Question:
A gas occupies a volume of $4.2~dm^3$ at $101~kPa$ pressure. What volume will the gas occupy if the pressure is increased to $235~kPa$ keeping the temperature constant?
A gas occupies a volume of $4.2~dm^3$ at $101~kPa$ pressure. What volume will the gas occupy if the pressure is increased to $235~kPa$ keeping the temperature constant?
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If pressure increases at constant temperature, volume decreases according to Boyle’s law.
Updated On: May 29, 2026
- $1.8~dm^3$
- $3.6~dm^3$
- $0.9~dm^3$
- $2.1~dm^3$
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The Correct Option is A
Solution and Explanation
Concept:
At constant temperature, Boyle’s law is used: \[ P_1V_1 = P_2V_2 \]
Step 1: Write the given values. \[ V_1 = 4.2~dm^3 \] \[ P_1 = 101~kPa \] \[ P_2 = 235~kPa \] \[ V_2 = ? \]
Step 2: Apply Boyle’s law. \[ P_1V_1 = P_2V_2 \] \[ (101)(4.2) = (235)V_2 \] \[ V_2 = \frac{101 \times 4.2}{235} \] \[ V_2 = \frac{424.2}{235} \] \[ V_2 \approx 1.8~dm^3 \]
Step 3: Conclusion. Hence, the required volume is: \[ \boxed{1.8~dm^3} \]
At constant temperature, Boyle’s law is used: \[ P_1V_1 = P_2V_2 \]
Step 1: Write the given values. \[ V_1 = 4.2~dm^3 \] \[ P_1 = 101~kPa \] \[ P_2 = 235~kPa \] \[ V_2 = ? \]
Step 2: Apply Boyle’s law. \[ P_1V_1 = P_2V_2 \] \[ (101)(4.2) = (235)V_2 \] \[ V_2 = \frac{101 \times 4.2}{235} \] \[ V_2 = \frac{424.2}{235} \] \[ V_2 \approx 1.8~dm^3 \]
Step 3: Conclusion. Hence, the required volume is: \[ \boxed{1.8~dm^3} \]
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