Coefficient of volume expansion of mercury is `0.18xx10^(-3)//.^(@)C`. If the density of mercury at `0^(@)C` is 13.6g/cc, then its density at `200^(@)C` is

Options

A. 13.11 g/cc
B. 52.11 g/cc
C. 16.11 g/cc
D. 26.11 g/cc

Answer

To find the density of mercury at \(200^\circ C\), we can use the formula for the change in density due to thermal expansion. The relationship between density and temperature can be expressed as follows:

1. **Identify the given values:**
- Coefficient of volume expansion of mercury, \(\beta = 0.18 \times 10^{-3} \, ^\circ C^{-1}\)
- Initial density of mercury at \(0^\circ C\), \(D_0 = 13.6 \, \text{g/cm}^3\)
- Change in temperature, \(\Delta T = 200^\circ C - 0^\circ C = 200^\circ C\)

2. **Calculate the change in volume due to thermal expansion:**
The volume expansion can be calculated using the formula:
\[ V = V_0 (1 + \beta \Delta T) \]
However, we are interested in density, so we will use the relationship between density and volume:
\[ D = \frac{D_0}{1 + \beta \Delta T} \]

3. **Substitute the values into the density formula:**
\[ D = \frac{13.6}{1 + (0.18 \times 10^{-3} \times 200)} \]

4. **Calculate the term in the denominator:**
\[ 0.18 \times 10^{-3} \times 200 = 0.036 \]
Thus,
\[ 1 + 0.036 = 1.036 \]

5. **Calculate the new density:**
\[ D = \frac{13.6}{1.036} \]

6. **Perform the division:**
\[ D \approx 13.12 \, \text{g/cm}^3 \]

So, the density of mercury at \(200^\circ C\) is approximately \(13.12 \, \text{g/cm}^3\).