Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A 100-kg mass drops 3 m, resulting in an increased volume in the cylinder of 0.002 m 3 (Fig.). The weight and the piston maintain a constant gage pressure of 100 kPa. Determine the net work done by the gas on the surroundings. Neglect all friction.

Text Solution
Verified by ExpertsThe correct answer is:
A
To determine the net work done by the gas on the surroundings, we can apply the formula for work done by a gas during expansion:
\( W = P \Delta V \)
where:
- \( W \) is the work done by the gas,
- \( P \) is the pressure (in Pa),
- \( \Delta V \) is the change in volume (in m³).
We know that:
- The pressure, \( P = 100 \text{ kPa} = 100,000 ext{ Pa} \)
- The volume increase, \( \Delta V = 0.002 ext{ m}^3 \)
Substituting these values into the formula:
\[ W = 100,000 ext{ Pa} \times 0.002 ext{ m}^3 = 200 \text{ Joules} \]
Therefore, the net work done by the gas on the surroundings is 200 Joules.
\( W = P \Delta V \)
where:
- \( W \) is the work done by the gas,
- \( P \) is the pressure (in Pa),
- \( \Delta V \) is the change in volume (in m³).
We know that:
- The pressure, \( P = 100 \text{ kPa} = 100,000 ext{ Pa} \)
- The volume increase, \( \Delta V = 0.002 ext{ m}^3 \)
Substituting these values into the formula:
\[ W = 100,000 ext{ Pa} \times 0.002 ext{ m}^3 = 200 \text{ Joules} \]
Therefore, the net work done by the gas on the surroundings is 200 Joules.
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