A container of capacity 700 mL is filled with two immiscible liquids of volume 200 mL and 500 mL with respective volume expansivities as `1.4xx10^-5//^(@)C` and `2.1xx10^(-5)//^(@)C`. During the heating of the vessel, it is observed that neither any liquid overflows nor any empty space is created. The volume expansivity of the container is

Options

A. `1.9xx10^(-5)//^(@)C`
B. `1.9xx10^(-6)//^(@)C`
C. `1.6xx10^(-5)//^(@)C`
D. `1.6xx10^(-6)//^(@)C`

Explanation

To solve the problem, we need to find the volume expansivity of the container filled with two immiscible liquids. We will follow these steps:

### Step 1: Understand the Given Data
We have:
- Total capacity of the container = 700 mL
- Volume of liquid 1 (V1) = 200 mL with volume expansivity (γ1) = 1.4 x 10^-5 /°C
- Volume of liquid 2 (V2) = 500 mL with volume expansivity (γ2) = 2.1 x 10^-5 /°C

### Step 2: Define the Volume Change
The volume change (ΔV) due to temperature change (ΔT) for each liquid can be expressed as:
- For liquid 1: ΔV1 = V1 * γ1 * ΔT
- For liquid 2: ΔV2 = V2 * γ2 * ΔT

### Step 3: Total Volume Change
Since the total volume does not change (no overflow or empty space), we can express the total volume change as:
ΔV_total = ΔV1 + ΔV2 = 0

### Step 4: Substitute the Volume Changes
Substituting the expressions for ΔV1 and ΔV2:
V1 * γ1 * ΔT + V2 * γ2 * ΔT = 0

### Step 5: Factor Out ΔT
Since ΔT is common in both terms, we can factor it out:
ΔT * (V1 * γ1 + V2 * γ2) = 0

Since ΔT cannot be zero (as we are heating), we can simplify to:
V1 * γ1 + V2 * γ2 = 0

### Step 6: Calculate the Resulting Volume Expansivity
The resulting volume expansivity (γ_container) of the container can be calculated using the formula:
\[ \gamma_{container} = \frac{V1 * γ1 + V2 * γ2}{V1 + V2} \]

### Step 7: Substitute the Values
Now we substitute the values:
\[ \gamma_{container} = \frac{(200 \, \text{mL} * 1.4 \times 10^{-5} /°C) + (500 \, \text{mL} * 2.1 \times 10^{-5} /°C)}{200 \, \text{mL} + 500 \, \text{mL}} \]

### Step 8: Calculate Each Term
Calculating the numerator:
- For liquid 1: \(200 \times 1.4 \times 10^{-5} = 2.8 \times 10^{-3}\)
- For liquid 2: \(500 \times 2.1 \times 10^{-5} = 1.05 \times 10^{-2}\)

Adding these:
\[ 2.8 \times 10^{-3} + 1.05 \times 10^{-2} = 1.33 \times 10^{-2} \]

### Step 9: Calculate the Denominator
The denominator is:
\[ 200 + 500 = 700 \, \text{mL} \]

### Step 10: Final Calculation
Now we can calculate the resulting volume expansivity:
\[ \gamma_{container} = \frac{1.33 \times 10^{-2}}{700} = 1.9 \times 10^{-5} /°C \]

### Conclusion
The volume expansivity of the container is:
\[ \gamma_{container} = 1.9 \times 10^{-5} /°C \]