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

Concept: The work done in a thermodynamic process on a \(p\text{–}V\) diagram is given by: \[ W = \oint p\,dV \] For a closed cycle, the work done equals the {area enclosed} by the loop. Clockwise cycles give positive work, while anticlockwise cycles give negative work.
Step 1: Understand the path of the cycle From the diagram:
The upper path \(A \to C\) is a horizontal line (constant pressure).
The lower path \(C \to B \to A\) is a curved path given by \[ (V-2)^2 = 4ap \Rightarrow p = \frac{(V-2)^2}{4a} \]
The cycle proceeds in the clockwise direction. The limits of volume are: \[ V=1 \quad \text{to} \quad V=3 \]
Step 2: Find the pressure of the top horizontal path At points \(A\) and \(C\), from the curve: \[ (1-2)^2 = 4ap \Rightarrow p = \frac{1}{4a} \] \[ (3-2)^2 = 4ap \Rightarrow p = \frac{1}{4a} \] Thus, the top path is at constant pressure: \[ p = \frac{1}{4a} \]
Step 3: Work done along the top path \(A \to C\) \[ W_{AC} = \int_{1}^{3} p\,dV = \frac{1}{4a}\int_{1}^{3} dV = \frac{1}{4a}(3-1) = \frac{1}{2a} \]
Step 4: Work done along the curved path \(C \to A\) \[ W_{CA} = \int_{3}^{1} \frac{(V-2)^2}{4a}\,dV = -\frac{1}{4a}\int_{1}^{3}(V-2)^2\,dV \] Evaluate the integral: \[ \int_{1}^{3}(V-2)^2\,dV = \int_{-1}^{1} x^2\,dx = \left[\frac{x^3}{3}\right]_{-1}^{1} = \frac{2}{3} \] So, \[ W_{CA} = -\frac{1}{4a}\cdot\frac{2}{3} = -\frac{1}{6a} \]
Step 5: Total work done in the closed cycle \[ W_{\text{total}} = W_{AC} + W_{CA} = \frac{1}{2a} - \frac{1}{6a} = \frac{1}{3a} \] Final Answer: \[ \boxed{\dfrac{1}{3a}} \]