Question

Derivation of relation between alpha beta and gamma in thermal expansion

Answer 1

We know that,

L =L• (1+αΔT)

α= coefficient of linear expansion

And,

A= A• (1+βΔT)

β= coefficient of aerial expansion

And,

V= V•(1+γΔT)

γ= coefficient of cubical expansion.

So, now

V= V• + γV•ΔT

V= V•(1+γΔT)

L³= L•³ (1+αΔT)³

L³= L•³(1+3αΔT + 3α²ΔT² +α³ΔT³)

L³= L•³(1+3αΔT)

{Neglecting 3α²ΔT² and α³ΔT³ because they are very smaller than 1}

L³= L•³(1+3αΔT)

V= L•³(1+3αΔT)

V•(1+γΔT) = V•(1+3αΔT)

1+γΔT = 1+3αΔT

γΔT = 3αΔT

γ=3α

And β=2α

A= A•(1+βΔT)

L²= L•²(1+αΔT)²

A= L•²(1+2αΔT+α²ΔT²)

A= A•(1+2αΔT)

A•(1+βΔT) = A•(1+2αΔT)

{α²ΔT² Neglecting them due to very smaller volume}

β=2α

α:β:γ=1:2:3

☺

Answer 2

Derivation of the relation between alpha, beta, and gamma in thermal expansion is α:β:γ=1:2:3.

Thermal expansion means the propensity of a body to switch its dimensions (volume, area, and length) when the temperature is switched.

The ratio between alpha, beta, and gamma is α:β:γ.

where,

*α is the linear expansion coefficient.

*β is the area thermal expansion coefficient.

*γ is the volume expansion coefficient.

We know,

*β=2α

*γ=3α

Therefore,

α:β:γ=1:2:3

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