Question

Write the relation between coefficient of volume expansion and coefficient of linear expansion.

Answer 1

Step 1: We have a coefficient of linear expansion $\left(\alpha \right)$

Let's have a cube having length $L$ and volume $V$

So, the volume of the cube $=V=L^3$

Let the cube is heated by $∆T$ then its new Length will be $L^{\text{'}}$

$L\text{'}=L\left(1+\alpha ∆T\right)...............\left(1\right)$

Where $\alpha =coefficientoflinearexpansion$

$L\text{'}=L\left(1+\alpha ∆T\right)$

$L\text{'}-L=\alpha L∆T$

$∆L=\alpha L∆T$

So, $\alpha =\frac{∆L}{L∆T}.........\left(2\right)$

Step 2: We have a coefficient of volume expansion $\left(\gamma \right)$

Let the cube is heated by $∆T$ then its new volume will be $V\text{'}$

$V\text{'}=V\left(1+\gamma ∆T\right)$

$V\text{'}-V=V\gamma ∆T$

$\gamma =\frac{∆V}{V∆T}...........\left(3\right)$

The new volume of the cube $=V\text{'}={\left(L\text{'}\right)}^3$

$V\text{'}={\left(L\left(1+\alpha ∆T\right)\right)}^3$

$V\text{'}=L^3\left(1+3\alpha ∆T\right)(Since\left[{\left(1+x\right)}^n=1+nx\right])$

$V\text{'}=V\left(1+3\alpha ∆T\right)$

$V\text{'}-V=3\alpha ∆T$

$∆V=3\alpha V∆T$

Step 3: To find the relation between coefficient of volume expansion and coefficient of linear expansion

$∆V=3\alpha V∆T$

$3\alpha =\frac{∆V}{V∆T}............\left(4\right)$

From equation $\left(3\right)and\left(4\right)$

$\gamma =3\alpha$

Hence, the relation between the coefficient of volume expansion and the coefficient of linear expansion is $\gamma =3\alpha$.