Density and Pressure — Question 1

Question 1

Equal volumes and equal masses give different mixtures

Two incompressible liquids have densities $800$ and $1200\,\mathrm{kg/m^3}$. They mix without volume change. Compare an equal-volume mixture with an equal-mass mixture.

1. Derive and evaluate both mixture densities.

2. Which mixture is denser, and why is an arithmetic mean inappropriate for equal masses?

Question 1 - Solution

Equal volumes and equal masses give different mixtures

(a) Derivation and calculation

For equal volumes $V$,$$\rho_v=\frac{\rho_1V+\rho_2V}{2V}=\frac{800+1200}{2}=\boxed{1000\,\mathrm{kg/m^3}}.$$For equal masses $m$,$$\rho_m=\frac{2m}{m/\rho_1+m/\rho_2}=\frac2{1/800+1/1200}=\boxed{960\,\mathrm{kg/m^3}}.$$

(b) Result and interpretation

$$\boxed{\rho_v-\rho_m=40\,\mathrm{kg/m^3}}.$$$$m_1=m_2:\quad V_1/V_2=\rho_2/\rho_1=1.50.$$$$\rho_{\mathrm{mix}}=\frac{\rho_1V_1+\rho_2V_2}{V_1+V_2}.$$The less dense liquid occupies more volume when masses are equal, so the arithmetic density mean is inappropriate.

Physical check

Both mixture densities lie between 800 and 1200 kg/m$^3$.