At constant temperature if the pressure of an ideal gas is increased by 10% then its volume must decrease by

Options

A. 0.0909
B. 0.1
C. 0.05
D. 0.2

Answer

To solve the problem step by step, we will use the ideal gas law and the concept of percentage change.

### Step-by-Step Solution:

1. **Understand the Problem**: We are given that the pressure of an ideal gas is increased by 10% at constant temperature. We need to find out how much the volume decreases in percentage.

2. **Use the Ideal Gas Law**: According to the ideal gas law, at constant temperature, the relationship between pressure (P) and volume (V) is given by:
\[ P_1 \times V_1 = P_2 \times V_2 \]
where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume.

3. **Define Initial Conditions**:
- Let the initial pressure \(P_1 = 100\) (arbitrary units for simplicity).
- Since the pressure increases by 10%, the final pressure \(P_2\) is:
\[ P_2 = P_1 + 0.10 \times P_1 = 100 + 10 = 110 \]

4. **Set Up the Equation**: Substitute the values into the ideal gas equation:
\[ 100 \times V_1 = 110 \times V_2 \]

5. **Solve for \(V_2\)**: Rearranging the equation gives:
\[ V_2 = \frac{100 \times V_1}{110} = \frac{10}{11} V_1 \]

6. **Calculate the Change in Volume**: The change in volume can be calculated as:
\[ \Delta V = V_1 - V_2 = V_1 - \frac{10}{11} V_1 = V_1 \left(1 - \frac{10}{11}\right) = V_1 \left(\frac{1}{11}\right) \]

7. **Calculate the Percentage Decrease**: The percentage decrease in volume is given by:
\[ \text{Percentage Decrease} = \left(\frac{\Delta V}{V_1}\right) \times 100 = \left(\frac{V_1 \left(\frac{1}{11}\right)}{V_1}\right) \times 100 = \frac{1}{11} \times 100 \]
\[ = \frac{100}{11} \approx 9.09\% \]

8. **Conclusion**: The volume must decrease by approximately 9.09%.

### Final Answer:
The volume must decrease by approximately **9.09%**.