Question

For a gas P-V curve is given as shown in the diagram. Curve path follows equations \((V - 2)^2 = 4aP\). Find work done by gas in given cyclic process. 

• A. \(-\frac{1}{3a}\)
• B. \(\frac{1}{3a}\)
• C. \(\frac{1}{5a}\)
• D. \(\frac{1}{2a}\)

Hint:

For a symmetric parabolic boundary like this, the area is exactly \(\frac{2}{3}\) of the enclosing rectangle.
Rectangle Area = \(\Delta V \times \Delta P = 2 \times \frac{1}{4a} = \frac{1}{2a}\). Enclosed Area = \(\frac{2}{3} \times \frac{1}{2a} = \frac{1}{3a}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Work done in a cyclic process is equal to the area enclosed by the cycle on a P-V diagram.
If the cycle is clockwise, work done is positive; if anti-clockwise, it is negative.
Step 2: Key Formula or Approach:
Area = \(\oint P dV\).
For the region between \(V=1\) and \(V=3\) bounded by a horizontal line \(P_0\) and a curve \(P(V)\):
\[ \text{Work} = \int_{V_1}^{V_2} (P_{top} - P_{bottom}) dV \]
Step 3: Detailed Explanation:
The curve is given by \(P = \frac{(V-2)^2}{4a}\).
At the boundaries \(V=1\) and \(V=3\), the pressure on the curve is:
At \(V=1 \Rightarrow P = \frac{(1-2)^2}{4a} = \frac{1}{4a}\).
At \(V=3 \Rightarrow P = \frac{(3-2)^2}{4a} = \frac{1}{4a}\).
Thus, the horizontal line \(P_0\) is at height \(\frac{1}{4a}\).
The area enclosed by the cycle (assuming it goes from \(1\) to \(3\) along the line and back along the curve) is:
\[ \text{Area} = \int_{1}^{3} \left( \frac{1}{4a} - \frac{(V-2)^2}{4a} \right) dV \]
\[ \text{Area} = \frac{1}{4a} \left[ V - \frac{(V-2)^3}{3} \right]_1^3 \]
\[ \text{Area} = \frac{1}{4a} \left[ \left(3 - \frac{1^3}{3}\right) - \left(1 - \frac{(-1)^3}{3}\right) \right] \]
\[ \text{Area} = \frac{1}{4a} \left[ \left(3 - \frac{1}{3}\right) - \left(1 + \frac{1}{3}\right) \right] = \frac{1}{4a} \left[ \frac{8}{3} - \frac{4}{3} \right] = \frac{1}{4a} \times \frac{4}{3} = \frac{1}{3a} \]
The diagram shows an anti-clockwise cycle, so the work done is negative: \(W = -\frac{1}{3a}\).
Step 4: Final Answer:
The work done by the gas is \(-\frac{1}{3a}\).