Question

Density ‘$\rho $’, mass ‘m’ and volume ‘V’ are related as $\rho =\dfrac{m}{v}$. What is the coefficient of volume expansion of liquid is
$\begin{align}
  & \text{A}\text{. }\gamma \text{=}\dfrac{d\rho }{dt} \\
 & \text{B}\text{. }\gamma =3\dfrac{d\rho }{dt} \\
 & \text{C}\text{. }\gamma =\dfrac{1}{\rho }\dfrac{d\rho }{dt} \\
 & \text{D}\text{. }\gamma =-\dfrac{1}{\rho }\dfrac{d\rho }{dT} \\
\end{align}$

Answer 1

Hint: We are given that density is mass by unit volume and are asked to find the coefficient of volume expansion. To find the coefficient of volume expansion we need to take the derivative of density with respect to temperature. By substituting the change in volume due to temperature in the differential equation we can find the coefficient of expansion of volume.
Formula used:
Change in volume due to temperature, ${{V}_{t}}=V\left( 1+\gamma \Delta T \right)$

Complete answer:
In the question we are given the relation between density, mass and volume as,
$\rho =\dfrac{m}{V}$, where ‘$\rho $’ is the density, ‘m’ is the mass and ‘V’ is the volume.
We need to find the coefficient of volume of expansion of the liquid.
For that let us differentiate the given relation with respect to temperature.
$\dfrac{d\rho }{dT}=\dfrac{d\left( \left( \dfrac{m}{V} \right) \right)}{dT}$
Since we are differentiating with respect to temperature, we know that mass of any substance is not affected by the change in temperature whereas density and volume will vary.
Hence the mass of the liquid will remain constant.
$\Rightarrow \dfrac{d\rho }{dT}=m\left( \dfrac{d\left( \dfrac{1}{V} \right)}{dT} \right)$
By differentiating we get,
$\Rightarrow \dfrac{d\rho }{dT}=m\left( -\dfrac{1}{{{V}^{2}}} \right)\left( \dfrac{dV}{dT} \right)$
We can rearrange the above equation as,
$\Rightarrow \dfrac{d\rho }{dT}=-\left( \dfrac{m}{V} \right)\left( \dfrac{dV}{VdT} \right)$
We know that change in volume due to temperature is expressed as,
$\begin{align}
  & {{V}_{t}}=V\left( 1+\gamma \Delta T \right) \\
 & \Rightarrow {{V}_{t}}=V+V\gamma \Delta T \\
\end{align}$
$\Rightarrow {{V}_{t}}-V=V\gamma \Delta T$
We know that ${{V}_{t}}-V=\Delta V$
Therefore,
$\begin{align}
  & \Rightarrow \Delta V=V\gamma \Delta T \\
 & \Rightarrow \gamma =\dfrac{\Delta V}{V\Delta T} \\
\end{align}$
Since $\Delta V$ is equivalent to $dV$ and $\Delta T$ is equivalent to $dT$ we can substitute the above equation in $\dfrac{d\rho }{dT}=-\left( \dfrac{m}{V} \right)\left( \dfrac{dV}{VdT} \right)$
$\Rightarrow \dfrac{d\rho }{dT}=-\left( \dfrac{m}{V} \right)\gamma $
It is given that $\dfrac{m}{V}=\rho $, hence
$\Rightarrow \dfrac{d\rho }{dT}=-\rho \gamma $
Therefore, we get coefficient of expansion of volume as,
$\gamma =-\dfrac{1}{\rho }\dfrac{d\rho }{dT}$

Hence the correct answer is option D.

Note:
Thermal expansion is the tendency of a body to change its volume in response to a change in temperature.
Volume expansion is a type of thermal expansion where the volume of a body expands with respect to temperature.
Linear expansion and area expansion are other two kinds of thermal expansion.
While solving the question, we take
$\dfrac{d\left( \dfrac{1}{V} \right)}{dT}=-\dfrac{1}{{{V}^{2}}}\left( \dfrac{dV}{dT} \right)$
This is because, we know that derivative of $\left( \dfrac{1}{x} \right)=-\dfrac{1}{{{x}^{2}}}$
Hence derivative of $\left( \dfrac{1}{V} \right)$ with respect to temperature will be, $\left(-\dfrac{1}{{{V}^{2}}}\dfrac{dV}{dT} \right)$.