Use equation `2H_(2)O(l) to 2H_(2)(g)` to answer the following
What is the final volume of `H_(2)` if the pressure is increased by 2.5 times at constant temperature?

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. This can be expressed mathematically 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 Initial Conditions:**
- Let \( P_1 \) be the initial pressure.
- Let \( V_1 \) be the initial volume of \( H_2 \).

2. **Determine the Final Pressure:**
- The problem states that the pressure is increased by 2.5 times. Thus, the final pressure \( P_2 \) can be expressed as:
\[ P_2 = 2.5 \times P_1 \]

3. **Apply Boyle's Law:**
- According to Boyle's Law, we can write:
\[ P_1 V_1 = P_2 V_2 \]
Substituting \( P_2 \) into the equation gives:
\[ P_1 V_1 = (2.5 \times P_1) V_2 \]

4. **Rearrange the Equation to Solve for \( V_2 \):**
- We can cancel \( P_1 \) from both sides (assuming \( P_1 \neq 0 \)):
\[ V_1 = 2.5 V_2 \]
Rearranging gives:
\[ V_2 = \frac{V_1}{2.5} \]

5. **Conclusion:**
- The final volume \( V_2 \) of hydrogen gas when the pressure is increased by 2.5 times at constant temperature is:
\[ V_2 = \frac{V_1}{2.5} \]