Question

If the pressure of an ideal gas is decreased by 10% isothermally, then its volume will

• A. decrease by 9%
• B. increase by 11.11%
• C. increase by 10%
• D. decrease by 11.11%

Hint:

Use the fractional inverse shortcut! A decrease of 10% means the pressure becomes $\frac{9}{10}$ of its original value. Since volume is inversely proportional to pressure ($V \propto \frac{1}{P}$), the volume must become the flipped fraction $\frac{10}{9}$ of its original value. An increase from 1 to $\frac{10}{9}$ is a gain of $\frac{1}{9}$, which equals 11.11% instantly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
An ideal gas undergoes an isothermal process (constant temperature $T$). The initial pressure $P_1$ is decreased by 10%, and we need to calculate the exact percentage change in its volume $V$.

Step 2: Key Formula or Approach:
For an isothermal process, Boyle's Law states that the product of pressure and volume remains constant: $$P_1 V_1 = P_2 V_2$$ The percentage change in volume is calculated using: $$\% \text{ change in } V = \left( \frac{V_2 - V_1}{V_1} \right) \times 100\%$$

Step 3: Detailed Explanation:
Let the initial pressure be $P_1 = P$ and the initial volume be $V_1 = V$. Since the pressure decreases by 10%, the final pressure $P_2$ becomes: $$P_2 = P - 0.10P = 0.90P$$ Apply Boyle's Law to find the new volume $V_2$: $$P \cdot V = (0.90P) \cdot V_2$$ $$V_2 = \frac{V}{0.90} = \frac{10}{9}V$$ Now, compute the fractional increase in volume: $$\Delta V = V_2 - V_1 = \frac{10}{9}V - V = \frac{1}{9}V$$ Calculate the percentage increase: $$\% \text{ increase} = \left( \frac{\frac{1}{9}V}{V} \right) \times 100\% = \frac{100}{9}\% \approx 11.11\%$$

Step 4: Final Answer:
The volume will increase by 11.11%, which corresponds to option (B).