Question 148: Thermodynamics & KTG

Question

The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure) is (in SI unit) :
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m6usssya/fa5f25c3-1aef-4c30-a37a-1bf2abd6d5d7/d7fdd720-e556-11ef-918b-9f97f398b68b/file-jaoe38c1m6usssyb.png?format=png)

Accepted Answer

Step 1: Find the Work Done ($\mathrm{W}$) in the Cycle
The work done in a cyclic process shown in a pressure-volume (P-V) graph is equal to the area enclosed by the cycle. For a circle, this area is $\frac{1}{2} \pi R^2$, where $R$ is the radius of the circle.
Step 2: Put in the Given Values
Radius $R$ on the graph is

$ R = \frac{200}{2} 	imes 10^3 	ext{ (for Pressure, P)} \ R = \frac{200}{2} 	imes 10^{-6} 	ext{ (for Volume, V)} $
Step 3: Calculate the Area (Work Done)
$ \mathrm{W} = \frac{1}{2} 	imes \pi 	imes \left(\frac{200}{2} 	imes 10^3ight) 	imes \left(\frac{200}{2} 	imes 10^{-6}ight) $
First, calculate $\frac{200}{2} = 100$, so:
$ \mathrm{W} = \frac{1}{2} 	imes \pi 	imes 100 	imes 10^3 	imes 100 	imes 10^{-6} $
Multiplying $100 	imes 100 = 10,000$, and $10^3 	imes 10^{-6} = 10^{-3}$, so:
$ \mathrm{W} = \frac{1}{2} 	imes \pi 	imes 10,000 	imes 10^{-3} = \frac{1}{2} 	imes \pi 	imes 10 = 5\pi~\mathrm{J} $
Step 4: Relate to Heat Exchanged
In a complete cycle, the change in internal energy is zero, so total heat exchanged is equal to the total work done.
So, the magnitude of heat exchanged is $5\pi~\mathrm{J}$.
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m794mukp/0344e68a-8ce6-4de5-b168-0d220ef9d200/1907fc80-ed38-11ef-b7a6-19e702ec54c4/file-jaoe38c1m794mukq.png?format=png)

Answer

zero

Answer

40$\pi$

Answer

5$\pi$

Answer

10$\pi$

Thermodynamics & KTG

The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure) is (in SI unit) :

Choose the correct answer: