Question

If the percentage increase in the pressure of a given mass of gas in a container at constant temperature is 100%, then its volume

• A. decreases by 100%
• B. decreases by 50%
• C. does not change
• D. increases by 100%

Hint:

Boyle’s Law: \(PV = \text{constant}\) — If pressure doubles, volume becomes half (50% decrease).

Answer 1

The Correct Option is B

This is a case of Boyle’s Law, which states that for a given mass of gas at constant temperature: \[ PV = \text{constant} \Rightarrow P_1V_1 = P_2V_2 \] Let initial pressure \(P_1 = P\), and final pressure \(P_2 = 2P\) (since 100% increase). Let initial volume \(V_1 = V\), and final volume = \(V_2\) Using: \[ P \cdot V = 2P \cdot V_2 \Rightarrow V_2 = \frac{V}{2} \] So, volume becomes half (i.e., decreased by 50%).

Answer 2

Step 1: Understand the problem
- Percentage increase in pressure \(P\) is 100%, so the new pressure \(P' = 2P\).
- Temperature is constant (isothermal process).

Step 2: Recall Boyle's law for isothermal processes
\[ P V = P' V' \]
where \(V\) and \(V'\) are initial and final volumes respectively.

Step 3: Calculate new volume
\[ P V = 2P \times V' \implies V' = \frac{V}{2} \]

Step 4: Percentage change in volume
Volume decreases to half, so decrease is:
\[ \frac{V - V'}{V} \times 100 = \frac{V - \frac{V}{2}}{V} \times 100 = 50\% \]

Step 5: Final answer
The volume decreases by 50%.