The pressure of a gas is increased by `50%` at constant temperature. The decrease in volume will be nearest to

Options

A. `66%`
B. `33%`
C. `17%`
D. `8%`

Explanation

To solve the problem, we will use the ideal gas law and the relationship between pressure and volume at constant temperature.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We are given that the pressure of a gas is increased by 50% at constant temperature. We need to find the decrease in volume.

2. **Initial and Final Pressure**:
- Let the initial pressure be \( P \).
- An increase of 50% means the final pressure \( P_2 \) is:
\[ P_2 = P + 0.5P = 1.5P \]

3. **Using the Ideal Gas Law**:
- According to the ideal gas law, at constant temperature, the product of pressure and volume is constant:
\[ P_1 V_1 = P_2 V_2 \]
- Here, \( P_1 = P \), \( V_1 \) is the initial volume, and \( V_2 \) is the final volume.

4. **Substituting Values**:
- Substitute \( P_1 \) and \( P_2 \) into the equation:
\[ P V_1 = (1.5P) V_2 \]

5. **Cancelling Pressure**:
- We can cancel \( P \) from both sides (assuming \( P \neq 0 \)):
\[ V_1 = 1.5 V_2 \]

6. **Rearranging for Final Volume**:
- Rearranging gives:
\[ V_2 = \frac{V_1}{1.5} = \frac{2}{3} V_1 \]

7. **Finding the Decrease in Volume**:
- The decrease in volume \( \Delta V \) can be calculated as:
\[ \Delta V = V_1 - V_2 = V_1 - \frac{2}{3} V_1 = \frac{1}{3} V_1 \]

8. **Calculating the Percentage Decrease**:
- The percentage decrease in volume is given by:
\[ \text{Percentage Decrease} = \left( \frac{\Delta V}{V_1} \right) \times 100 = \left( \frac{\frac{1}{3} V_1}{V_1} \right) \times 100 = \frac{100}{3} \approx 33.33\% \]

### Final Answer:
The decrease in volume will be nearest to **33.33%**.

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