A sample of gas at `20^(@)C` occupies a volume of 2 L at a pressure of `0.867xx10^(5)Pa`. Calculate its volume at 100 kPa atmospheric pressure.

Explanation

To solve the problem, we will use the ideal gas law and the relationship between the initial and final states of the gas. The equation we will use is derived from the ideal gas law, which states that for a given amount of gas at constant temperature, the product of pressure and volume is constant.

### Step-by-Step Solution:

1. **Convert the Initial Temperature to Kelvin**:
\[ T_1 = 20^\circ C + 273 = 293 \, K \]

2. **Identify the Given Values**:
- Initial Volume, \( V_1 = 2 \, L \)
- Initial Pressure, \( P_1 = 0.867 \times 10^5 \, Pa \)
- Final Pressure, \( P_2 = 100 \, kPa = 100 \times 10^3 \, Pa = 1.0 \times 10^5 \, Pa \)
- Since the temperature remains constant, \( T_2 = T_1 = 293 \, K \)

3. **Use the Relationship Between Initial and Final States**:
Since the temperature and the number of moles of gas remain constant, we can use the equation:
\[ P_1 V_1 = P_2 V_2 \]

4. **Rearrange the Equation to Solve for \( V_2 \)**:
\[ V_2 = \frac{P_1 V_1}{P_2} \]

5. **Substitute the Known Values**:
\[ V_2 = \frac{(0.867 \times 10^5 \, Pa) \times (2 \, L)}{(1.0 \times 10^5 \, Pa)} \]

6. **Calculate \( V_2 \)**:
- First, calculate the numerator:
\[ 0.867 \times 10^5 \times 2 = 1.734 \times 10^5 \]
- Now divide by \( P_2 \):
\[ V_2 = \frac{1.734 \times 10^5}{1.0 \times 10^5} = 1.734 \, L \]

### Final Answer:
The volume of the gas at 100 kPa atmospheric pressure is \( V_2 = 1.734 \, L \).