Question:medium

A certain mass of a gas occupies a volume of \(2.5\ \text{dm}^3\) at NTP. Calculate the change in volume of gas at the same temperature if pressure of gas is changed to \(1.25\ \text{atm}\).

Show Hint

For Boyle's law: \[ P \propto \frac{1}{V} \] This means:
• Increase in pressure causes decrease in volume.
• Decrease in pressure causes increase in volume. Always remember: \[ P_1V_1 = P_2V_2 \] Temperature must remain constant for Boyle's law to be applicable.
Updated On: May 29, 2026
  • \(3.0\ \text{dm}^3\)
  • \(0.5\ \text{dm}^3\)
  • \(4.5\ \text{dm}^3\)
  • \(1.5\ \text{dm}^3\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Boyle's Law states that for a fixed amount of an ideal gas at a constant temperature, the volume (\( V \)) is inversely proportional to the pressure (\( P \)).
\[ P_1V_1 = P_2V_2 \]
At NTP (Normal Temperature and Pressure), the pressure is standardly taken as 1 atm.
Step 2: Detailed Explanation:
Initial state:
\( P_1 = 1 \text{ atm} \)
\( V_1 = 2.5 \text{ dm}^3 \)

Final state:
\( P_2 = 1.25 \text{ atm} \)
\( V_2 = ? \)

Using Boyle's Law:
\[ 1 \times 2.5 = 1.25 \times V_2 \]
\[ V_2 = \frac{2.5}{1.25} \]
\[ V_2 = 2.0 \text{ dm}^3 \]

Calculating the change in volume:
The question asks for the "change in volume", not the final volume.
\[ \text{Change in volume } (\Delta V) = |V_1 - V_2| \]
\[ \Delta V = |2.5 - 2.0| = 0.5 \text{ dm}^3 \]
Step 3: Analyzing the Result:
The pressure increased (from 1 to 1.25), which caused the volume to decrease (from 2.5 to 2.0).
The magnitude of this decrease is 0.5 \( \text{dm}^3 \).
Step 4: Final Answer:
The change in volume is 0.5 \( \text{dm}^3 \).
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