= A_1(1+\alpha\Delta\theta)^2 \dots\dots (1)[/latex]

Area of the plate at [latex]θ_2[/latex] is also given by:

[latex]A_2 = A_1(1+\beta\Delta\theta) \dots\dots\dots\dots (2)[/latex]

Where, β is the coefficient of superficial expansion.

From eqns. (1) and (2), we get,

[latex]A_1(1+\beta \Delta \theta) = A_1(1+\alpha \Delta \theta)^2[/latex]

Or, [latex](1+\beta \Delta \theta) = 1 + 2\alpha\Delta\theta + \alpha^2\Delta\theta^2[/latex]

Or, [latex]\beta\Delta\theta = 2\alpha\Delta\theta + \alpha^2\Delta\theta^2[/latex]

Since, α is very small, the term containing higher power of α can be neglected. Thus,

[latex]\beta\Delta\theta = 2\alpha\Delta\theta[/latex]

[latex]\beta = 2\alpha \dots\dots\dots\dots\dots\dots (3)[/latex]

Hence, the coefficient of superficial expansion of a solid is twice the value of linear expansion.

Relation between α and γ:

Let us consider, a cubical plate of side [latex]l_1\ at\ θ_1\ \&\ l_2\ at\ θ_2[/latex] as shown in the given figure. Also, let, volume [latex]V_1\ at\ θ_1\ \&\ V_2[/latex] at [latex]θ_2[/latex]. Then,

[latex]l_2 = l_1(1+\alpha\Delta\theta)[/latex]

Where, α is the coefficient of linear expansion.

Volume of the cube at [latex]θ_1 = V_1 = l_1^3[/latex]

Volume of the cube at [latex]θ_2 = V_2 = l_2^3[/latex]

Or, [latex]V_2 = l_1^3(1+\alpha\Delta\theta)^3[/latex]

∴ [latex]V_2 = V_1(1+\alpha\Delta\theta)^3 \dots\dots\dots.\dots (1)[/latex]

Also, the volume of the cube at [latex]θ_2[/latex],

[latex]V_2 = V_1(1+\gamma\Delta\theta) \dots\dots\dots\dots\dots\dots\dots\dots\dots. (2)[/latex]

From eqns. (1) and (2), we get,

[latex]V_1(1+\alpha\Delta\theta)^3 = V_1(1+\gamma\Delta\theta)[/latex]

[latex]1 + 3\alpha\Delta\theta + 3\alpha^2\Delta\theta^2 + \alpha^3\Delta\theta^3 = 1+\gamma\Delta\theta[/latex]

Or, [latex]3\alpha\Delta\theta + 3\alpha^2\Delta\theta^2 + \alpha^3\Delta\theta^3 = \gamma\Delta\theta[/latex]

Since, α is very small, the term containing higher power of α can be neglected. Thus,

[latex]3\alpha\Delta\theta = \gamma\Delta\theta[/latex]

Or, [latex]3\alpha = \gamma[/latex]

Hence, coefficient of cubical expansion of a solid is three times the coefficient of linear expansion.

Relation between α, β and γ:

We have, [latex]\beta = 2\alpha[/latex]

Or, [latex]\alpha = \dfrac{\beta}{2}[/latex]

Also, [latex]\gamma = 3\alpha[/latex]

[latex]\alpha = \dfrac{\gamma}{3}[/latex]

Hence, [latex]\alpha = \dfrac{\beta}{2} = \dfrac{\gamma}{3}[/latex]

[latex]\alpha : \beta : \gamma[/latex] = 1:2:3

Coefficient of Linear Expansion of a solid by Pullinger’s Apparatus

The experimental method to determine coefficient of linear expansion is as shown in figure. It consists of steam chamber having three holes for steam inlet, steam outlet and thermometer. An experimental rod is placed vertically at the middle of steam chamber. From the two ends of the rod, a galvanometer is connected along with the cell as shown in figure. And, spherometer is placed at the top of rod. The galvanometer shows deflection when central leg of spherometer touches the rod. Now, steam is passed from steam inlet and recorded the temperature by thermometer. And the expansion of metal rod is measured by spherometer.
Let,
Initial length of rod = l₁
Initial temperature of rod = [latex]\theta_1^oC[/latex]
Final temperature of rod = [latex]\theta_2^oC[/latex]
Initial reading of spherometer = R₁
Final reading of spherometer = R₂
Now, from the definition of linear expansion
[latex]\alpha = \frac{Change\ in\ length}{Original\ length\ \times\ change\ in\ temperature}[/latex]

[latex]\alpha = \frac{R_2 – R_1}{l_1(\theta_2 – \theta_1)}[/latex]
This is the required expression for coefficient of linear expansivity from Pullinger’s apparatus.

Differential Expansion

Different metals have different coefficient of expansion. The difference in expansion of the bodies when they are heated by the same range of temperature is called differential expansion.

Let us consider, two rods A & B having the lengths L₁ & L₁^‘ respectively at temperature [latex]\theta_1[/latex]. They are heated upto temperature [latex]\theta_2[/latex] such that their length becomes L₂ & L₂^‘ respectively. Then, we have,

At temperature [latex]\theta_1, d_1 = L_1 – L_1^{‘}[/latex] …….. (i)

Similarly, at temperature [latex]\theta_2, d_2 = L_2 – L_2^{‘}[/latex] ………… (ii)

We also have,

L₂ = L₁ [latex](1+\alpha \Delta \theta)[/latex] ………………… (iii)

And, L₂^‘= L₁^‘ [latex](1+\alpha^{‘} \Delta \theta)[/latex] ……………….. (iv) Where, α and α^‘ are the coefficients of linear expansivities of A and B.

Now, From eq^n. (ii),

d₂ = L₂ – L₂^‘

     = [latex]L_1(1+\alpha \Delta \theta) – L_1^{‘}(1+\alpha^{‘} \Delta \theta)[/latex]

            = L₁ + L₁ α[latex]\Delta \theta[/latex] – L₁^‘ – L₁^‘ α^‘[latex]\Delta \theta[/latex]

            = (L₁ – L₁^‘) + (L₁α – L₁^‘ α^‘)[latex]\Delta \theta[/latex]

            = d₁ + (L₁α – L₁^‘ α^‘)[latex]\Delta \theta[/latex]

[latex]\therefore[/latex] d₂ – d₁ = (L₁α – L₁^‘ α^‘)[latex]\Delta \theta[/latex]

This is an expression for differential expansion.

If the original lengths & materials of the bodies are so chosen that, they always have the same expansion, then,

[latex]d_2 – d_1 = 0[/latex]

(L₁α – L₁^‘ α^‘)[latex]\Delta \theta[/latex] = 0

Then, [latex]\frac{L_1}{L_1^{‘}} = \frac{\alpha^{‘}}{\alpha}[/latex]

i.e., if the lengths of the rods are inversely proportional to their linear expansivity, then, the difference between the lengths is constant at all temperature.

Force or Stress set up in rod due to expansion or contraction:

Consider, a rod R of length [latex]l_1[/latex] fixed between two rigid supports P & Q at temperature [latex]θ_1[/latex]. When the rod is heated upto [latex]θ_2[/latex], it expands and its length increases & becomes [latex]l_2[/latex]. Due to increase in temperature, the rod tries to expand but will not be able to expand due to rigid ends, as a result, a force or tension is produced, which would compress the rod.

The corresponding length [latex](l_2)[/latex] is given by:

[latex]l_2 = l_1(1+\alpha\Delta\theta)[/latex]

Where, α = Coeff. Of linear expansion of material of rod.

Or, [latex]l_2 – l_1 = l_1\alpha\Delta\theta[/latex]

∴ Increase in length (e) = [latex]l_1\alpha\Delta\theta[/latex] ……………………….. (1)

If Y be the Young’s modulus of elasticity of the material of rod, then,

[latex]Y = \dfrac{Stress}{Strain}[/latex]

Stress: The ratio of force to the area is called stress.

Strain: The ratio of increase in length to the original length is called strain. i.e.

[latex]Stress = \dfrac{Force\ (F)}{Area\ (A)} = \dfrac{F}{A} = \dfrac{Tension}{Area}[/latex]

[latex]Strain = \dfrac{Increase\ in\ length\ (e)}{Original\ length\ (l_1)} = \dfrac{e}{l_1}[/latex]

[latex]= \dfrac{l_1\alpha\Delta\theta}{l_1}[/latex] [∵ From eqn. (1)]

[latex]= \alpha\Delta\theta[/latex]

Young’s Modulus (Y) = [latex]\dfrac{Stress}{Strain}[/latex]

∴ [latex]Y = \dfrac{F/A}{\alpha\Delta\theta}[/latex]

[latex]\dfrac{F}{A} = Y\alpha\Delta\theta[/latex]

∴ [latex]F = YA\alpha\Delta\theta[/latex]

Expansion of Liquid

Liquids, just like solids expands on heating & contracts on cooling. The expansion on liquid is more than that of the solid. It is because, the intermolecular force of attraction on liquid is less than that of solid. So, the molecules vibrate quickly on liquid than solid.
For the study of expansion of liquid, we have to study about the cubical expansion. We know, liquid have no shape but have definite volume. When it is kept in a vessel & then heated, the volume of liquid as well as the vessel increases.

Real & Apparent expansion of a liquid

The expansion of liquid on rise in temperature is always associated with the expansion of vessel. The expansion, which we observe is called apparent expansion. It is always less than real expansion of the liquid.

Let us take a round bottom flask made of glass with long narrow stem. Let, it be filled with liquid upto A. If the bulb is immersed in a hot liquid bath, the liquid level comes down to B. This is because the vessel is heated before the liquid. So, it expands first and the level of liquid decreases. After this, the liquid is heated and its level rises to C.
So, real expansion of liquid = increase in volume from B to C
Or, real expansion of liquid = increase in volume from B to A + increase in volume from A to C
Or, real expansion of liquid = expansion of vessel + apparent expansion of liquid
i.e., V_BC = V_AB + V_AC ………………… (i)

Relation between Coefficient of Real and Apparent Expansion

Let us consider, a liquid contained in a vessel at room temperature [latex]\theta_1[/latex]. If V be the original volume, then, suppose, the liquid is heated to temperature [latex]\theta_2[/latex]. So,
Coefficient of real expansion of liquid [latex](\gamma_r) = \frac{Real\ increase\ in\ volume}{Original\ volume \times rise\ in\ temperature}[/latex]

= [latex]\frac{V_{BC}}{V\times \Delta \theta}[/latex]

[latex]V_{BC} = V\gamma_r \Delta \theta[/latex]

Also, coefficient of apparent expansion of liquid [latex](\gamma_a)[/latex]

= [latex]\frac{\text{Apparent increase in volume}}{Original\ volume \times rise\ in\ temperature}[/latex]

= [latex]\frac{V_{AC}}{V\times \Delta \theta}[/latex]

Or, V_AC = [latex]V\gamma_a \Delta \theta[/latex]
Now, coefficient of volume expansion of glass vessel [latex](\gamma_g)[/latex]

= [latex]\frac{\text{Increase in volume of vessel}}{Original\ volume \times rise\ in\ temperature}[/latex]

= [latex]\frac{V_{AB}}{V\times \Delta \theta}[/latex]

Or, V_AB = [latex]V\gamma_g \Delta \theta[/latex]
So, from equation (i),
V_BC = V_AB + V_AC
Or, [latex]V\gamma_r \Delta = V\gamma_g \Delta \theta + V\gamma_a \Delta \theta[/latex]

Or, [latex]\gamma_r = \gamma_g + \gamma_a[/latex]

This is the required relation between coefficients of real and apparent expansivity of a liquid.

Determination of the co-efficient of real expansion of a liquid by Dulong and Petit’s method

The apparatus consists of a U- shaped tube which contains liquid whose coefficient of real expansion is to be determined. The vertical limbs are surrounded by glass jackets. One jacket is placed with ice water and another jacket is placed at steam temperature by passing steam continuously through it. The horizontal part of the tube is continuously cooled by spraying cold water on it to prevent the flow of heat due to conduction across it. Ice water and steam is passed for a long time until it attains steady state separately and the temperatures are noted.

Calculation

Let, the temperatures of the cold and hot limbs be [latex]\theta_1^oC[/latex] and [latex]\theta_2^oC[/latex] respectively. For the liquid to be in steady state, the pressure at B must be equal to the pressure at C.

If h₁ and h₂ be the heights of the liquid columbs and [latex]\rho_1[/latex] and [latex]\rho_2[/latex] be the densities of the liquid at temperatures [latex]\theta_1^oC[/latex] and [latex]\theta_2^oC[/latex] respectively, then,

[latex]h_1\rho_1 g = h_2\rho_2 g[/latex] ………………….. (i)

 [latex]\frac{h_1}{h_2} = \frac{\rho_2}{\rho_1}[/latex] ……………….. (ii)

But, density at [latex]\theta_2[/latex],

[latex]\rho_2 = \frac{\rho_1}{1+\gamma \Delta \theta}[/latex]   ………………….. (iii)

And from eq^n. (ii), we get,
[latex]\frac{h_1}{h_2} = \frac{\rho_1}{\rho_1(1+\gamma \Delta \theta)}[/latex]

 Or, [latex]h_2 = h_1 + h_1 \gamma \Delta \theta[/latex]

[latex]\therefore h_2 – h_1 = h_1 \gamma \Delta \theta[/latex]

Or, [latex]\gamma = \frac{h_2 – h_1}{h_1\Delta \theta}[/latex]

By knowing the values of RHS, the coefficient of real expansion of liquid can be determined.