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Question

A liquid has a density ρ\rho.

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

Δρ/ρ=βΔT\Delta \rho / \rho=-\beta \Delta T

Solution

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Answered 1 month ago
Answered 1 month ago

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

ρ=mVΔρ=mV2ΔVΔρ=mV2(V(1+βΔT)V)Δρ=mVβΔTΔρρ=βΔT\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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