Question

Figure 18-27 shows two closed cycles on p-Vdiagrams for a gas. The three parts of cycle 1 are of the same length and shape as those of cycle 2. For each cycle, should the cycle be traversed clockwise or counterclockwise if (a) the net work done by the gas is to be positive and (b) the net energy transferred by the gas as heatis to be positive?

Short answer

1. For each cycle, if the net work done W by the gas is to be positive, then the cycle should be traversed both clockwise.
2. For each cycle, if the net energy transferred by the gas as heat Q is to be positive, then the cycle should be traversed both clockwise.

Step-by-step solution

The given data

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).

Understanding the concept of work and heat

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, $\mathrm{W}={\int }_{{\mathrm{V}}_{\mathrm{i}}}^{{\mathrm{V}}_{\mathrm{f}}}\mathrm{p}d\mathrm{v}$ …(i)

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

Where,

Q is heat transfer and W is work done

(a) Calculation of the cycle direction for net work done to be positive

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:

$\mathrm{\Delta V}={\mathrm{V}}_{\mathrm{f}}-{\mathrm{V}}_{\mathrm{i}}$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:

$\begin{array}{c}\Delta V=V_a-V_f\\ =0\end{array}$

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:

$\mathrm{\Delta V}={\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{i}}$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:

$\begin{array}{c}\mathrm{\Delta V}={\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{i}}\\ =0\end{array}$

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:

$\mathrm{\Delta V}={\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{i}}$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:

$\Delta V=V_f-V_i$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.

(b) Calculation of the ranking according to their heat transferred by the gas

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.