Question

A gas expands from volume 2 m\(^3\) to 6 m\(^3\) at constant pressure 100 Pa. Work done is:

• A. 200 J
• B. 300 J
• C. 400 J
• D. 600 J

Hint:

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 \( \Delta V \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Work done in a thermodynamic process is related to the change in volume of the gas.
In an isobaric process (where pressure remains constant), the work done can be calculated simply.
Physically, work done is the area under the curve in a Pressure-Volume (P-V) diagram.
If the pressure is constant, the area is simply a rectangle.
Key Formula or Approach:
The formula for work done during an isobaric process is:
\[ W = P \cdot \Delta V \]
Where:
- \(P\) is the constant pressure.
- \(\Delta V = V_{\text{final}} - V_{\text{initial}}\) is the change in volume.
Step 2: Detailed Explanation:
Given values:
Constant Pressure (\(P\)) = 100 Pa.
Initial Volume (\(V_1\)) = 2 m\(^3\).
Final Volume (\(V_2\)) = 6 m\(^3\).
First, calculate the change in volume (\(\Delta V\)):
\[ \Delta V = V_2 - V_1 = 6 \text{ m}^3 - 2 \text{ m}^3 = 4 \text{ m}^3 \]
Now, calculate the work done:
\[ W = 100 \text{ Pa} \cdot 4 \text{ m}^3 \]
Since 1 Pa = 1 N/m\(^2\), the unit becomes (N/m\(^2\)) \(\cdot\) m\(^3\) = N \(\cdot\) m = Joules (J).
\[ W = 400 \text{ J} \]
Step 3: Final Answer:
The work done by the gas is 400 J.
This matches option (C).