Question:
A certain mass of a gas occupies a volume of $2.5~dm^3$ at NTP. Calculate the change in volume of gas at the same temperature, if pressure of gas is changed to $1.25~atm$.
A certain mass of a gas occupies a volume of $2.5~dm^3$ at NTP. Calculate the change in volume of gas at the same temperature, if pressure of gas is changed to $1.25~atm$.
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According to Boyle’s law:
\[
P \propto \frac{1}{V}
\]
at constant temperature.
Updated On: May 29, 2026
- $3.0~dm^3$
- $0.5~dm^3$
- $4.5~dm^3$
- $1.5~dm^3$
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The Correct Option is B
Solution and Explanation
Concept:
At constant temperature, Boyle’s law is applicable: \[ P_1V_1 = P_2V_2 \]
Step 1: Write given values. At NTP: \[ P_1 = 1~atm \] \[ V_1 = 2.5~dm^3 \] New pressure: \[ P_2 = 1.25~atm \] \[ V_2 = ? \]
Step 2: Apply Boyle’s law. \[ P_1V_1 = P_2V_2 \] \[ (1)(2.5) = (1.25)V_2 \] \[ V_2 = \frac{2.5}{1.25} \] \[ V_2 = 2.0~dm^3 \]
Step 3: Find change in volume. \[ \Delta V = V_1 - V_2 \] \[ \Delta V = 2.5 - 2.0 \] \[ \Delta V = 0.5~dm^3 \]
Step 4: Conclusion. Hence, the change in volume is: \[ \boxed{0.5~dm^3} \]
At constant temperature, Boyle’s law is applicable: \[ P_1V_1 = P_2V_2 \]
Step 1: Write given values. At NTP: \[ P_1 = 1~atm \] \[ V_1 = 2.5~dm^3 \] New pressure: \[ P_2 = 1.25~atm \] \[ V_2 = ? \]
Step 2: Apply Boyle’s law. \[ P_1V_1 = P_2V_2 \] \[ (1)(2.5) = (1.25)V_2 \] \[ V_2 = \frac{2.5}{1.25} \] \[ V_2 = 2.0~dm^3 \]
Step 3: Find change in volume. \[ \Delta V = V_1 - V_2 \] \[ \Delta V = 2.5 - 2.0 \] \[ \Delta V = 0.5~dm^3 \]
Step 4: Conclusion. Hence, the change in volume is: \[ \boxed{0.5~dm^3} \]
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