Total Work Done in a Cyclical Process Equals the Area Inside the Closed Loop on a PV Diagram

To find the work along any path on a $\text{PV}$ diagram, you use the fact that work is pressure times change in volume, or $W=P\mathrm{\Delta }V$. So in part (a), this value is calculated for each leg of the path around the closed loop.

The work along path AB is

Since the path BC is isochoric, $\mathrm{\Delta }{V}_{\text{BC}}=0$, and so $W_{\text{BC}}=0$. The work along path CD is negative, since $\mathrm{\Delta }{V}_{\text{CD}}$ is negative (the volume decreases). The work is

Again, since the path DA is isochoric, $\mathrm{\Delta }{V}_{\text{DA}}=0$, and so $W_{\text{DA}}=0$. Now the total work is

The area inside the rectangle is its height times its width, or

Thus,

The result, as anticipated, is that the area inside the closed loop equals the work done. The area is often easier to calculate than is the work done along each path. It is also convenient to visualize the area inside different curves on $\text{PV}$ diagrams in order to see which processes might produce the most work. Recall that work can be done to the system, or by the system, depending on the sign of $W$. A positive $W$ is work that is done by the system on the outside environment; a negative $W$ represents work done by the environment on the system.

Figure 15.13(a) shows two other important processes on a $\text{PV}$ diagram. For comparison, both are shown starting from the same point A. The upper curve ending at point B is an isothermal process—that is, one in which temperature is kept constant. If the gas behaves like an ideal gas, as is often the case, and if no phase change occurs, then $\text{PV}=\text{nRT}$. Since $T$ is constant, $\text{PV}$ is a constant for an isothermal process. We ordinarily expect the temperature of a gas to decrease as it expands, and so we correctly suspect that heat transfer must occur from the surroundings to the gas to keep the temperature constant during an isothermal expansion. To show this more rigorously for the special case of a monatomic ideal gas, we note that the average kinetic energy of an atom in such a gas is given by

The kinetic energy of the atoms in a monatomic ideal gas is its only form of internal energy, and so its total internal energy $U$ is

where $N$ is the number of atoms in the gas. This relationship means that the internal energy of an ideal monatomic gas is constant during an isothermal process—that is, $\mathrm{\Delta }U=0$. If the internal energy does not change, then the net heat transfer into the gas must equal the net work done by the gas. That is, because $\mathrm{\Delta }U=Q-W=0$ here, $Q=W$. We must have just enough heat transfer to replace the work done. An isothermal process is inherently slow, because heat transfer occurs continuously to keep the gas temperature constant at all times and must be allowed to spread through the gas so that there are no hot or cold regions.

Also shown in Figure 15.13(a) is a curve AC for an adiabatic process, defined to be one in which there is no heat transfer—that is, $Q=0$. Processes that are nearly adiabatic can be achieved either by using very effective insulation or by performing the process so fast that there is little time for heat transfer. Temperature must decrease during an adiabatic expansion process, since work is done at the expense of internal energy:

(You might have noted that a gas released into atmospheric pressure from a pressurized cylinder is substantially colder than the gas in the cylinder.) In fact, because $Q=\mathrm{0, }\mathrm{\Delta }U=–W$ for an adiabatic process. Lower temperature results in lower pressure along the way, so that curve AC is lower than curve AB, and less work is done. If the path ABCA could be followed by cooling the gas from B to C at constant volume (isochorically), Figure 15.13(b), there would be a net work output.

Figure 15.13 (a) The upper curve is an isothermal process ($\text{Δ}T=0$), whereas the lower curve is an adiabatic process ($Q=0$). Both start from the same point A, but the isothermal process does more work than the adiabatic because heat transfer into the gas takes place to keep its temperature constant. This keeps the pressure higher all along the isothermal path than along the adiabatic path, producing more work. The adiabatic path thus ends up with a lower pressure and temperature at point C, even though the final volume is the same as for the isothermal process. (b) The cycle ABCA produces a net work output.