# Fractional change of density

I'm asked to prove that the fractional change of density of a fluid ($$\frac{\Delta\rho}{\rho_0}$$) is so that $$\frac{\Delta\rho}{\rho_0}=-\beta\,\Delta{T},$$ where $$\beta$$ is the volumetric coefficient of expansion and given that $$\Delta{V}=\beta\,V_0\,\Delta{T}$$. However, my attempt at a solution, which starts from $$\Delta{\rho}$$:

$$\Delta{\rho}=m\,\left[\frac{1}{V_0+V_0\,\beta\,\Delta{T}}-\frac{1}{V_0}\right]=\frac{-\beta\,\Delta{T}}{1+\beta\,\Delta{T}}\,\rho_0 \implies\frac{\Delta\rho}{\rho_0}=\frac{-\beta\,\Delta{T}}{1+\beta\,\Delta{T}}.$$ ends up in something different. Did I do something wrong? or is it the problem that's poorly written?

Your analysis is not wrong. Note that the general definition of the coefficient of expansion is $$\beta \equiv \frac{1}{V} \frac{dV}{dT}$$; so when you write it as $$\beta \simeq \frac{1}{V} \frac{\Delta V}{\Delta T}$$, you're implicitly assuming that the temperature variation $$\Delta T$$, and hence $$\Delta V/V$$ is small$${}^*$$.

With that in mind, using $$\Delta V = \beta V \Delta T$$, your final equation can be rewritten as: $$\frac{\Delta \rho}{\rho_0} = - \frac{\beta \Delta T}{1+\Delta{V}/V}$$ As mentioned previously, the change in volume due to expansion is much less than the initial volume of the object so $$\Delta V/V << 1$$, meaning that you can neglect $$\Delta V/V$$ in the denomenator in comparison with $$1$$; resulting in: $$\frac{\Delta \rho}{\rho}\simeq-\beta \Delta T$$

* To be more precise, from the definition of $$\beta$$ we have: $$\frac {dV}{V}=\beta dT$$ so: $$\int_{V_1}^{V_2}\frac {dV}{V}=\int_{T_1}^{T_2}\beta dT$$ Now provided that the temperature variation $$\Delta T = T_2 - T_1$$ is sufficiently small so that we can neglect the variations of $$\beta$$ in the $$[T_1,T_2]$$ interval, $$\beta$$ (and $$V$$) can be approximately pulled out of the integral, resulting in: $$\frac{\Delta V}{V} \simeq \beta \Delta T$$ which is the formula you were using.

Note that you can also rearrange the definition of $$\beta$$ in terms of the density $$\rho=\frac m V$$ to get that approximation:

$$\beta \equiv \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{p} = \frac{1}{m / \rho}\left(\frac{\partial (\frac {m}{\rho})}{\partial T}\right)_{p} = \frac {\rho}{m}\left(-\frac{m}{\rho^2}\right)\left(\frac{\partial \rho}{\partial T}\right)_{p}=-\frac{1}{\rho}\left(\frac{\partial \rho}{\partial T}\right)_{p}$$

So

$$\beta \simeq-\frac{1}{\rho} \frac{\Delta \rho}{\Delta T} \implies \frac{\Delta\rho}{\rho}=-\beta\,\Delta{T}$$

• Thanks! It was during a course where one-variable calculus wasn't an option though, so partial derivatives wouldn't have even been an option :( Commented Dec 7, 2021 at 11:03