Question

Define the coefficient of expansion of solids and deduce the relation between them, In what way if at all, do the coefficients of expansions depend on the unit of length used and the scale of temperature employed ?

please answer with proper explaination !

Class - 11 :- Thermal properties of matter.​

Answer 1

Solution -

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Coefficient Of Expansion In Solids

Say when a solid is heated or cooled , it expands or contracts . Overall , the solid changes its dimensions on being applied a heating/cooling effect / change in temperature .

This can be stated as , a solid changes its dimensions with respect to the change in temperature. This new dimension is proportional to the original dimension .

For solids , generally speaking there exists three types of expansions .

These are -

Linear

Areal

Cubical

The coefficient of these expansions are , α ( Linear ) , β ( Areal ) and γ ( Cubical ) .

The relation between them is -

6α = 3β = 2γ

Derivation of this relation :

The concepts needed to understand :

Basic Differenciation and Logarithms

Let us take a cube of length x units.

As mentioned before , the expansion depends on the coefficient of the type of expansion and the coefficient of expansion .

If this cube expands to a length ,

$\overline{x}$ = x₀ ( 1 + α ∆σ) [ # ∆σ is the change in temperature ]

Let is keep this here and move on to areal expansion .

Suppose that the cube expands such an areal dimension .

Then , $\overline{ \overline{ {x} }$ = $\underbar{x}$ ( 1 + β∆σ )

Now , for any cube ;

Area of one surface = [ Length ]².

So

$\overline{ \overline{ {x} }$ = [ $\overline{x}$ ]²

${\underbar{x}}^2={x_{0}}^2$

So

${\underbar{x}}^2={x_{0}}^2$ ( 1 + β∆σ )

Hence

x₀² ( 1 + α ∆σ) ² = x₀²( 1 + β∆σ )

Now x₀² gets cancelled from both sides.

> 1 + 2α ∆σ = 1 + β∆σ

Cancelling the rest

> β = 2α

For the second part

[ $\overline{x}$ ]³ = x₀( 1 + γ∆σ)

[ $\overline{x}$ ]³ = (x₀)³ [ 1 + γ∆σ ]

> (x₀)³ [ 1 + α∆σ ] ³ = (x₀)³ [ 1 + γ∆σ ]

> 1 + 3α∆σ = 1 + γ∆σ

> γ = 3α

We have got the following relations

β = 2α > α = ß/2

γ = 3α > α = γ/3

> α = ß/2 = γ/3

> 6α = 3β = 2γ

Hence Derived !

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Answer 2

Coefficient of Linear Expansion:

• It is numerically equal to the change in length of a solid of unit length for unit change in temperature.

Coefficient of Areal Expansion:

• It is numerically equal to the change in area of a solid of unit area for unit change in temperature.

Coefficient of Volume Expansion:

• It is numerically equal to the change in volume of a solid of unit volume for unit change in temperature.

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Relationship:

• Let's consider a solid with length, breadth and height each equal to 'l_(0)' :

For ∆T change in temperature:

$l = l_{0}(1 + \alpha \Delta T)$

Similarly for area :

$A = A_{0}(1 + \beta \Delta T)$

$\implies {l}^{2} = {(l_{0})}^{2} (1 + \beta \Delta T)$

$\implies { \bigg \{l_{0}(1 + \alpha \Delta T) \bigg \}}^{2} = {(l_{0})}^{2} (1 + \beta \Delta T)$

$\implies {(1 + \alpha \Delta T)}^{2} = (1 + \beta \Delta T)$

Applying Approximations as $\alpha$ is very small:

$\implies 1 + 2\alpha \Delta T = 1 + \beta \Delta T$

$\implies 2\alpha \Delta T = \beta \Delta T$

$\boxed{ \implies \beta = 2\alpha }$

Similarly, we can prove:

• $\gamma = 3\alpha$ , using volume as l³ and then applying approximations.

So, final relationship is :

$\boxed{ \alpha = \dfrac{ \beta }{2} = \dfrac{ \gamma }{3} }$

Hope It Helps.