Question

In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 

• A. \( -\dfrac{1}{3a} \)
• B. \( +\dfrac{1}{3a} \)
• C. \( \dfrac{1}{2a} \)
• D. \( -\dfrac{1}{a} \)

Hint:

For closed \(p\text{–}V\) cycles, always compute the area between the upper and lower curves; this directly gives the work done.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Work done in a cyclic process is equal to the area enclosed by the loop in the \(p\text{-}V\) plane.
Step 2: Key Formula or Approach:
Area under a curve: \(\int p dV\).
The process is a clockwise cycle, so the net work done is positive.
Step 3: Detailed Explanation:
The loop consists of a straight line \(p_{constant}\) and a curve \(p = \frac{(V-2)^2}{4a}\).
From the diagram, the volume range is from \(V=1\) to \(V=3\).
At \(V=1\), \(p = \frac{(1-2)^2}{4a} = \frac{1}{4a}\).
At \(V=3\), \(p = \frac{(3-2)^2}{4a} = \frac{1}{4a}\).
The straight line connects these points at \(p = \frac{1}{4a}\).
Net work = Area of the region between the horizontal line and the parabola:
\[ W = \int_{1}^{3} \left[ \frac{1}{4a} - \frac{(V-2)^2}{4a} \right] dV \] \[ W = \frac{1}{4a} \left[ V - \frac{(V-2)^3}{3} \right]_1^3 \] \[ W = \frac{1}{4a} \left[ \left(3 - \frac{1}{3}\right) - \left(1 - \frac{-1}{3}\right) \right] \] \[ W = \frac{1}{4a} \left[ \frac{8}{3} - \frac{4}{3} \right] = \frac{1}{4a} \cdot \frac{4}{3} = \frac{1}{3a} \] The cycle is clockwise, hence \(W = +\frac{1}{3a}\).
Step 4: Final Answer:
The total work done is \(+\frac{1}{3a}\).