A gas occupies `0.6 dm^(3)` under a pressure of `0.92` bar. Find under what pressure, the volume of gas will be reduced by 20 percent of its original volume, temperature remaining constant.

Explanation

To solve the problem, we will use Boyle's Law, which states that for a given mass of gas at constant temperature, the product of pressure and volume is a constant. Mathematically, this can be expressed as:

\[ P_1 V_1 = P_2 V_2 \]

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( P_2 \) = final pressure
- \( V_2 \) = final volume

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume \( V_1 = 0.6 \, \text{dm}^3 \)
- Initial pressure \( P_1 = 0.92 \, \text{bar} \)

2. **Calculate the Final Volume \( V_2 \):**
- The volume is reduced by 20% of its original volume.
- Calculate 20% of \( V_1 \):
\[ 20\% \text{ of } 0.6 = \frac{20}{100} \times 0.6 = 0.12 \, \text{dm}^3 \]
- Therefore, the final volume \( V_2 \) is:
\[ V_2 = V_1 - 0.12 = 0.6 - 0.12 = 0.48 \, \text{dm}^3 \]

3. **Apply Boyle's Law:**
- According to Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
- Rearranging the equation to find \( P_2 \):
\[ P_2 = \frac{P_1 V_1}{V_2} \]

4. **Substitute the Values:**
- Substitute \( P_1 = 0.92 \, \text{bar} \), \( V_1 = 0.6 \, \text{dm}^3 \), and \( V_2 = 0.48 \, \text{dm}^3 \):
\[ P_2 = \frac{0.92 \times 0.6}{0.48} \]

5. **Calculate \( P_2 \):**
- First, calculate the numerator:
\[ 0.92 \times 0.6 = 0.552 \]
- Now divide by \( V_2 \):
\[ P_2 = \frac{0.552}{0.48} \approx 1.15 \, \text{bar} \]

### Final Answer:
The pressure at which the volume of gas will be reduced by 20% is approximately \( 1.15 \, \text{bar} \).