Understanding the Problem: Mixing Substances and Density

This problem involves two substances with unknown densities, \(\rho_1\) and \(\rho_2\). We are given information about the average relative density of the mixture when the substances are combined in two different ways: first, by mixing equal volumes, and second, by mixing equal masses. Our goal is to determine the individual values of \(\rho_1\) and \(\rho_2\) based on this information.

Relative density is the ratio of the density of a substance to the density of a reference substance (usually water, which has a density of approximately 1 g/cm³ or 1000 kg/m³). In this problem, the given values (4 and 3) likely represent the average density of the mixture relative to water, or simply the average density value in specific units where the density of water is 1. We will proceed by setting up equations based on the definition of average density for each mixing scenario.

Average Density when Mixing by Equal Volumes

Let's consider mixing equal volumes of the two substances. Let the volume of substance 1 be \(V\) and the volume of substance 2 also be \(V\).

• The mass of substance 1 is \(m_1 = \rho_1 \times V\).
• The mass of substance 2 is \(m_2 = \rho_2 \times V\).
• The total mass of the mixture is \(M_{total} = m_1 + m_2 = \rho_1 V + \rho_2 V = (\rho_1 + \rho_2)V\).
• The total volume of the mixture is \(V_{total} = V + V = 2V\).
• The average density of the mixture is \(\rho_{avg, V} = \frac{M_{total}}{V_{total}} = \frac{(\rho_1 + \rho_2)V}{2V} = \frac{\rho_1 + \rho_2}{2}\).

We are given that the relative density (average density) when mixed in equal volumes is 4. So, we can write our first equation:

\(\frac{\rho_1 + \rho_2}{2} = 4\)

\(\rho_1 + \rho_2 = 8\) (Equation 1)

Average Density when Mixing by Equal Masses

Now, let's consider mixing equal masses of the two substances. Let the mass of substance 1 be \(M\) and the mass of substance 2 also be \(M\).

• The volume of substance 1 is \(V_1 = \frac{M}{\rho_1}\).
• The volume of substance 2 is \(V_2 = \frac{M}{\rho_2}\).
• The total mass of the mixture is \(M_{total} = M + M = 2M\).
• The total volume of the mixture is \(V_{total} = V_1 + V_2 = \frac{M}{\rho_1} + \frac{M}{\rho_2} = M \left( \frac{1}{\rho_1} + \frac{1}{\rho_2} \right) = M \left( \frac{\rho_2 + \rho_1}{\rho_1 \rho_2} \right)\).
• The average density of the mixture is \(\rho_{avg, M} = \frac{M_{total}}{V_{total}} = \frac{2M}{M \left( \frac{\rho_1 + \rho_2}{\rho_1 \rho_2} \right)} = \frac{2 \rho_1 \rho_2}{\rho_1 + \rho_2}\).

We are given that the relative density (average density) when mixed in equal masses is 3. So, we can write our second equation:

\(\frac{2 \rho_1 \rho_2}{\rho_1 + \rho_2} = 3\) (Equation 2)

Solving for Densities \(\rho_1\) and \(\rho_2\)

We now have a system of two equations with two unknowns (\(\rho_1\) and \(\rho_2\)):

1. \(\rho_1 + \rho_2 = 8\)
2. \(\frac{2 \rho_1 \rho_2}{\rho_1 + \rho_2} = 3\)

We can use Equation 1 to substitute \((\rho_1 + \rho_2)\) in Equation 2:

\(\frac{2 \rho_1 \rho_2}{8} = 3\)

\(\frac{\rho_1 \rho_2}{4} = 3\)

\(\rho_1 \rho_2 = 12\) (Equation 3)

Now we need to find two numbers (\(\rho_1\) and \(\rho_2\)) whose sum is 8 and whose product is 12. We can think of \(\rho_1\) and \(\rho_2\) as the roots of a quadratic equation. A quadratic equation with roots \(x_1\) and \(x_2\) is given by \(x^2 - (x_1 + x_2)x + x_1 x_2 = 0\).

Substituting our values, the quadratic equation is:

\(x^2 - 8x + 12 = 0\)

We can solve this quadratic equation by factoring:

\((x - 6)(x - 2) = 0\)

The roots are \(x = 6\) and \(x = 2\).

This means that \(\{\rho_1, \rho_2\}\) is either \(\{6, 2\}\) or \(\{2, 6\}\). The question asks for the values of \(\rho_1\) and \(\rho_2\) respectively. Checking the options, the pair (6, 2) is available.

Verifying the Solution

Let's verify if \(\rho_1 = 6\) and \(\rho_2 = 2\) satisfy both original conditions:

• Equal Volume Mixing: Average density = \(\frac{\rho_1 + \rho_2}{2} = \frac{6 + 2}{2} = \frac{8}{2} = 4\). This matches the given information.
• Equal Mass Mixing: Average density = \(\frac{2 \rho_1 \rho_2}{\rho_1 + \rho_2} = \frac{2 \times 6 \times 2}{6 + 2} = \frac{24}{8} = 3\). This also matches the given information.

The values \(\rho_1 = 6\) and \(\rho_2 = 2\) satisfy both conditions.

Final Answer

The values of \(\rho_1\) and \(\rho_2\) are 6 and 2, respectively.

Revision Table: Key Concepts

Additional Information: Density and Mixtures

Density is a fundamental physical property of a substance that relates its mass to its volume. It is an intensive property, meaning it does not depend on the amount of substance present. When different substances are mixed, the density of the resulting mixture depends on the densities of the components and the proportions in which they are mixed, either by volume or by mass.

Mixing by equal volumes leads to a simple arithmetic average of densities because the total volume is the sum of the individual volumes, and the total mass is the sum of individual masses, each proportional to density times volume. The formula \(\rho_{avg, V} = \frac{\rho_1 V_1 + \rho_2 V_2}{V_1 + V_2}\) simplifies to \(\frac{\rho_1 + \rho_2}{2}\) when \(V_1 = V_2 = V\).

Mixing by equal masses is slightly more complex because the volumes occupied by equal masses of different substances will be different if their densities are different. The total volume is the sum of individual volumes, which are calculated as mass divided by density (\(V = \frac{M}{\rho}\)). The formula \(\rho_{avg, M} = \frac{M_1 + M_2}{V_1 + V_2} = \frac{M_1 + M_2}{\frac{M_1}{\rho_1} + \frac{M_2}{\rho_2}}\) simplifies to \(\frac{2M}{\frac{M}{\rho_1} + \frac{M}{\rho_2}} = \frac{2}{\frac{1}{\rho_1} + \frac{1}{\rho_2}} = \frac{2 \rho_1 \rho_2}{\rho_1 + \rho_2}\) when \(M_1 = M_2 = M\). This average is related to the harmonic mean.