Animated Solution for Physics - Thermodynamics: For the given cyclic process $CAB$ as shown for a gas, the work done is

Imagine you are looking at the heartbeat of a thermodynamic engine. The $p-V$ diagram on your screen is exactly that—a visual representation of a gas undergoing a series of transformations.

We are given a cyclic process $CAB$. The gas starts at state $C$, expands to state $A$, drops its pressure to reach state $B$, and finally compresses back to its initial state $C$.

The $y$-axis represents the pressure $p$ in Pascals ($\text{Pa}$), and the $x$-axis represents the volume $V$ in cubic meters (${\text{m}}^3$).

Whenever you face a cyclic process on a $p-V$ diagram, there is one golden rule you must remember: the net work done by the gas is exactly equal to the area enclosed by the loop.

Why? Because the work done during expansion is the area under the upper curve, and the work done on the gas during compression is the area under the lower curve. Subtract the two, and you are left with the area inside the cycle!

Before we calculate the area, we must determine the sign of the work. Look closely at the arrows. The cycle $C\to A\to B\to C$ flows in a clockwise direction.

By our standard sign convention, a clockwise cycle on a $p-V$ diagram means the gas does more work expanding than the work done on it compressing. Therefore, the net work done is positive.

The shape enclosed by our cycle is a perfect right-angled triangle. The formula for the area of a triangle is beautifully simple:

Let's find the dimensions of this triangle. The base of the triangle corresponds to the horizontal line segment $CA$, which represents the change in volume $\Delta V$.

Looking at the $x$-axis, the volume at $A$ is $5{\text{ m}}^3$ and the volume at $C$ is $1{\text{ m}}^3$.

Next, we find the height of the triangle. The height corresponds to the vertical line segment $AB$, which represents the change in pressure $\Delta p$.

Looking at the $y$-axis, the pressure at $A$ is $6.0\text{ Pa}$ and the pressure at $B$ is $1\text{ Pa}$.

We have our raw values ready. Let's substitute them into our area formula:

Executing the final multiplication, half of $4$ is $2$, and $2$ times $5$ is $10$.

The net work done in the cyclic process is exactly $10\text{ J}$. Since it is a clockwise cycle, the work is positive. This matches option (b) perfectly.

Always remember to check the direction of the cycle and the labels on the axes. A simple swap of the axes or a reversal of the arrows can completely change the sign of your answer!