Question

What would be the amount of heat absorbed in the cyclic process shown below? 
 

• A. 5\(\pi\) J
• B. 15\(\pi\) J
• C. 25\(\pi\) J
• D. 100\(\pi\) J

Hint:

In a cyclic process, the work done is the area enclosed by the path in the pressure-volume diagram, and it equals the heat absorbed if no internal energy change occurs.

Answer 1

The Correct Option is D

Step 1: {Understanding the Cyclic Process} 
In a cyclic process, the internal energy change (\(\Delta U\)) is zero. According to the first law of thermodynamics: \[ \Delta U = q + W \] where \(q\) is heat absorbed and \(W\) is the work done. 
Step 2: {Work Done in a Cyclic Process} 
The work done in a cyclic process is equal to the area enclosed by the cycle in the \(p-v\) diagram. For a process with a circular path, the work done is the area of the circle, which is given by: \[ W = \pi r^2 \] where \(r\) is the radius of the circle. 
Step 3: {Calculating the Work Done} 
From the diagram, the radius \(r = 25 - 5 = 20\). Therefore: \[ W = \pi (20)^2 = 400\pi \, {J} \] Since there is no change in internal energy, the heat absorbed \(q = -W\). Thus, the heat absorbed is \(100\pi\) J. 
 

Answer 2

Step 1: Understand the nature of the cyclic process
The process is a closed loop on a PV diagram, which implies a cyclic thermodynamic process.
For any cyclic process, the **net heat absorbed (q)** over the cycle equals the **net work done (W)** by the system, because internal energy change \( \Delta U = 0 \).

Step 2: Determine the shape and dimensions
The diagram shows a circular loop in the PV graph bounded within:
- Pressure (P): 5 to 25 units
- Volume (V): 5 to 25 units
This is a full circle with diameter = 20 units in both pressure and volume axes.
So, radius \( r = \frac{25 - 5}{2} = 10 \) units

Step 3: Calculate the area of the circle
Work done in a PV cycle is numerically equal to the **area enclosed** in the loop:
\[ W = \text{Area} = \pi r^2 = \pi (10)^2 = 100\pi \, \text{J} \]

Step 4: Apply first law of thermodynamics
Since it’s a cyclic process:
\[ \Delta U = 0 \Rightarrow q = W \]
So, heat absorbed = work done = \( 100\pi \) J

Final Answer: 100π J