A flask of mercury is sealed off at `20^@C` and is completely filled with mercury. If the bulk modulus for mercury is 250 Mpa and the coefficient of volume expansion of mercury is `1.82xx10^(-4)//^(@)C` and the expansion of glass is ignored, the pressure of mercury within flask at `100^@C` will be

Options

A. 100MPa
B. 72MPa
C. 36MPa
D. 24MPa

Explanation

To find the pressure of mercury within the flask at \(100^\circ C\), we can follow these steps:

### Step 1: Identify the given values
- Initial temperature, \(T_1 = 20^\circ C\)
- Final temperature, \(T_2 = 100^\circ C\)
- Change in temperature, \(\Delta T = T_2 - T_1 = 100 - 20 = 80^\circ C\)
- Bulk modulus of mercury, \(B = 250 \text{ MPa} = 250 \times 10^6 \text{ Pa}\)
- Coefficient of volume expansion of mercury, \(\beta = 1.82 \times 10^{-4} \text{ }^\circ C^{-1}\)

### Step 2: Calculate the change in volume
The change in volume \(\Delta V\) can be calculated using the formula:
\[ \Delta V = V_0 \cdot \beta \cdot \Delta T \]
Where \(V_0\) is the original volume of mercury in the flask. Since we are looking for the pressure change, we will keep \(V_0\) as a variable.

Substituting the values:
\[ \Delta V = V_0 \cdot (1.82 \times 10^{-4}) \cdot (80) \]
Calculating this gives:
\[ \Delta V = V_0 \cdot 1.456 \times 10^{-2} \]

### Step 3: Use the bulk modulus to find the pressure change
The pressure change \(\Delta P\) is given by the formula:
\[ \Delta P = -B \cdot \frac{\Delta V}{V_0} \]
Substituting \(\Delta V\) from the previous step:
\[ \Delta P = -250 \times 10^6 \cdot \frac{1.456 \times 10^{-2} V_0}{V_0} \]
Here, \(V_0\) cancels out:
\[ \Delta P = -250 \times 10^6 \cdot 1.456 \times 10^{-2} \]
Calculating this gives:
\[ \Delta P = -250 \times 1.456 \times 10^4 \text{ Pa} \]
\[ \Delta P = -364000 \text{ Pa} = -364 \text{ kPa} \]
Since pressure is typically expressed as a positive value, we take the absolute value:
\[ \Delta P = 364 \text{ kPa} = 0.364 \text{ MPa} \]

### Step 4: Final pressure calculation
The initial pressure at \(20^\circ C\) is atmospheric pressure, which is approximately \(101.3 \text{ kPa}\). Therefore, the final pressure at \(100^\circ C\) will be:
\[ P_{final} = P_{initial} + \Delta P \]
Assuming \(P_{initial} \approx 0\) for the change:
\[ P_{final} = 0 + 0.364 \text{ MPa} = 0.364 \text{ MPa} \]

### Final Answer
The pressure of mercury within the flask at \(100^\circ C\) will be approximately \(0.364 \text{ MPa}\).

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