Finding coefficient of thermal expansion

  • Thread starter Junkwisch
  • Start date
  • #1
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Homework Statement



"see attachment" "q1"

Homework Equations



V=[itex]\frac{-A}{r}+\frac{B}{r^{10}}[/itex]
A=5*10^-30
B=8*10^-121

V=potential energy r=interatomic separation distance

Coefficient of thermal expansion = [itex]\frac{change in L}{L*change in T}[/itex]

The Attempt at a Solution



I have tried making graphs for V vs r however it give me a very weird curve. Furthermore, how do I find the coefficient of thermal expansion from potential energy vs interatomic seperation distance? All I know is that in a V vs r, if the energy well to the right of the absolute zero is larger than that of the left, the CTE is positive and negative for vice versa.

"see graph 1"

I also try to draw this graph on my graphic calculator but it gives me result similar to y= -1/x
 

Attachments

  • q1.png
    q1.png
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  • graph 1.jpg
    graph 1.jpg
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Answers and Replies

  • #2
31
3
In your plot V(r) you also consider [itex]x\leq0[/itex], which is not correct, because distance r must be r>0. The coefficient of thermal (linear) expansion alpha relates the change in longitude L and the change in temperature T, [itex]\frac{\Delta L}{L}=\alpha \Delta T[/itex], in units K-1. So now you have an input for the variation of T...

PS. The energy on the 'left' of the absolute zero is like saying the north of the north pole, or the time before the Big Bang... :O
 

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