Question

An ideal gas undergoes cyclic process ABCDA as shown in given p-V diagram. What is the magnitude of amount of work done by the gas?

• A. \(6P_0V_0\)
• B. \(4P_0V_0\)
• C. \(2P_0V_0\)
• D. \(P_0V_0\)

Hint:

In a cycle the work equals the area enclosed on the p-V graph. Find the width and height of the rectangle and multiply.

Answer 1

The Correct Option is B

Step 1: Understand the figure
The path ABCDA on the p-V diagram is a closed rectangle. The lower side is at pressure \(2P_0\) and the upper side at pressure \(4P_0\). The left side is at volume \(3V_0\) and the right side at \(5V_0\). The arrows show the gas moves up the right side and back along the top, so the loop is traced as a cycle.

Step 2: Use the cyclic process rule
In a cycle the gas returns to its starting state, so the net change in internal energy is zero. The net work done by the gas equals the area enclosed by the loop on the p-V diagram. The sign depends on the direction of the loop, but the question asks only for the magnitude.

Step 3: Find the sides of the rectangle
Width (change in volume): \(\Delta V=5V_0-3V_0=2V_0\).
Height (change in pressure): \(\Delta P=4P_0-2P_0=2P_0\).

Step 4: Calculate the area
\[ |W|=\Delta P\times\Delta V=(2P_0)(2V_0)=4P_0V_0 \]

Step 5: Why the other options are wrong
\(6P_0V_0\) would come from a wrong width or height, such as \(2P_0\times3V_0\). \(2P_0V_0\) and \(P_0V_0\) are too small because they leave out part of the rectangle. Only \(4P_0V_0\) equals width times height.

Final Answer:
The enclosed area is \(4P_0V_0\), so the magnitude of work is option (B). \[ \boxed{4P_0V_0} \]