Expansion Of Solid And Liquids – Coefficient Of Volume Expansion
Coefficient Of Volume Expansion Definition: The increase in volume for unit rise in temperature for a unit volume of a solid is called the coefficient of volume expansion of the material of that solid.
The coefficient of volume expansion is denoted by γ.
Let V1 and V2 be the volumes of a solid at temperatures t1 and t2 respectively, where t2 > t1.
Proceeding in a way similar, we get, the coefficient of volume expansion,
⇒ \(\gamma=\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}\)
= \(\frac{\text { increase in volume }}{\text { initial volume } \times \text { rise in temperature }}\)….(2)
Hence, \(V_2-V_1\)=\(\gamma V_1\left(t_2-t_1\right)\)
or, \(V_2=V_1\left\{1+\gamma\left(t_2-t_1\right)\right\}\)
Read and Learn More: Class 11 Physics Notes
When initial temperature = 0 and the final temperature = t, then Vt = V0{1 + γt}, where V0 = volume at zero temperature.
- Coefficient of volume expansion γ is not a constant For precise measurements, the volume at 0 C is to be taken as the initial volume.
- Value of γ does not depend on the unit of volume.
- value of γ depends on the unit of temperature. Unit of γ is °C-1 or °F-1. The change in temperature by 1°F = 5/9 °C change in temperature.
∴ \(\gamma_F=\frac{5}{9} \gamma_C\)
where γF = value of γ in Fahrenheit scale, and γC = value of γ in Celsius scale.
Properties Of Bulk Matter – Expansion Of Solid And Liquids Relation Among The Three Coefficients of Expansion
Relation between α and β: Consider a square metal plate with each side of length l0 at its initial temperature. So the surface area of the plate at that temperature, \(S_0=l_0^2\)
Let the temperature of the plate be increased by t so that the length of each side changes to lt. If the coefficient of linear expansion of the material of the plate is a then, lt = l0(l+αt).
The surface area of the plate now becomes, \(S_t=l_t^2=\left\{l_0(1+\alpha t)\right\}^2\)
= \(l_0^2\left(1+2 \alpha t+\alpha^2 t^2\right)=S_0(1+2 \alpha t)\)…..(1)
[neglecting α² t² as a <<1]
But from the definition of the coefficient of surface expression β for the material of the plate,
⇒ \(S_t=S_0(1+\beta t)\)….(2)
∴ Comparing equation (1) and (2) we get, = 2 ….(3)
Relation between α and γ: Let us consider, a metal cube with each side of length l0 at the initial temperature. So, the volume of the cube at that temperature, \(V_0=l_0^3 \text {. }\)
Let the temperature of the cube be increased by t, so that the length of each side changes to lt. If the coefficient of linear expansion of the material of the cube is α, then, \(l_t=l_0(1+\alpha t)\)
The volume of the cube now becomes, \(V_t =l_t^3=l_0^3(1+\alpha t)^3=V_0\left(1+3 \alpha t+3 \alpha^2 t^2+\alpha^3 t^3\right)\)
= \(V_0(1+3 \alpha t) \quad \text { [neglecting } \alpha^2 t^2 \text { and } \alpha^3 t^3 \text { as } \alpha \ll 1]\)….(4)
But from the definition of the coefficient of volume expression γ for the material of the cube,
⇒ \(V_t=V_0(1+\gamma t)\)…(5)
∴ From equations (4) and (5),
γ = 3α ….(6)
So, for the same material,
α = \(\frac{\beta}{2}=\frac{\gamma}{3}\)….(7)
This is the relation among α, β and γ. The relation can also be expressed as α:β:γ= 1:2:3 ….(8)
To find the relation among α, β and γ using calculus: Suppose the length of a solid changes from l to (l+dl) when it is heated from a temperature θ to a temperature (θ + dθ). From the definition of coefficient of linear expansion,
α = \(\frac{\text { increase in length }}{\text { initial length } \times \text { rise in temperature }}=\frac{1}{l} \cdot \frac{d l}{d \theta}\)
Similarly, if the initial surface area and initial volume are S and V respectively,
coefficient of surface expansion, \(\beta=\frac{1}{S} \cdot \frac{d S}{d \theta}\)
and coefficient of volume expansion, \(\gamma=\frac{1}{V} \cdot \frac{d V}{d \theta}\)
Let us consider a cube of length l at temperature θ. Therefore, area of each surface of the cube, S= l² ….(9)
and volume of the cube, v = l³ …..(10)
Differentiating equation (9) with respect to θ, we get
⇒ \(\frac{d S}{d \theta}=2 l \frac{d l}{d \theta}\)
Now, \(\beta=\frac{1}{S} \frac{d S}{d \theta}=\frac{1}{l} \cdot 2 l \cdot \frac{d l}{d \theta}=2 \cdot \frac{1}{l} \frac{d l}{d \theta}=2 \alpha\)
∴ \(\beta=2 \alpha\)
Differentiating equation (10) with respect to θ,
⇒ \(\frac{d V}{d \theta}=3 l^2 \frac{d l}{d \theta}\)
⇒ \(\gamma=\frac{1}{V} \frac{d V}{d \theta}=\frac{1}{l^3} \cdot 3 l^2 \frac{d l}{d \theta}=3 \cdot \frac{1}{l} \frac{d l}{d \theta}=3 \alpha\)
∴ \(\gamma =3 \alpha\)…(12)
Hence, from equation (1) and (12) \(\alpha=\frac{\beta}{2}=\frac{\gamma}{3}\)….(13)