Question

Consider the liquid in a barometer whose coefficient of volume expansion is$\mathbf{6}\mathbf{.6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{\text{/}}}^{\mathbf{\text{0}}}\mathbf{\text{C.}}$Find the relative change in the liquid’s height if the temperature changes by$\mathbf{12}{\mathbf{}}^{\mathbf{\text{0}}}\mathbf{\text{C}}$while the pressure remains constant. Neglect the expansion of the glass tube.

Short answer

The relative change in height of a liquid is$7.92\times {10}^{-3}.$

Step-by-step solution

Stating thegiven data

1. The coefficient of volume expansion of liquid is.$\beta =6.6\times {10}^{-4}\text{/°C}$
2. The temperature rise is$\Delta T=12\text{°C.}$

Understanding the concept of thermal expansion

Thermal radiation is the process of transferring heat through electromagnetic radiation that is produced by the thermal motion of matter particles. We use the concept of volume thermal expansion. As the expansion of the glass tube is neglected, the relative change in height will be equal to the relative change in volume.

Formulae:

Volume expansion of the body due to thermal radiation,$\Delta V=V\times \beta \times \Delta T$ …(i)

Volume of the body in terms of area and length, $V=A\times L$.…(ii)

Calculation of height change

Now, as the expansion of the glass tube is to be neglected, the area of cross-section gets cancelled.

So, the relative change in height,$\Delta H$of a liquid becomes

$\frac{\Delta L}{L}=\frac{\Delta V}{V}$ …(iii)

Using equation (i), the relative change in volume can be given as

$\frac{\Delta V}{V}=\beta \Delta T$ …(iv)

So, the relative change in height using equations (iii) and (iv) can be given as

$\begin{array}{c}\Delta H=\frac{\Delta L}{L}\\ =\beta \Delta T\\ =6.6\times {10}^{-4}\text{/°C}\times 12\text{°C(usinggivenvalues)}\\ =7.92\times {10}^{-3}\end{array}$

Hence, the relative change in the value of height is.$7.92\times {10}^{-3}$