Question:
A gas expands from volume 2 m\(^3\) to 6 m\(^3\) at constant pressure 100 Pa. Work done is:
A gas expands from volume 2 m\(^3\) to 6 m\(^3\) at constant pressure 100 Pa. Work done is:
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On a Pressure-Volume (P-V) diagram, the work done is equal to the area under the curve. For constant pressure, this area is simply a rectangle with height \( P \) and width \( \Delta V \).
Updated On: Jun 3, 2026
- 200 J
- 300 J
- 400 J
- 600 J
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The Correct Option is C
Solution and Explanation
Concept:
For an isobaric process (a process occurring at constant pressure), the work done by a gas during expansion or compression is given by the product of the constant pressure and the change in volume.
• Formula: \( W = P \Delta V \)
• \( P \): Constant pressure.
• \( \Delta V = V_f - V_i \): Change in volume (Final volume - Initial volume).
Step 1: Identifying the given parameters.
Constant Pressure (\( P \)) = 100 Pa
Initial Volume (\( V_i \)) = 2 m\(^3\)
Final Volume (\( V_f \)) = 6 m\(^3\)
Step 2: Calculating the change in volume.
\[ \Delta V = V_f - V_i \] \[ \Delta V = 6 \text{ m}^3 - 2 \text{ m}^3 = 4 \text{ m}^3 \]
Step 3: Substituting values into the work formula.
\[ W = P \times \Delta V \] \[ W = 100 \text{ Pa} \times 4 \text{ m}^3 \] \[ W = 400 \text{ J} \]
• Formula: \( W = P \Delta V \)
• \( P \): Constant pressure.
• \( \Delta V = V_f - V_i \): Change in volume (Final volume - Initial volume).
Step 1: Identifying the given parameters.
Constant Pressure (\( P \)) = 100 Pa
Initial Volume (\( V_i \)) = 2 m\(^3\)
Final Volume (\( V_f \)) = 6 m\(^3\)
Step 2: Calculating the change in volume.
\[ \Delta V = V_f - V_i \] \[ \Delta V = 6 \text{ m}^3 - 2 \text{ m}^3 = 4 \text{ m}^3 \]
Step 3: Substituting values into the work formula.
\[ W = P \times \Delta V \] \[ W = 100 \text{ Pa} \times 4 \text{ m}^3 \] \[ W = 400 \text{ J} \]
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