Integrating a non-function - Work Done using P-V graph

Question

The above figure was given to us, with the question saying

An ideal gas is taken from state $A$ to state $B$ via the above three processes. Then for heat absorbed by the gas, which of the following is true?

(A) Greater in $ACB$ than in $ADB$

(B) Least in $ADB$

(C) Same in $ACB$ and $AEB$

(D) Lesser in $AEB$ than in $ADB$

The answer is (D), but I'm not sure how.

A little background for those who have lost contact with Physics:

Heat absorbed is represented by $\Delta Q$, Change in Internal Energy by $\Delta U$, and Work Done by $\Delta W$, with the equation $\Delta Q=\Delta U+\Delta W$ always valid.

Here $\Delta U$ is same for all three processes owing to the fact that they all have the same starting and ending point.

And $\Delta W=\int P\ dV$.

Hence, to compare Heat Absorbed, simply comparing Work Done suffices.

My Question:

In process $ADB$ it is easy to see that the Work done is zero, owing to $dV=0$, but I'm unsure how to calculate Work Done for the other two processes, as they are not functions (of $P$ in $V$)?

Edit: I had missed out on writing that the gas is ideal, I have now added that in.

Answer 1

To describe the three processes, we could as follows

1.Process ADB - Constant volume process. For an ideal gas, this would be a zero work process, and hence all heat is used to change the internal energy

2. Process AEB, ACB - Follows a circular process

$$P = P^* + K\sin \theta \\V = V^* + K\cos \theta$$

Hence

$$dV = -K\sin \theta d\theta$$

Now, for AEB

$$W = -\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(KP^*\sin \theta +K^2\sin^2\theta )d\theta $$

$$\implies W_{AEB} = -\frac{K^2\pi}{2} < 0$$

Can you approach ACB in a similar fashion?

Answer 2

While the other answer gives a great method, there is a more conceptual approach to this problem.

The work for a given process is the signed area under the PV curve. If we split AEB into the AE curve and the EB curve, we can see that the area under EB is greater than that under AE. We can also tell that the area under AE will be positive and EB will be negative (since we are traversing the curve from right to left). Combining the two, the overall work for AEB will be negative.

And in fact, we can arrive at the actual result if we recognize that geometrically that $W_{EB}=-(\frac{\pi r^2}{2}+W_{AE})$ where $r$ is the radius of the circle. So $W_{AEB}=W_{EB}+W_{AE}=-\frac{\pi r^2}{2}$