A 1-L flask contains some mercury. It is found that at different temperature, the volume of air inside the flask remains the same. What is the volume of mercury in the flask, given that the coefficient of linear expansion of glass`=9xx10^(-6)//^(@)C` and the coefficient of volume expansion of `Hg=1.8xx10^(-4)//^(@)C` ?

Answer

To solve the problem, we need to find the volume of mercury in a 1-L flask, given that the volume of air inside the flask remains constant at different temperatures. We are provided with the coefficients of linear expansion for glass and the coefficient of volume expansion for mercury.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a 1-L flask (which is equivalent to 1000 cm³) containing mercury.
- The volume of air inside the flask remains constant despite temperature changes, which implies that the volume expansion of the glass flask and the volume expansion of mercury must balance each other.

2. **Volume Expansion Formula**:
- The change in volume due to thermal expansion can be expressed as:
\[ \Delta V = V \cdot \gamma \cdot \Delta \theta \]
- Where \( \Delta V \) is the change in volume, \( V \) is the original volume, \( \gamma \) is the coefficient of volume expansion, and \( \Delta \theta \) is the change in temperature.

3. **For Glass Flask**:
- The coefficient of volume expansion for glass (\( \gamma_g \)) is related to the coefficient of linear expansion (\( \alpha_g \)) by the formula:
\[ \gamma_g = 3 \alpha_g \]
- Given \( \alpha_g = 9 \times 10^{-6} \, ^\circ C^{-1} \), we can calculate \( \gamma_g \):
\[ \gamma_g = 3 \times (9 \times 10^{-6}) = 27 \times 10^{-6} \, ^\circ C^{-1} \]

4. **For Mercury**:
- The volume expansion of mercury can be expressed as:
\[ \Delta V_{Hg} = V_{Hg} \cdot \gamma_{Hg} \cdot \Delta \theta \]
- Given \( \gamma_{Hg} = 1.8 \times 10^{-4} \, ^\circ C^{-1} \).

5. **Setting Up the Equation**:
- Since the volume of air remains constant, the change in volume of the glass flask must equal the change in volume of mercury:
\[ \Delta V_g = \Delta V_{Hg} \]
- This leads to the equation:
\[ V_g \cdot \gamma_g \cdot \Delta \theta = V_{Hg} \cdot \gamma_{Hg} \cdot \Delta \theta \]
- The \( \Delta \theta \) cancels out, giving:
\[ V_g \cdot \gamma_g = V_{Hg} \cdot \gamma_{Hg} \]

6. **Solving for Volume of Mercury**:
- Rearranging the equation to find \( V_{Hg} \):
\[ V_{Hg} = \frac{V_g \cdot \gamma_g}{\gamma_{Hg}} \]
- Substituting \( V_g = 1 \, L = 1000 \, cm^3 \):
\[ V_{Hg} = \frac{1000 \cdot (27 \times 10^{-6})}{1.8 \times 10^{-4}} \]

7. **Calculating the Volume**:
- Performing the calculation:
\[ V_{Hg} = \frac{1000 \cdot 27 \times 10^{-6}}{1.8 \times 10^{-4}} = \frac{27 \times 10^{-3}}{1.8 \times 10^{-4}} = \frac{27}{1.8} \times 10^{1} = 15 \times 10^{1} = 150 \, cm^3 \]

### Final Answer:
The volume of mercury in the flask is **150 cm³**.