Question:easy
Iron powder produced by gas atomization has a tap density and an apparent density of 3.26 g/cm\(^3\) and 2.66 g/cm\(^3\), respectively. The density of solid iron is 7.85 g/cm\(^3\).
Which ONE of the following options is the closest approximation to the degree of densification (in percentage) of this iron powder?
Iron powder produced by gas atomization has a tap density and an apparent density of 3.26 g/cm\(^3\) and 2.66 g/cm\(^3\), respectively. The density of solid iron is 7.85 g/cm\(^3\).
Which ONE of the following options is the closest approximation to the degree of densification (in percentage) of this iron powder?
Which ONE of the following options is the closest approximation to the degree of densification (in percentage) of this iron powder?
Show Hint
Degree of densification = tap density divided by solid density, times 100. Apparent density is not needed for this ratio.
Updated On: Aug 5, 2026
- 41.52
- 81.59
- 33.89
- 75.41
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Reframing the problem with a unit volume:
Imagine we tap-fill exactly $1\ cm^3$ of powder. Since the tap density is $3.26\ g/cm^3$, this $1\ cm^3$ of loosely packed powder contains a mass of $3.26$ g of iron.
The apparent density value of $2.66\ g/cm^3$ describes the powder before tapping and does not enter this calculation, we only need the tap state and the fully solid state.
Step 2: Finding the true solid volume of that same mass:
If that $3.26$ g of iron were melted and cast with zero porosity, its density would be the solid density $7.85\ g/cm^3$.
\[ V_{solid} = \frac{mass}{\rho_{solid}} = \frac{3.26}{7.85} = 0.4153\ cm^3 \]
So the same mass of iron that fills $1\ cm^3$ as tapped powder would occupy only $0.4153\ cm^3$ if it were fully dense with no pores at all.
Step 3: Converting the volume ratio into percent densification:
The degree of densification is the fraction of the tapped volume that is actually solid metal, expressed as a percentage.
\[ \text{Densification} = \frac{V_{solid}}{V_{tapped}} \times 100 = \frac{0.4153}{1} \times 100 = 41.53\% \]
This is the same number as the direct density ratio, because $V_{solid}/V_{tapped}$ and $\rho_{tap}/\rho_{solid}$ are algebraically the same expression, just built up from a physical picture instead of a formula.
Final Answer:
Whether we compare densities directly or trace the volume of a fixed mass of powder, the iron powder is about 41.5 percent as dense as solid iron. \[ \boxed{41.52\%} \]
Imagine we tap-fill exactly $1\ cm^3$ of powder. Since the tap density is $3.26\ g/cm^3$, this $1\ cm^3$ of loosely packed powder contains a mass of $3.26$ g of iron.
The apparent density value of $2.66\ g/cm^3$ describes the powder before tapping and does not enter this calculation, we only need the tap state and the fully solid state.
Step 2: Finding the true solid volume of that same mass:
If that $3.26$ g of iron were melted and cast with zero porosity, its density would be the solid density $7.85\ g/cm^3$.
\[ V_{solid} = \frac{mass}{\rho_{solid}} = \frac{3.26}{7.85} = 0.4153\ cm^3 \]
So the same mass of iron that fills $1\ cm^3$ as tapped powder would occupy only $0.4153\ cm^3$ if it were fully dense with no pores at all.
Step 3: Converting the volume ratio into percent densification:
The degree of densification is the fraction of the tapped volume that is actually solid metal, expressed as a percentage.
\[ \text{Densification} = \frac{V_{solid}}{V_{tapped}} \times 100 = \frac{0.4153}{1} \times 100 = 41.53\% \]
This is the same number as the direct density ratio, because $V_{solid}/V_{tapped}$ and $\rho_{tap}/\rho_{solid}$ are algebraically the same expression, just built up from a physical picture instead of a formula.
Final Answer:
Whether we compare densities directly or trace the volume of a fixed mass of powder, the iron powder is about 41.5 percent as dense as solid iron. \[ \boxed{41.52\%} \]
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