# Thermal Expansion

When substances are heated, their kinetic energy increases. Due to the gain in kinetic energy, the amplitude and frequency of vibration of their molecules increases. Hence, the intermolecular spacing increases and the substances expand.

## Linear Expansion

Consider a metallic rod of length L₁ is initially at the temperature θ₁. On heating to θ₂, its length changes to L₂. It has been found that the change in length ΔL (L₂-L₁) is directly proportional to the;

• Original length i.e. $ΔL∝L_1 \text{ ____(1)}$
• ​Change in temperature i.e. $ΔL∝Δθ \text{ ____(2)}$

Combining (1) and (2), $ΔL∝L_1Δθ$ $∴ΔL=αL_1Δθ \text{ ___(3)}$ where, α is known as coefficient of linear expansion or linear expansivity. and, $α=\frac{ΔL}{L_1Δθ}$ Thus, linear expansivity (α) is defined as the ratio of change in length per unit original length per degree change in temperature. From (3), $ΔL=αL_1Δθ$ $L_2-L_1=αL_1Δθ$ $L_2=L_1+αL_1Δθ$ $L_2=L_1(1+αΔθ)$ This gives the final length of the rod.

## Superficial Expansion

Consider a metallic sheet of area A₁ is initially at the temperature θ₁. On heating to θ₂, its area changes to A₂. It has been found that the change in area ΔA (A₂-A₁) is directly proportional to the;

• Original area i.e. $ΔA∝A_1 \text{ ____(i)}$
• Change in temperature i.e. $ΔA∝Δθ \text{ ____(ii)}$

Combining (i) and (ii), $ΔA∝A_1Δθ$ $∴ΔA=βA_1Δθ \text{ ___(iii)}$ where, β is known as coefficient of superficial expansion or superficial expansivity. and, $β=\frac{ΔA}{A_1Δθ}$ Thus, superficial expansivity (β) is defined as the ratio of change in area per unit original area per degree change in temperature. From (iii), $ΔA=βA_1Δθ$ $A_2-A_1=βA_1Δθ$ $A_2=A_2+βA_1Δθ$ $A_2=A_1(1+βΔθ)$ This gives the final area of the sheet.

## Cubical Expansion

Consider a metallic cube of volume V₁ is initially at the temperature θ₁. On heating to θ₂, its volume changes to V₂. It has been found that the change in volume ΔV (V₂-V₁) is directly proportional to the;

• Original volume i.e. $ΔV∝V_1 \text{ ____(a)}$
• Change in temperature i.e. $ΔV∝Δθ \text{ ____(b)}$

Combining (a) and (b), $ΔV∝V_1Δθ$ $∴ΔV=γV_1Δθ \text{ ___(c)}$ where, γ is known as coefficient of cubical expansion or cubical expansivity. and, $γ=\frac{ΔV}{V_1Δθ}$ Thus, cubical expansivity (γ) is defined as the ratio of change in volume per unit original volume per degree change in temperature. From (c), $ΔV=γV_1Δθ$ $V_2-V_1=γV_1Δθ$ $V_2=V_1+γV_1Δθ$ $V_2=V_1(1+γΔθ)$ This gives the final volume of the cube.

## Relation between Linear, Superficial and Cubical Expansivities

### Relation between Linear and Superficial Expansivities

​Consider a metallic sheet of area A₁ and Length L₁ at temperature θ₁. $∴A_1=L_1^2$ On heating to θ₂, its area becomes A₂ and length becomes L₂. $∴A_2=L_2^2$ We have, $L_2=L_1(1+αΔθ)$ Here, α = Linear Expansivity $L_2^2=L_1^2(1+αΔθ)^2$ $L_2^2=L_1^2(1+2αΔθ+α^2Δθ^2)$ Here, α is a small quantity and higher power of it is very small which can be neglected. $∴L_2^2=L_1^2(1+2αΔθ)$ $A_2=A_1(1+2αΔθ) \text{ ___(1)}$ Also, $A_2=A_1(1+βΔθ) \text{ ___(2)}$ Here, β = Superficial Expansivity
​Comapring (1) and (2), we get, $β=2α$ $∴α=\frac{β}{2}$

### Relation between Linear and Cubical Expansivities

Consider a metallic cube of volume V₁ and Length L₁ at temperature θ₁. $∴V_1=L_1^3$ On heating to θ₂, its volume becomes V₂ and length becomes L₂. $∴V_2=L_2^3$ We have, $L_2=L_1(1+αΔθ)$ $L_2^3=L_1^3(1+αΔθ)^3$ $L_2^3=L_1^3(1+3αΔθ+3α^2Δθ^2+α^3Δθ^3)$ Here, α is a small quantity and higher power of it is very small which can be neglected. $∴L_2^3=L_1^3(1+3αΔθ)$ $V_2=V_1(1+3αΔθ) \text{ ___(3)}$ Also, $V_2=V_1(1+γΔθ) \text{ ___(4)}$ Here, γ = Cubical Expansivity
​Comapring (3) and (4), we get, $γ=3α$ $∴α=\frac{γ}{3}$

​Thus, we have the relation, $α=\frac{β}{2}=\frac{γ}{3}$

## Change of Density of a Substance with Temperature

​Consider a metallic cube of volume V₁ at temperature θ₁. On heating to θ₂, its volume becomes V₂. Then, we have, $V_2=V_1(1+γΔθ)$ where, γ is cubical expansivity and Δθ is the change in temperature. also, $V(Volume)=\frac{m(mass)}{ρ(density)}$ Therefore, $\frac{m}{ρ_2}=\frac{m}{ρ_1}(1+γΔθ)$ where, m is the mass of the cube and ρ₁ and ρ₂ are its densities at θ₁ and θ₂ temperature respectively. $∴\frac{1}{ρ_2}=\frac{1}{ρ_1}(1+γΔθ)$ $ρ_2=\frac{ρ_1}{1+γΔθ}$