## 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 V_{1} and V_{2} be the volumes of a solid at temperatures t_{1 }and t_{2} respectively, where t_{2 }> t_{1}.

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 V_{t} = V_{0}{1 + γ_{t}}, where V_{0} = 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 γ_{F} = value of γ in Fahrenheit scale, and γ_{C} = 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 l_{0} 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 l_{t}. If the coefficient of linear expansion of the material of the plate is a then, l_{t} = l_{0}(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 l_{0} 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 l_{t}. 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)