91Ó°ÊÓ

1. The p-V diagram for cycle 1 and cycle 2.
2. The net workdone W by the gas should be positive for (a).
3. The net energy transferred by the gas as heat Q should be positive for (b).

Work is the exchange of mechanical energy between two systems, whereas heat is the exchange of thermal energy between systems. Using the formula for work done for gas and the first law of thermodynamics for the P-V diagram (a) and (b) of figures 18-27, we can find whether the cycle should be traversed clockwise or counterclockwise for the given conditions of work and heat for each cycle.

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

For each cycle, if the net work done$\mathrm{W}$ by the gas is to be positive, then in diagram

We can see that in the cycle 1, pressure$p$increases as the volumeincreases from the curve between the point i to f of the cycle, that is the net volume enclosed within from I to f is given as:

$\mathrm{Δ³Õ}={\mathrm{V}}_{\mathrm{f}}−{\mathrm{V}}_{\mathrm{i}}$is positive.

Hence, the work done is positive $W$considering equation (i).

From the curve between the point$f$to $a$, the volume$V$remains constant, that is the net volume enclosed from a to f is given by:

$\begin{array}{c}\mathrm{Δ}V=V_a−V_f\\ =0\end{array}$

And the pressure$p$ decreases, and hence, the work done $W=0$considering equation (i).

From the curve between the points a to i in the cycle, the pressure p in constant and volume V is decreasing, that is the net volume enclosed from a to i is given by:

$\mathrm{Δ³Õ}={\mathrm{V}}_{\mathrm{a}}−{\mathrm{V}}_{\mathrm{i}}$is negative.

Hence, the work done W is negative considering equation (i).

To get the net work done positive, the cycle 1 should be traversed clockwise since the area under the curve between the points i and f of the cycle is greater than the area under the curve between the points i and a of the cycle.

Similarly, we can see that in the cycle, From the curve between the points i to a, the volume V remains constant, that is the net volume enclosed from a to i is given by:

$\begin{array}{c}\mathrm{Δ³Õ}={\mathrm{V}}_{\mathrm{a}}−{\mathrm{V}}_{\mathrm{i}}\\ =0\end{array}$

And the pressure p increases, and hence, the work done $W=0$considering equation (i).

From the curve between the points f to i in the cycle, the pressure p remains constant and volume V increases, that is the net volume enclosed from a to f is given by:

$\mathrm{Δ³Õ}={\mathrm{V}}_{\mathrm{a}}−{\mathrm{V}}_{\mathrm{i}}$is positive.

Hence, the work done$W$is positive considering equation (i).

From the curve between points to of the cycle, pressure p decreases as the volume V decreases, that is the net volume enclosed from f to i is given by:

$\mathrm{Δ}V=V_f−V_i$is negative. Hence, the work done is negative W considering equation (i).

To get the net work done positive, the cycle 1 should be traversed clockwise since the area under the curve between the points a and f of the cycle is greater than the area under the curve between the points i and f of the cycle.

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

$∆{\mathrm{E}}_{\mathrm{int}}=0$

Hence, we get the following condition using equation (i) as:

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

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

Therefore, if the net energy transformed by the gas as heat Q is to be positive, then both the cycles are traversed in clockwise direction.