Question

1. study the matter in figure 2. assume the particles are uniformly distributed throughout each object, and particles of the same size have the same mass.

a. in the table below show how the masses, volumes, and densities
compare by adding the symbol <, >, or = to the statement in the second
column.
b. explain your reasoning for each

property relationship reasoning
mass a __ b
a __ c
volume a __ b
a __ c
density a __ b
a __ c

Answer:

Part a: Filling the Relationship Table

Mass

• A vs B: Count the number of particles in A and B. A has fewer particles than B. Since same - sized particles have the same mass, mass of A < mass of B.
• A vs C: Count the number of particles in A and C. A has fewer particles than C. So, mass of A < mass of C.

Volume

• A vs B: Visually, the size (volume) of A is smaller than B (B is taller and wider). So, volume of A < volume of B.
• A vs C: Visually, the size (volume) of A is smaller than C (C is wider). So, volume of A < volume of C.

Density

Density formula: $
ho=\frac{m}{V}$ (mass per unit volume).

• A vs B: Let the mass of one particle be $m_0$, volume of A be $V_A$, volume of B be $V_B$. Let number of particles in A be $n_A$, in B be $n_B$. $m_A = n_A m_0$, $m_B=n_B m_0$. From the figure, the particle density (particles per unit volume) in A and B: Let's assume the volume of a “unit cell” or the packing is such that the ratio $\frac{n_A}{V_A}$ and $\frac{n_B}{V_B}$: Looking at the figures, A and B have particles of the same size (so same $m_0$) and the particle distribution density (particles per unit volume) seems the same? Wait, no, wait. Wait, the problem says “particles of the same size have the same mass” and “uniformly distributed”. Wait, maybe A and B have the same particle density? Wait, no, let's re - examine. Wait, A and B: the particles in A and B are of the same size? Wait, the circles in A and B: are they the same size? Looking at the figure, A's particles and B's particles: maybe same size. Then, density $

ho=\frac{m}{V}=\frac{n\times m_0}{V}$. If the particle density (n/V) is the same for A and B, then $
ho_A=
ho_B$. Wait, but wait, maybe I made a mistake. Wait, let's check the number of particles and volume. Let's say A has volume $V_A$, number of particles $n_A$; B has volume $V_B$, number of particles $n_B$. From the figure, B is larger in volume than A, and has more particles. Let's assume that the particle density (n/V) is the same for A and B. So $
ho_A=
ho_B$.

• A vs C: C has more particles per unit volume (denser packing) than A. So, $

ho_A <
ho_C$? Wait, no. Wait, C's particles are more tightly packed. So, for the same volume, C has more mass. So density of A is less than density of C. Wait, let's do it properly.

Wait, let's define:

• Let the volume of A be $V_A$, mass $m_A$; B be $V_B$, mass $m_B$; C be $V_C$, mass $m_C$.
• Mass: $m = n\times m_0$, where $n$ is number of particles, $m_0$ is mass per particle (same for same - sized particles).
• Volume: from the figure, $V_A < V_B$, $V_A < V_C$.
• Number of particles: $n_A < n_B$, $n_A < n_C$.

Density: $
ho=\frac{m}{V}=\frac{n\times m_0}{V}$.

For A and B: Let's count the number of particles. Let's say in A: let's count the dots. A has, say, 12 particles (3 rows: 4 + 5+3? Wait, the figure: A has 3 rows, first row 4, second 5, third 4? Wait, maybe A: 4 + 5+4 = 13? B: let's see, B has more rows. Wait, maybe it's better to look at the volume and particle count ratio.

Alternatively, the key is:

• Mass:
• A vs B: A has fewer particles than B → $m_A < m_B$
• A vs C: A has fewer particles than C → $m_A < m_C$

• Volume:
• A vs B: A is smaller (shorter and narrower) than B → $V_A < V_B$
• A vs C: A is smaller (narrower) than C → $V_A < V_C$

• Density:
• A vs B: The particle density (particles per unit volume) is the same (since same - sized particles, uniform distribution, and the “packing” looks similar in terms of particles per unit volume). So $

ho_A=
ho_B$ (because $\frac{m_A}{V_A}=\frac{n_A m_0}{V_A}$ and $\frac{m_B}{V_B}=\frac{n_B m_0}{V_B}$, and $\frac{n_A}{V_A}=\frac{n_B}{V_B}$ as the particle distribution density is same).

• A vs C: C has more particles per unit volume (denser) than A → $

ho_A <
ho_C$ (because $\frac{n_A}{V_A}<\frac{n_C}{V_C}$, so $\frac{n_A m_0}{V_A}<\frac{n_C m_0}{V_C}$)

Part b: Reasoning

Mass

• A < B: Mass depends on the number of particles (same - sized particles have same mass). A has fewer particles than B.
• A < C: A has fewer particles than C, so its mass is less.

Volume

• A < B: By visual inspection, the object A is smaller in size (length, width, height) than object B, so its volume is less.
• A < C: Object A is smaller in size than object C, so its volume is less.

Density

• A = B: Density is mass per unit volume ($

ho=\frac{m}{V}$). Both A and B have the same particle density (particles per unit volume) (same - sized particles, uniform distribution, and the ratio of mass to volume is the same as $\frac{n\times m_0}{V}$, and $\frac{n}{V}$ is same for A and B).

• A < C: Density is mass per unit volume. C has more particles per unit volume (denser packing) than A, so for the same volume, C has more mass, leading to a higher density for C than A.

Filled Table

Property	Relationship	Reasoning
Volume	A < B <br> A < C	A is smaller in size than B and C
Density	A = B <br> A < C	A and B have same particle density; C has higher particle density than A