Thermo Thursday: Finishing Compression Work

It’s the end of the semester and I’ve got other projects I need to be working on, so this week I’m just going to do the last problem from the compression work section and leave the adiabats and isotherms for next time.

Problem 1.34: An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangular cyclic process shown in the figure to the right. Assume that the temperature is always such that rotational degrees of freedom are active, but vibrational modes are “frozen out.” Also assume that the only type of work done on the gas is quasistatic compression-expansion work. (a) For each of the four steps A through D, compute the work done on the gas, the heat added to the gas, and the change in the energy content of the gas. Express all answers in terms of {P}_{1}, {P}_{2}, {V}_{1}, and {V}_{2}. (Hint: Compute \Delta U before Q, using the ideal gas law and the equipartition theorem.)

Step A: This is a diatomic gas with no vibration modes, so for this system the equipartition theorem says that U = \frac {5}{2}PV. For this step, \Delta U = \frac{5}{2}({P}_{2}-{P}_{1}){V}_{1}. Since the volume of the gas doesn’t change in this step, there is no work being done on the gas. Recalling that \Delta U = W + Q, that means that Q = \frac{5}{2}({P}_{2}-{P}_{1}){V}_{1}.

Step B: here \Delta U = \frac{5}{2}{P}_{2}({V}_{2}-{V}_{1}). The volume is increasing, so the work done on the gas is negative, W=-{P}_{2}({V}_{2}-{V}_{1}). Since W is negative, that means that the heat added must be the total energy change plus the amount of energy subtracted by negative work. Q=\Delta U - W = \frac{7}{2}{P}_{2}({V}_{2}-{V}_{1}).

Step C: this time \Delta U = \frac{5}{2}({P}_{1}-{P}_{2}){V}_{2}. Note that this is a negative quantity, the system has lost energy. Again, there is no volume change, so no work is being done on the gas. Thus the entire energy change is due to the system losing heat energy, so Q=\frac{5}{2}({P}_{1}-{P}_{2}){V}_{2}.

Step D: in the last step, \Delta U = \frac{5}{2}{P}_{1}({V}_{1}-{V}_{2}). This is another negative quantity. But this time, the work being done on the gas is positive, W=-{P}_{1}({V}_{1}-{V}_{2}). Since the work done on the gas is positive, and the energy change is negative, it must be losing even more heat energy than it is gaining in work energy. Here Q= \Delta U - W = \frac{7}{2}{P}_{1}({V}_{1}-{V}_{2}).

(b) Describe in words what is physically being done during each of the four steps; for example, during step A, heat is added to the gas (from an external flame or something) while the piston is held fixed.

The author went ahead and did step A for me. In step B, the volume of the gas increases, so the piston is being drawn out (or, more likely, pushed out by internal pressure). But this doesn’t result in any decrease in pressure, which means that the gas is also being heated as the volume increases. In step C, the system just loses heat energy, so the system is being cooled while the piston is held fixed. In step D the piston is compressed, but the cylinder is still being cooled so the pressure doesn’t change.

(c) Compute the net work done on the gas, the net heat added to the gas, and the net change in energy of the gas during the entire cycle. Are the results as you expected? Explain briefly.

The net change in energy is zero, that’s what makes it a cycle. For the work over the whole cycle, that’s {W}_{cycle}=-({P}_{2}-{P}_{1})({V}_{2}-{V}_{1}). That’s a negative quantity, meaning that over the course of the whole cycle work was done on the environment. Since the energy change is zero, the heat added perfectly balances the work, so {Q}_{cycle}=({P}_{2}-{P}_{1})({V}_{2}-{V}_{1}). That’s a positive quantity. As mentioned at the end of the last Thermo Thursday, this cycle takes in heat energy and converts it to work done on the environment. So these results are as I expected.