Solution for (a)
The work along path AB is
W_{ AB }=P_{ AB } \Delta V_{ AB } (15.14)
=\left(1.50 \times 10^{6} N / m ^{2}\right)\left(5.00 \times 10^{-4} m ^{3}\right)=750 J.
Since the path BC is isochoric, \Delta V_{ BC }=0, and so W_{ BC }=0. The work along path CD is negative, since \Delta V_{ CD } is negative (the volume decreases). The work is
W_{ CD }=P_{ CD } \Delta V_{ CD } (15.15)
=\left(2.00 \times 10^{5} N / m ^{2}\right)\left(-5.00 \times 10^{-4} m ^{3}\right)=-100 J.
Again, since the path DA is isochoric, \Delta V_{ DA }=0, and so W_{ DA }=0. Now the total work is
W=W_{ AB }+W_{ BC }+W_{ CD }+W_{ DA } (15.16)
=750 J +0+(-100 J )+0=650 J.
Solution for (b)
The area inside the rectangle is its height times its width, or
\text { area }=\left(P_{ AB }-P_{ CD }\right) \Delta V (15.17)
=\left[\left(1.50 \times 10^{6} N / m ^{2}\right)-\left(2.00 \times 10^{5} N / m ^{2}\right)\right]\left(5.00 \times 10^{-4} m ^{3}\right)
= 650 J.
Thus,
area = 650 J = W. (15.18)
Discussion
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 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 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 P V=n R T. Since T is constant, 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
\frac{1}{2} m \bar{v}^{2}=\frac{3}{2} k T. (15.19)
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
U=N \frac{1}{2} m \bar{v}^{2}=\frac{3}{2} N k T, (monatomic ideal gas), (15.20)
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, ΔU = 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 ΔU = 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 process, since work is done at the expense of internal energy:
U=\frac{3}{2} N k T. (15.21)
(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 = 0, ΔU = – 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.
