A glass flask of volume one liter at `0^(@)C` is filled, level full of mercury at this temperature. The flask and mercury are now heated to `100^(@)C`. How much mercury will spill out if coefficient of volume expansion of mercury is `1.82 xx 10^(-4)//^(@)C` and linear expansion of glass is `0.1 xx 10^(-4)//^(@)C` respectively?

Options

A. `21.2 cc`
B. `15.2 cc`
C. `1.52 cc`
D. `2.12 cc`

Explanation

To solve the problem step by step, we need to calculate the change in volume of both the mercury and the glass flask when they are heated from \(0^\circ C\) to \(100^\circ C\).

### Step 1: Determine the change in volume of the glass flask.

The formula for the change in volume due to thermal expansion is given by:

\[ \Delta V = V_0 \cdot \beta \cdot \Delta T \]

Where:
- \(\Delta V\) = change in volume
- \(V_0\) = initial volume
- \(\beta\) = coefficient of volume expansion
- \(\Delta T\) = change in temperature

For glass, the coefficient of linear expansion \(\alpha\) is given as \(0.1 \times 10^{-4} / ^\circ C\). The coefficient of volume expansion \(\beta\) for glass is approximately \(3\alpha\).

Calculating \(\beta\) for glass:

\[ \beta_{\text{glass}} = 3 \cdot (0.1 \times 10^{-4}) = 0.3 \times 10^{-4} / ^\circ C \]

Now, substituting the values into the volume change formula:

\[ V_0 = 1 \text{ liter} = 1000 \text{ cm}^3 \]
\[ \Delta T = 100^\circ C - 0^\circ C = 100^\circ C \]

Now we calculate \(\Delta V_g\):

\[ \Delta V_g = 1000 \cdot (0.3 \times 10^{-4}) \cdot 100 = 3 \times 10^{-3} \text{ cm}^3 \]

### Step 2: Determine the change in volume of the mercury.

The coefficient of volume expansion for mercury is given as \(1.82 \times 10^{-4} / ^\circ C\).

Using the same formula for the change in volume:

\[ \Delta V_m = V_0 \cdot \beta_{\text{mercury}} \cdot \Delta T \]

Substituting the values:

\[ \Delta V_m = 1000 \cdot (1.82 \times 10^{-4}) \cdot 100 = 1.82 \times 10^{-2} \text{ cm}^3 \]

### Step 3: Calculate the volume of mercury that spills out.

The volume of mercury that spills out is the difference between the change in volume of the mercury and the change in volume of the glass:

\[ V_{\text{spilled}} = \Delta V_m - \Delta V_g \]

Substituting the values we calculated:

\[ V_{\text{spilled}} = (1.82 \times 10^{-2}) - (3 \times 10^{-3}) = 1.52 \times 10^{-2} \text{ cm}^3 \]

### Step 4: Convert the volume from cm³ to mL.

Since \(1 \text{ cm}^3 = 1 \text{ mL}\):

\[ V_{\text{spilled}} = 15.2 \text{ mL} \]

### Final Answer:

The amount of mercury that will spill out is **15.2 mL**.

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