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A two-compartment vessel contains a fluid in each section. For the first compartment the fluid density is 1200kg/m3 and for the second is 500 kg/m3. The partition is removed and the fluids mix. Determine the difference in densities of the mixture for fluid with equal masses to that of fluid with equal volumes.    a.) the fluid mass are equal  b.) the fluid volume are equal

To solve this problem, we need to consider the two cases separately: when the masses of the fluids are equal and when the volumes of the fluids are equal.

### Given:
- Density of fluid in the first compartment, \(\rho_1 = 1200 \, \text{kg/m}^3\)
- Density of fluid in the second compartment, \(\rho_2 = 500 \, \text{kg/m}^3\)

### Case 1: **Equal Masses of the Fluids**

Let the mass of each fluid be \(m\).

- **Volume of Fluid 1**: \(V_1 = \frac{m}{\rho_1}\)
- **Volume of Fluid 2**: \(V_2 = \frac{m}{\rho_2}\)

Total mass of the mixture \(M = m + m = 2m\).

Total volume of the mixture \(V = V_1 + V_2 = \frac{m}{\rho_1} + \frac{m}{\rho_2}\).

The density of the mixture \(\rho_m\) can be found by:

\[
\rho_m = \frac{\text{Total Mass}}{\text{Total Volume}} = \frac{2m}{\frac{m}{\rho_1} + \frac{m}{\rho_2}}
\]

Simplifying:

\[
\rho_m = \frac{2m}{m \left( \frac{1}{\rho_1} + \frac{1}{\rho_2} \right)} = \frac{2}{\frac{1}{\rho_1} + \frac{1}{\rho_2}}
\]

\[
\rho_m = \frac{2}{\frac{1}{1200} + \frac{1}{500}} = \frac{2}{\frac{5 + 12}{6000}} = \frac{2}{\frac{17}{6000}} = \frac{2 \times 6000}{17} \approx 705.88 \, \text{kg/m}^3
\]

### Case 2: **Equal Volumes of the Fluids**

Let the volume of each fluid be \(V\).

- **Mass of Fluid 1**: \(m_1 = \rho_1 \times V\)
- **Mass of Fluid 2**: \(m_2 = \rho_2 \times V\)

Total mass of the mixture \(M = m_1 + m_2 = \rho_1 V + \rho_2 V = V(\rho_1 + \rho_2)\).

Total volume of the mixture \(V_{\text{total}} = V + V = 2V\).

The density of the mixture \(\rho_m'\) can be found by:

\[
\rho_m' = \frac{\text{Total Mass}}{\text{Total Volume}} = \frac{V(\rho_1 + \rho_2)}{2V} = \frac{\rho_1 + \rho_2}{2}
\]

\[
\rho_m' = \frac{1200 + 500}{2} = \frac{1700}{2} = 850 \, \text{kg/m}^3
\]

### Difference in Densities

- For equal masses: \(\rho_m \approx 705.88 \, \text{kg/m}^3\)
- For equal volumes: \(\rho_m' = 850 \, \text{kg/m}^3\)

The difference in densities is:

\[
\Delta \rho = \rho_m' - \rho_m \approx 850 - 705.88 \approx 144.12 \, \text{kg/m}^3
\]

### Final Answer:
- **(a)** When the fluid masses are equal: \(\rho_m \approx 705.88 \, \text{kg/m}^3\)
- **(b)** When the fluid volumes are equal: \(\rho_m' = 850 \, \text{kg/m}^3\)
- **Difference**: \(\Delta \rho \approx 144.12 \, \text{kg/m}^3\)

Would you like more details or have any questions?  
Here are 5 related questions for further exploration:

1. How does changing the mass ratio affect the final density of the mixture?
2. What would happen if the fluids had different temperatures when mixed?
3. How would the density change if one fluid were a gas and the other a liquid?
4. Can you derive a general formula for the density of a mixture with arbitrary fluid densities and mass/volume ratios?
5. What assumptions are made in the calculation of mixture densities, and how might they affect real-world scenarios?

**Tip:** Always consider the conservation of mass when dealing with fluid mixtures.

A two-compartment vessel contains a fluid in each section. For the first compartment, the fluid density is 1200kg/m3 and for the second is 500 kg/m3. The partition is removed and the fluids mix. Determine the difference in densities of the mixture for fluid with equal masses to that of fluid with equal volumes. a.) the fluid mass are equal b.) the fluid volume are equal

Explore how to determine the density difference in a mixture of fluids when masses are equal versus when volumes are equal. This problem involves key concepts in fluid mechanics, including density calculations and the conservation of mass. Suitable for students in Grades 11-12.

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