Why Do Solids Expand When Heated?

  • Thread starter Thread starter Amith2006
  • Start date Start date
  • Tags Tags
    Expansion Solids
Click For Summary
Solids expand when heated primarily due to an increase in the kinetic energy of atoms, which leads to greater atomic vibrations. The potential energy curve of neighboring atoms is asymmetric around their equilibrium distance, which is crucial for understanding thermal expansion. The harmonic approximation fails to account for this asymmetry, resulting in a zero thermal expansion coefficient, as it assumes purely quadratic interactions. In contrast, anharmonicity allows for the expansion of solids as energy increases, leading to a change in the mean positions of atoms. Some solids exhibit negligible thermal expansion due to nearly parabolic atomic interactions over a range of energies.
Amith2006
Messages
416
Reaction score
2
Sir,
Why does solid expand on heating?
The answer to this question given in my book is “The potential energy curve is asymmetric about the equilibrium distance between neighbouring atoms. Is it right?
I think solid expands due to increase in kinetic energy of atoms.
 
Physics news on Phys.org
Amith2006 said:
Sir,
Why does solid expand on heating?
The answer to this question given in my book is “The potential energy curve is asymmetric about the equilibrium distance between neighbouring atoms. Is it right?
Yes it is. The most eggregious flaw of the harmonic approximation to the treatment of solids is that it leads to zero thermal coefficient of expansion. Without anharmonicity (or asymmetry, or deviation from purely quadratic interaction terms), you can't solve this problem.
I think solid expands due to increase in kinetic energy of atoms.
The classical oscillator turning points are separated by distances smaller than typical interatomic spacings. In the harmonic approximation, it's possible to make atoms oscillate much faster (irrespective of phase differences) about their mean positions, without having to change the distance between these mean positions. Alternatively, one can see that with a symmetric potential, no matter what the energy (and hence turning points), the mean position is always that of the minimum.

Now there are a few solids with virtually zero thermal expansion coefficient. These are solids where the interaction between atoms is almost exactly parabolic over a significant range of energies.

Edit : Look here for a picture - http://www.mse.uiuc.edu/info/mse182/t79.html
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 33 ·
2
Replies
33
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
7K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K