# 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. This is known as Thermal Expansion.

## Linear Expansion

Consider a metallic rod of length $L_1$ is initially at the temperature $θ_1$. On heating to $θ_2$, its length changes to $L_2$. It has been found that the change in length $ΔL$ $(L_2-L_1)$ 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 $A_1$ is initially at the temperature $θ_1$. On heating to $θ_2$, its area changes to $A_2$. It has been found that the change in area $ΔA$ $(A_2-A_1)$ 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 $V_1$ is initially at the temperature $θ_1$. On heating to $θ_2$, its volume changes to $V_2$. It has been found that the change in volume $ΔV$ $(V_2-V_1)$ 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 $A_1$ and Length $L_1$ at temperature $θ_1$. $∴A_1=L_1^2$ On heating to $θ_2$, its area becomes $A_2$ and length becomes $L_2$. $∴A_2=L_2^2$ 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

Comparing $(1)$ and $(2)$, we get, $β=2α$ $∴α=\frac{β}{2}$

### Relation between Linear and Cubical Expansivities

Consider a metallic cube of volume $V_1$ and Length $L_1$ at temperature $θ_1$. $∴V_1=L_1^3$ On heating to $θ_2$, its volume becomes $V_2$ and length becomes $L_2$. $∴V_2=L_2^3$ 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

Comparing $(3)$ 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 $V_1$ at temperature $θ_1$. On heating to $θ_2$, its volume becomes $V_2$. Then, we have, $V_2=V_1(1+γΔθ)$ where, $γ$ is cubical expansivity and $Δθ$ is the change in temperature.

Also, $V\;\text{(Volume)}=\frac{m;\text{(mass)}}{ρ\text{(density)}}$ Therefore, $\frac{m}{ρ_2}=\frac{m}{ρ_1}(1+γΔθ)$ where, $m$ is the mass of the cube and $ρ_1$ and $ρ_2$ are its densities at $θ_1$ and $θ_2$ temperatures respectively. $∴\frac{1}{ρ_2}=\frac{1}{ρ_1}(1+γΔθ)$ $ρ_2=\frac{ρ_1}{1+γΔθ}$