Question

Density $\rho$, mass m and volume V are related as $\rho = \dfrac{m}{V}$. Prove that $\gamma = - \dfrac{1}{\rho }\dfrac{{d\rho }}{{dT}}$.

Answer 1

Differentiate the equation of density with respect to temperature. Recall the expression for the volume expansion of the material with small change in the temperature. Solving these two equations, you will get the desired relation.

Formula used:
Volume expansion,$dV = V\gamma dT$
Here, $\gamma$ is the coefficient of volume expansion and $dV$ is the small change in the volume with small change in temperature $dT$.

Complete step by step answer:
We have given that $\rho = \dfrac{m}{V}$. …… (1)
Differentiating both sides with respect to temperature T, we get,
$\dfrac{{d\rho }}{{dT}} = m\dfrac{d}{{dT}}\left( {\dfrac{1}{V}} \right)$
$\Rightarrow \dfrac{{d\rho }}{{dT}} = m\left( { - \dfrac{1}{{{V^2}}}} \right)\dfrac{{dV}}{{dT}}$
$\Rightarrow \dfrac{{d\rho }}{{dT}} = - \dfrac{m}{{{V^2}}}\dfrac{{dV}}{{dT}}$
Rearranging the above equation as,
$\dfrac{{d\rho }}{{dT}} = - \dfrac{m}{V}\dfrac{{dV}}{{V \cdot dT}}$ …… (2)
We have the expression for the volume expansion of the material with change in temperature,
$dV = V\gamma dT$
Here, $\gamma$ is the coefficient of volume expansion and $dV$ is the small change in the volume with small change in temperature $dT$.
Rearranging the above equation as,
$\dfrac{{dV}}{V} = \gamma dT$
Using the above equation and equation (1) in equation (2), we get,
$\dfrac{{d\rho }}{{dT}} = - \rho \gamma$
$\therefore \gamma = - \dfrac{1}{\rho }\dfrac{{d\rho }}{{dT}}$

Hence, it proved that $\gamma = - \dfrac{1}{\rho }\dfrac{{d\rho }}{{dT}}$.

Additional information:
If the temperature of material changes from ${T_1}$ to ${T_2}$, it undergoes change in volume from ${V_1}$ to ${V_2}$. The relationship for the change in the volume of the material with change in the temperature is given as,
${V_2} = {V_1} + {V_1}\gamma \left( {{T_2} - {T_1}} \right)$
$\Rightarrow {V_2} - {V_1} = {V_1}\gamma \left( {{T_2} - {T_1}} \right)$
$\Rightarrow \Delta V = {V_1}\gamma \,\Delta T$
For the small change in the temperature, the above equation is expressed as,
$dV = {V_1}\gamma \,dT$
The coefficient of volume expansion of the material is three times the coefficient of linear expansion of the material. Therefore, we can also write the above equation as, $dV = 3{V_1}\alpha \,dT$, where, $\alpha$ is the coefficient of the linear expansion.

Note: One can also write the density $\rho = \dfrac{m}{V}$as, $\rho = m{V^{ - 2}}$ and take the derivative using the formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = \left( {n - 1} \right){x^{n - 1}}$, where, n is the number. In the solution, the derivative of$\dfrac{1}{V}$ is $- \dfrac{1}{{{V^2}}}$ and not $\dfrac{1}{{{V^2}}}$.