Why do the substances expand on heating? Explain using atomic theory.

In matter, interatomic/molecular forces bind the atoms or molecules. The potential energy due to binding force depends on the interatomic separation between the atoms and varies as shown in figure.

In matter, interatomic/molecular forces bind the atoms or molecules.

Let us consider two atoms A1, and A2 at O and P. At a point of minimum energy, the atoms are at a distance r0 apart and temperature is minimum. If we increase the temperature, the atoms absorb the energy and energy of each atom increases. Let at temperature T1, the energy be U1 There are two positions r1(=OL) and r2(=OK) corresponding to energy U1. As both the positions are equal probable, therefore the atom A2 starts vibrating between positions L and K. At temperature T1, the position of atom A2is taken at G (mean position of L and K). Being the potential energy curve unsymmetrical, G is on the right of P and hence at temperature T1, the average distance between A1 and A2is OG = r0 +∆r > r0. Thus with the increase in temperature, the distance between two atoms increases and hence substance expands on heating.

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The density of a solid whose temperature coefficient of cubical expansion is γ is ρ0 at 0°C. What will be its density if it is heated to 6°C?


Let, the mass of the solid be m. 

If V
0 is the volume of solid and ρ0 is density at 0°C, then

                                               ...(1) 

If we increase the temperature then the volume will increase.  

Density will decrease because mass remains constant. 

Let V be the volume and ρ be the density of the same mass of solid at temperature θ°C.

                                                  ...(2) 

From (1) and (2), we have 

                 

If γ is the temperature coefficient of volume expansion of solid then,

                   

Thus,      

                  
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A rod is clamped between two rigid supports so that it neither bends nor expands. Find the thermal compression force and thermal stress set up in the rod when it is heated through 9°C.

Consider a rod of length L, area of cross-section A, clamped between the two rigid supports so that it neither expands nor bends.

Heat the rod so that its temperature increases by θ°C.

As the rod is neither allowed to expand nor bend, therefore a compressive force or stress is set up in the rod. This force or stress is called thermal force or thermal stress. 

Let  be the coefficient of linear expansion and Y be the Young’s modulus of material.

If the ends of rod were free, then its length would increase by, 

                         

The rod being clamped, does not expand but a strain is produced in the rod.

The strain in the rod is given by, 

                     

The stress in the wire is, 

        Thermal stress = Y x strain 

                               =  

Thermal force developed in the rod is, 

Thermal force = Thermal Stress x Area 

                    

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Derive the relation between a and γ. Assume that coefficient of linear expansions are same in all directions.

Let a rectangular cuboid be of sides L0, B0 and H0 be at 0°C.

Therefore the volume of cuboid at 0°C is V
Q = L0B0H0

When cuboid is heated, all of its sides will expand and hence volume.

If straight alpha is the coefficient of linear expansion and γ is coefficient of cubical expansion, then at temperature θ°C, 

                  straight L space equals space straight L subscript 0 left parenthesis 1 plus αθ right parenthesis comma

straight B equals straight B subscript 0 left parenthesis 1 plus αθ right parenthesis comma

straight H space equals space straight H subscript 0 left parenthesis 1 plus αθ right parenthesis comma 

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But,            straight V equals LBH 

Therefore,

 straight V subscript 0 left parenthesis 1 plus γθ right parenthesis space equals space straight L subscript 0 left parenthesis 1 plus αθ right parenthesis straight B subscript 0 left parenthesis 1 plus αθ right parenthesis space straight H subscript 0 left parenthesis 1 plus αθ right parenthesis 

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Since a<<1,  therefore using binomial expansion, 

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rightwards double arrow                  straight gamma equals 3 straight alpha

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Two metal strips A and B, each of length L0 and thickness d at temperature Tbe fastened together so that their ends coincide. The temperature coefficient of linear expansion of A is αA and that of B is αA> αB). Find the radius of curvature of strip when it is heated.

Let us consider two strips A and B each of length L0 are fastened together so that their ends coincide. Let the bimetallic strip be heated through θ°C. Temperature coefficient of linear expansion of A being greater than that of B, the bimetallic strip will bend in the form of arc of circle of radius R with A on convex side and B on concave side.

Let us consider two strips A and B each of length L0 are fastened to
If the strip is heated through 9°C, the length of A and B respectively is,

Let us consider two strips A and B each of length L0 are fastened to
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