## Related questions with answers

Question

A liquid has a density $\rho$.

Show that the fractional change in density for a change in temperature $\Delta T$ is

$\Delta \rho / \rho=-\beta \Delta T$

Solution

VerifiedAnswered 1 month ago

Answered 1 month ago

The given relation is obtained using the basic relation for density and for the expansion of volume:

$\begin{aligned} &\rho=\dfrac{m}{V}\\ &\Delta \rho=-\dfrac{m}{V^{2}}\Delta V\\ &\Delta \rho=-\dfrac{m}{V^{2}}(V(1+\beta \Delta T)-V)\\ &\Delta \rho=-\dfrac{m}{V}\beta \Delta T\\ &\dfrac{\Delta \rho}{\rho}=\boxed{-\beta \Delta T} \end{aligned}$

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