Question

Net heat energy absorbed by the system in going through a cyclic process shown in figure is ( 1 atm = ${10}^5$ pa):

• A. ${-10}^2\pi J$
• B. $\pi J$
• C. ${10}^2\pi J$
• D. $-\pi J$

Answer 1

The correct option is A ${-10}^2\pi J$

Since, the system returns to its initial state, the net change in the internal energy will be 0
Since, the process is cyclic, the work done in the process will be, the area under the curve in the PV diagram.
W = $\pi \times 1\times {10}^5\left(Pa\right)\times 1\times {10}^{-3}\left(m^3\right)$
W = 100 $\pi$ J
Now, U = Q + W
Since, U = 0,
$Q=-W=-100\pi$ J