A gas occupies a certain volume at `600` torr pressure and `27^(@)C`. If pressure is reduced to `500' torr, what temperature should be employed to keep the volume constant?

Explanation

To solve the problem, we will use Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature when the volume is held constant. The formula is given by:

\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial pressure, \( P_1 = 600 \) torr
- Final pressure, \( P_2 = 500 \) torr
- Initial temperature, \( T_1 = 27^\circ C \)

2. **Convert Temperature to Kelvin**:
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- Thus,
\[ T_1 = 27 + 273 = 300 \text{ K} \]

3. **Apply Gay-Lussac's Law**:
- Substitute the known values into the formula:
\[ \frac{600 \text{ torr}}{300 \text{ K}} = \frac{500 \text{ torr}}{T_2} \]

4. **Cross-Multiply to Solve for \( T_2 \)**:
- Rearranging gives:
\[ 600 \text{ torr} \cdot T_2 = 500 \text{ torr} \cdot 300 \text{ K} \]
- Thus,
\[ T_2 = \frac{500 \text{ torr} \cdot 300 \text{ K}}{600 \text{ torr}} \]

5. **Calculate \( T_2 \)**:
- Simplifying the equation:
\[ T_2 = \frac{150000 \text{ torr K}}{600 \text{ torr}} = 250 \text{ K} \]

6. **Convert \( T_2 \) Back to Celsius**:
- To convert Kelvin back to Celsius:
\[ T(°C) = T(K) - 273 \]
- Therefore,
\[ T_2 = 250 - 273 = -23^\circ C \]

### Final Answer:
The temperature that should be employed to keep the volume constant when the pressure is reduced to 500 torr is \( -23^\circ C \).