Thermal expansion — linear, area, volume with coefficient relationships

Question

What are the three types of thermal expansion? Derive the relationship between the coefficients of linear, area, and volume expansion ( $\alpha$ , $\beta$ , $\gamma$ ). Why do railway tracks have gaps between segments? Solve a numerical example.

(CBSE 11 + JEE Main + NEET — derivation + application)

Solution — Step by Step

Type	Applies To	Formula	Coefficient
Linear	Length of a rod/wire	$\Delta L = L_0 \alpha \Delta T$	$\alpha$ (per °C or per K)
Superficial (Area)	Surface area of a plate	$\Delta A = A_0 \beta \Delta T$	$\beta$ (per °C or per K)
Cubical (Volume)	Volume of a solid, liquid, or gas	$\Delta V = V_0 \gamma \Delta T$	$\gamma$ (per °C or per K)

$L_0$ , $A_0$ , $V_0$ are the original length, area, and volume at initial temperature.

For an isotropic solid (expands equally in all directions):

\alpha : \beta : \gamma = 1 : 2 : 3

Or equivalently: $\beta = 2\alpha$ and $\gamma = 3\alpha$

Derivation: Consider a cube of side $L_0$ . After heating by $\Delta T$ :

New side: $L = L_0(1 + \alpha \Delta T)$

New volume: $V = L^3 = L_0^3(1 + \alpha \Delta T)^3 \approx L_0^3(1 + 3\alpha \Delta T)$

(using binomial approximation since $\alpha \Delta T \ll 1$ )

So: $\Delta V = V_0 \cdot 3\alpha \cdot \Delta T$ , which means $\gamma = 3\alpha$ .

Similarly, for area: $A = L^2 = L_0^2(1 + 2\alpha \Delta T)$ , giving $\beta = 2\alpha$ .

Steel has $\alpha \approx 12 \times 10^{-6}$ per °C. A 10 m rail segment experiences a temperature range of 40°C (from winter to summer).

\Delta L = L_0 \alpha \Delta T = 10 \times 12 \times 10^{-6} \times 40 = 0.0048 \text{ m} = 4.8 \text{ mm}

Without gaps, adjacent rails would push against each other and buckle (creating dangerous track distortions). The gap allows for this expansion. Modern continuous welded rails handle this differently — they pre-stress the rail at an intermediate temperature.

A brass rod is 1 m long at 20°C. Find its length at 120°C. ( $\alpha_{brass} = 19 \times 10^{-6}$ per °C)

L = L_0(1 + \alpha \Delta T) = 1.0 \times (1 + 19 \times 10^{-6} \times 100)

L = 1.0 \times 1.0019 = \mathbf{1.0019 \text{ m}}

The expansion is only 1.9 mm — tiny, but significant in precision engineering and large structures.

graph TD
    A["Thermal Expansion"] --> B["Linear: length change"]
    A --> C["Area: surface change"]
    A --> D["Volume: body change"]
    B --> E["alpha"]
    C --> F["beta = 2 alpha"]
    D --> G["gamma = 3 alpha"]
    E --> H["alpha : beta : gamma = 1 : 2 : 3"]
    F --> H
    G --> H
    style A fill:#fbbf24,stroke:#000,stroke-width:2px
    style H fill:#86efac,stroke:#000,stroke-width:2px

Why This Works

When a material is heated, its atoms vibrate more vigorously. The average inter-atomic distance increases because the potential energy curve of atomic interaction is asymmetric (anharmonic). This increase in average spacing manifests as expansion at the macroscopic level.

The 1:2:3 ratio follows from geometry: area is proportional to the square of length, and volume to the cube. Each dimension expands independently by factor $\alpha$ , so two dimensions give $2\alpha$ and three dimensions give $3\alpha$ .

Common Mistake

The classic error: using $\gamma = \alpha$ instead of $\gamma = 3\alpha$ . Students forget the factor of 3 when converting between linear and volume expansion coefficients. Also common: applying $\Delta T$ in Celsius but forgetting that $\alpha$ values are often given per Kelvin. Since a change of 1°C equals a change of 1 K, this actually does not matter for temperature DIFFERENCES — but students waste time converting unnecessarily.

For JEE: the “apparent expansion of a liquid” concept is tested frequently. When a liquid is heated in a container, the container also expands. The apparent expansion = actual expansion of liquid - expansion of container. This is why $\gamma_{apparent} = \gamma_{liquid} - \gamma_{container}$ . This subtlety catches many students off guard.

Question

Solution — Step by Step

Why This Works

Common Mistake

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