I was studying $$\text{Thermal Expansion}$$ from H.C. Verma - Concepts of Physics 2 (if you have the book, page 6 - topic 23.10). If you don't have the book, this is the concerned matter.

Note: If you know this stuff, you can jump to the end.

Consider a rod at temperature $$T$$ and suppose its length at this temperature is $$L$$. As the temperature is changed to $$T+\Delta T$$, the length is changed to $$L+\Delta L$$. We define average coefficient of linear expansion in the temperature range $$\Delta T$$ as $\overline{\alpha}=\frac{1}{L}\frac{\Delta L}{\Delta T}$ The coefficient of linear expansion at temperature $$T$$ is limit of average coefficient as $$\Delta T\rightarrow0$$, i.e., $\alpha=\displaystyle\lim_{\Delta T\rightarrow0}\frac{1}{L}\frac{\Delta L}{\Delta T}=\frac{1}{L}\frac{dL}{dT}$ Suppose the length of a rod is $$L_{0}$$ at $$0^{\circ}C$$ and $$L_{\theta}$$ at temperature $$\theta$$ measured in Celsius. If $$\alpha$$ is small and constant over the given temperature interval, $\alpha=\frac{L_{\theta}-L_{0}}{L_{0}\theta}$ or $L_{\theta}=L_{0}(1+\alpha\theta)$ The coefficient of volume expansion $$\gamma$$ is defined in a similar way. If $$V$$ is the volume of a body at temperature $$T$$, the coefficient of volume expansion at temperature $$T$$ is $\gamma=\frac{1}{V}\frac{dV}{dT}$ It is also known as coefficient of cubical expansion.

If $$V_{0}$$ and $$V_{\theta}$$ denote the volumes at $$0^{\circ}C$$ and $$\theta$$ (measured in Celsius) respectively and $$\gamma$$ is small and constant over the given temperature range, we have $V_{\theta}=V_{0}(1+\gamma\theta)$

It is easy to show that $$\gamma=3\alpha$$.

My question is: How do you prove that $$\gamma=3\alpha$$?

Note for those who jumped: $$\gamma$$ is the coefficient of cubical expansion, and $$\alpha$$ is the coefficient of linear expansion.

Note by Omkar Kulkarni
3 years, 2 months ago

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We consider a cube of metal with each side equal to L.

If we heat it up, the length of each side will change from L to (L+ΔL), where ΔL = αLΔT.

In other words, the new length is:

L′ = L + αLΔT = L(1 + αΔT)

So the new volume is:

Vo′ = (L′)^3 = [L(1 + αΔT)]^3 = (L^3)(1 + 3αΔT + 3(αΔT)² + (αΔT)^3) = Vo(1 + 3αΔT + 3(αΔT)² + (αΔT)^3)

Now we make an approximation. Since αΔT is quite small compared to 1, we can safely say that the "3(αΔT)²" term and the "(αΔT)^3" term are negligible compared to the "3αΔT" term. Therefore, to a good approximation,

Vo′ = Vo(1 + 3αΔT)

But we already know, from the definition of γ, that:

Vo′ = Vo(1 + γΔT)

So this means that γ = 3α.

I could have written the solution in Latex but I am short on time. (NTSE upcoming)

- 3 years, 2 months ago

Wait did you mean $$\overline{\alpha}$$ instead of $$\alpha$$?

- 3 years, 2 months ago

Ohh okay! Makes sense now :) Thanks! All the best for NTSE.

- 3 years, 2 months ago

Y=3a

- 3 years, 1 month ago

What?

- 3 years, 1 month ago