# 11.6.2 Cyclic Process

If a system returns to its original state after a number of processes, we have what is called a cyclic process. On the P-V diagram, a cyclic process traces out a closed loop.

In the above P-V diagram, an ideal (monatomic) gas undergoes a cyclic process of 1→2→3→4→1. Since it returns to state 1, it must have returned to its original internal energy. So for the complete cycle $\displaystyle \Delta U=0$ .

There is work by done by the gas during 1→2 (the blue rectangle), and work done on the gas during 3→4 (the red rectangle). So overall, the area enclosed by the cyclic loop (the yellow rectangle in the top graph, which is the blue rectangle minus the red rectangle) represents the net work done by the gas during this cyclic process. In other words, WBY is positive and WON is negative for this particular cyclic process.

Applying the first law for the complete cycle, we can deduce that Q is positive for this cyclic process.

\displaystyle \displaystyle \begin{aligned}(\Delta U&=Q+{{W}_{{ON}}})\\\overset{0}{\mathop{{\Delta U}}}\,&=\overset{{}}{\mathop{Q}}\,+\overset{{\text{-ve}}}{\mathop{{{{W}_{{ON}}}}}}\,\end{aligned}

Now let’s try a different cyclic process.

Notice that state 2 has been removed so the cyclic process is 1→4→3→1. Notice that the direction of the loop has also been reversed. Since it is still a cyclic loop, we still have $\displaystyle \Delta U=0$ . With the switch in direction, the work done on the gas (during 3→1) is higher than the work done by the gas (during 4→3). So the enclosed area represents work done on the gas during this cyclic process. In other words, WON is positive and WBY is negative for this particular cyclic process.

Applying the first law for the complete cycle, we can deduce that Q is negative for this cyclic process.

$\displaystyle \displaystyle \overset{0}{\mathop{{\Delta U}}}\,=\overset{{}}{\mathop{Q}}\,+{{\overset{{\text{+ve}}}{\mathop{W}}\,}_{{ON}}}$

Heat Engines and Heat Pumps

Firstly, note that for all cyclic processes,

$\displaystyle \Delta U=0$

From the first law,

\displaystyle \displaystyle \begin{aligned}\overset{0}{\mathop{{\Delta U}}}\,&=\overset{{}}{\mathop{Q}}\,+{{\overset{{}}{\mathop{W}}\,}_{{ON}}}\\Q&=-{{W}_{{ON}}}\end{aligned}

So net Q and net WON are always opposite in signs to each other for a cyclic process. The net work done and the amount of net heat transfer is equal to the area of the enclosed loop.

For a clockwise cyclic process, since the gas expands at high pressure but is compressed at low pressure, there is net work done by the gas. So WON is negative and Q is positive. This means that in one complete cycle, the net heat supplied is equal to the net work done by the system. This is the basis of a heat engine (e.g. diesel engine, steam engine)

For an anti-clockwise cyclic process, since the gas is compressed at high pressure but expands at low pressure, there is net work done on the gas. So WON is positive and Q is negative. This means that in one complete cycle, the net work done on the system is equal to the net heat lost to the surrounding. This is the basis for a heat pump (e.g. refrigerator, freezer).

Demonstration

Sterling Engine

Concept Test

1660