Chapter 2, Problem 2.7P

Interpretation Introduction

Interpretation:

The buoyancy correction equal to 1, when density of an object is $\text{(8}\text{.0 g/mL)}$ should be explained.

Concept introduction:

Buoyancy Correction:

The weight of the object in air is less then weight of the object in vacuum because of buoyancy, this weight variation is corrected using Buoyancy equation it is,

$\text{m}\text{=}\frac{\text{m'}\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{{\text{d}}_{\text{w}}}\right)}{\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{\text{d}}\right)}$

Where,

$\text{m}$ is mass of object in vacuum.

$\text{m'}$ is mass of object in air (read on a balance).

${\text{d}}_{\text{a}}$ is the density of air $\left(\text{0}\text{.001 2 g/mL near I bar and 25°C}\right)$.

${\text{d}}_{\text{w}}$ is the density of calibration weights $\text{(8}\text{.0 g/mL)}$.

$\text{d}$ is the density of the object being weighed.