Question

If the system goes around the cycle in a counter-clock wise direction, does it absorb or liberate heat in one cycle ? What is the magnitude of the heat absorbed or liberates in one counterclockwise cycle ?

Answer 1

As $\Delta U$ is a state function, thus $\Delta U_{cycle}=0$

Area enclosed by the cyclic process in P-V graph gives the work done by the gas during that cycle.

For the process to be in counter-clock wise direction, the work of expansion (a to b) is less than that of compression (b to a).

$⟹$ Net work done $W_{cycle}<0$

Using 1st law, $Q_{cycle}=\Delta U_{cycle}+W_{cycle}$

$\therefore$ $Q_{cycle}=0+W_{cycle}$ $⟹Q_{cycle}<0$

Hence the system liberates heat.

As the magnitude of work done for the clock wise and anti-clockwide directions is same, thus the amount of heat liberated is also same i.e $7200$ J