91Ó°ÊÓ

1. The p-V diagram for cycle 1 and cycle 2.
2. Both cycles should be traversedclockwise.
3. The three parts of the cycle 1 are of the same length and shape as those of cycle 2.

Work is the exchange of mechanical energy between two systems, whereas heat is the exchange of thermal energy between systems. By using the formula for work done for gas and the first law of thermodynamics for P-V diagrams (1) and (2) of Figures 18-27, we can find which traversed clockwise cycle has greater work and heat.

Formulae:

The work done per cycle,$\mathrm{W}={∫}_{{\mathrm{V}}_{\mathrm{i}}}^{{\mathrm{V}}_{\mathrm{f}}}\mathrm{p}d\mathrm{V}$ …(¾±)

According to the first law of thermodynamics, the change in internal energy$∆{\mathrm{E}}_{\mathrm{int}}$ is given by, $∆{\mathrm{E}}_{\mathrm{int}}=\mathrm{Q}-\mathrm{W}$ …(¾±¾±)

Where,

Q is heat transfer and W is work done

The cycle be traversed both clockwise; then the net work done W by the gas is to be positive in both cycles.

In cycle, we can see that the net work done W₁ is given by the area under the curve between the points i to f of the cycle and the area under the curve between the points i to a of the cycle considering equation (i).

Similarly, in cycle, we can see that the net work done W₂ is given by the area under the curve between the points f to i of the cycle and the area under the curve between the points a to f of the cycle considering equation (i).

We are given that the three parts of the cycle 1 are of the same length and shape as those of cycle 2.

Then, by comparing the net work done W₁ of cycle 1 with the net work done W₂ of cycle 2, we can say that the net work done W₂ of cycle 2 is greater than that of cycle 1 because, the area between the curves if and af of cycle 2 is greater than that of cycle 2.

From pV diagram (1) and (2) of Figure 18-27, we can see that the process in which there is no change inthe internal energy and process is closed cyclic. That is,

$∆{\mathrm{E}}_{\mathrm{int}}=0$

Hence, we get the following condition from equation (ii) as follows:

$\begin{array}{rcl}\mathrm{Q}-\mathrm{W}& =& 0\\ \mathrm{Q}& =& \mathrm{W}\end{array}$

Hence, here the heat Q is completely transformed into work done W.

If both the cycles be traversed clockwise, then the net energy transformed by the gas as heat Q is to be positive. Both the curves show cyclic processes.

Hence, we have $\mathrm{Q}=\mathrm{W}$. Also, the net work done W₂ of cycle 2 is greater than that of cycle 1. Then, we can say that the net energy transferred by the gas as heat Q₂ of cycle 2 is also greater than the net energy transformed by the gas as heat Q₁ of cycle 1.