A glass beaker holds exactly `1 L "at" 0^@ C`
(a) What is its volume at `50^@ C` ?
(b) If the beaker is filled with mercury at `0^@ C`, what volume of mercury overflows ehen the temperature is `50^@ C` ? `alpha _ g = 8.3 xx 10^-6 per ^@ C` and` gamma_Hg = 1.82 xx 10^-4 per ^@ C`.

Explanation

To solve the problem step by step, we will address both parts (a) and (b) of the question.

### Part (a): Volume of the Glass Beaker at 50°C

1. **Identify the initial conditions**:
- Initial volume \( V_0 = 1 \, \text{L} \)
- Initial temperature \( T_0 = 0^\circ C \)
- Final temperature \( T = 50^\circ C \)

2. **Use the formula for volume expansion**:
The formula for the final volume \( V \) after thermal expansion is given by:
\[ V = V_0 \left(1 + \gamma (T - T_0)\right) \]
where \( \gamma \) is the coefficient of volume expansion.

3. **Substitute the known values**:
Given \( \gamma \) for glass is \( 3 \alpha \), and \( \alpha = 8.3 \times 10^{-6} \, \text{per} \, ^\circ C \):
\[ \gamma = 3 \times 8.3 \times 10^{-6} = 2.49 \times 10^{-5} \, \text{per} \, ^\circ C \]
Now plug in the values:
\[ V = 1 \, \text{L} \left(1 + 2.49 \times 10^{-5} \times (50 - 0)\right) \]
\[ V = 1 \, \text{L} \left(1 + 2.49 \times 10^{-5} \times 50\right) \]
\[ V = 1 \, \text{L} \left(1 + 0.001245\right) \]
\[ V \approx 1.001245 \, \text{L} \]

4. **Final result for part (a)**:
The volume of the beaker at \( 50^\circ C \) is approximately \( 1.001 \, \text{L} \).

### Part (b): Volume of Mercury Overflowing at 50°C

1. **Identify the initial conditions for mercury**:
- Initial volume of mercury \( V_0 = 1 \, \text{L} \)
- Initial temperature \( T_0 = 0^\circ C \)
- Final temperature \( T = 50^\circ C \)

2. **Use the volume expansion formula for mercury**:
The final volume of mercury at \( 50^\circ C \) is given by:
\[ V_{\text{Hg}} = V_0 \left(1 + \gamma_{\text{Hg}} (T - T_0)\right) \]
where \( \gamma_{\text{Hg}} = 1.82 \times 10^{-4} \, \text{per} \, ^\circ C \).

3. **Substitute the known values**:
\[ V_{\text{Hg}} = 1 \, \text{L} \left(1 + 1.82 \times 10^{-4} \times (50 - 0)\right) \]
\[ V_{\text{Hg}} = 1 \, \text{L} \left(1 + 1.82 \times 10^{-4} \times 50\right) \]
\[ V_{\text{Hg}} = 1 \, \text{L} \left(1 + 0.0091\right) \]
\[ V_{\text{Hg}} \approx 1.0091 \, \text{L} \]

4. **Calculate the volume of mercury that overflows**:
The volume of mercury that overflows is the difference between the final volume of mercury and the final volume of the beaker:
\[ \text{Volume overflow} = V_{\text{Hg}} - V \]
\[ \text{Volume overflow} = 1.0091 \, \text{L} - 1.001 \, \text{L} \]
\[ \text{Volume overflow} \approx 0.0081 \, \text{L} \]

5. **Final result for part (b)**:
The volume of mercury that overflows is approximately \( 8.1 \, \text{mL} \).

### Summary of Results:
- (a) Volume of the beaker at \( 50^\circ C \) is approximately \( 1.001 \, \text{L} \).
- (b) Volume of mercury that overflows is approximately \( 8.1 \, \text{mL} \).