# Density of a porous particle knowing the mass of N particles

I know the mass of N porous particles and their mean diameter. The volume of the particles is N*(PI*Dp^3)/6, where N is the number of
porous particles, PI is 3.14159... and Dp is the mean particle diameter. To which kind of density the ratio mass/volume so obtained corresponds? I need to determine the porous particle density (including pores) and I know the pore volume (in cc/gram).
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Porosity wrote:

Walks like homework, talks like homework........
Ok, It is a "Homework" density.
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Porosity wrote:

V = N*pi*(Dp^3)/6. Assuming 'mass' is total mass (particles + whatever's in the pores):
density including pores: mass/V
mass excluding pores: mass - V/(pore volume)
density excluding pores: (mass - V/(pore volume))/V
Assuming 'mass' is just for the particles, not the pores (closely approximated if only air is filling the pores, i. e. mass_pores << mass_solid)
density including pores: (mass + V/(pore volume))/V
density excluding pores: mass/V
Think.
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