The pressure of an ideal gas undergoin isothermal change is increased by 10%. The volume of the gas must decrease by about

Options

A. `0.1%`
B. `9%`
C. `10%`
D. `0.9%`

Explanation

To solve the problem, we will use Boyle's Law, which states that for a given mass of an ideal gas at constant temperature (isothermal conditions), the product of pressure (P) and volume (V) 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 Initial and Final Pressures**:
- Let the initial pressure \( P_1 \) be \( P \).
- The pressure is increased by 10%, so:
\[ P_2 = P_1 + 0.1 P_1 = 1.1 P_1 \]

2. **Set Up the Boyle's Law Equation**:
- According to Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
- Substitute \( P_2 \) with \( 1.1 P_1 \):
\[ P_1 V_1 = 1.1 P_1 V_2 \]

3. **Cancel \( P_1 \) from Both Sides**:
- Since \( P_1 \) is common on both sides, we can cancel it out (assuming \( P_1 \neq 0 \)):
\[ V_1 = 1.1 V_2 \]

4. **Rearranging to Find \( V_2 \)**:
- Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{V_1}{1.1} \]

5. **Calculate the Change in Volume**:
- The change in volume \( \Delta V \) can be calculated as:
\[ \Delta V = V_1 - V_2 = V_1 - \frac{V_1}{1.1} \]
- Simplifying this gives:
\[ \Delta V = V_1 \left(1 - \frac{1}{1.1}\right) = V_1 \left(\frac{1.1 - 1}{1.1}\right) = V_1 \left(\frac{0.1}{1.1}\right) \]

6. **Express the Decrease in Volume as a Percentage**:
- The percentage decrease in volume can be calculated as:
\[ \text{Percentage Decrease} = \left(\frac{\Delta V}{V_1}\right) \times 100 = \left(\frac{0.1}{1.1}\right) \times 100 \approx 9.09\% \]

### Conclusion:
The volume of the gas must decrease by approximately 9.09% when the pressure is increased by 10% under isothermal conditions.