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 that relates the density of a substance to its initial density, the coefficient of volume expansion, and the change in temperature.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Coefficient of volume expansion of mercury, \(\gamma = 0.18 \times 10^{-3} \, ^\circ C^{-1}\)
- Density of mercury at \(0^\circ C\), \(D_0 = 13.6 \, \text{g/cc}\)
- Change in temperature, \(\Delta T = 200^\circ C - 0^\circ C = 200^\circ C\)

2. **Use the Density Change Formula:**
The density at a new temperature can be calculated using the formula:
\[ D = \frac{D_0}{1 + \gamma \Delta T} \]
where \(D\) is the density at the new temperature, \(D_0\) is the initial density, \(\gamma\) is the coefficient of volume expansion, and \(\Delta T\) is the change in temperature.

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

4. **Calculate the Change in Volume:**
First, calculate \((0.18 \times 10^{-3}) \times 200\):
\[ 0.18 \times 10^{-3} \times 200 = 0.036 \]

5. **Calculate the Denominator:**
Now substitute this value back into the equation:
\[ D = \frac{13.6}{1 + 0.036} = \frac{13.6}{1.036} \]

6. **Perform the Final Calculation:**
Now, calculate \(D\):
\[ D \approx \frac{13.6}{1.036} \approx 13.12 \, \text{g/cc} \]

### Final Answer:
The density of mercury at \(200^\circ C\) is approximately \(13.12 \, \text{g/cc}\).

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