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Homework Help: Thermodynamics- thermal expansion coefficient

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three directions. [itex]\beta[/itex]=[itex]\alpha[/itex]x +[itex]\alpha[/itex]y+[itex]\alpha[/itex]z

    For isotopic solid when [itex]\beta[/itex] = 3[itex]\alpha[/itex]

    2. Relevant equations

    [itex]\beta[/itex]=[1/v][dv/dt]p= [itex]\alpha[/itex]x +[itex]\alpha[/itex]y+[itex]\alpha[/itex]z

    3. The attempt at a solution

    I have looked at several websites but can't seem to get it.
    http://www.ami.ac.uk/courses/topics/0197_cte/index.html [Broken]

    I thought about just plugging in the [itex]\beta[/itex] but I don't think that is correct.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 29, 2011 #2


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    Consider a cube, and calculate β in terms of changing edge lengths, and then in terms of α. What do you get? What approximations can you make if α is very small?
  4. Sep 1, 2011 #3
    Thanks for your help.

    I know see what you mean.

    I did of course only understand it until I found this solution online, I wanted to post it here again just in case someone else wants it.

    Let's start with that cube of side 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 β, that:

    Vo′ = Vo(1 + βΔT)

    So this means that β = 3α.
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