Thermo Thursday: Chapter 1.5: Compression Work, continued

I never got around to finishing the problem set last week, so instead of editing the last entry, I’ll just finish it for this week.

Problem 1.32: By applying a pressure of 200 atm, you can compress water to 99% of its usual volume. Sketch this process (not necessarily to scale) on a PV diagram, and estimate the work required to compress a liter of water by this amount. Does the result surprise you?

I’m just going to assume a linear function here and say that the work done is the area under the PV curve. Here that ends up being about 100 J to compress water by 1%.

Problem 1.33: An ideal gas is made to undergo a cyclic process shown in the figure to the right. For each of the steps A, B, C, determine whether each of the following is positive, negative, or zero: (a) the work done on the gas; (b) the change in the energy content added to the gas; (c) the heat added to the gas. Then determine the sign of each of these three quantities for the whole cycle. What does this process accomplish?

For A, the work being done on the gas is (a) negative, by the equation W = -P\Delta V. The energy added to the gas is (b) positive, which I could demonstrate with the equipartition of energy principle, but would rather show by observing that if the gas increases in volume but maintains the same pressure, that means that the frequency of collisions with the container must remain the same as when the gas was at lower volume. The only way for that to happen is if the molecules are moving faster, so energy must have been added to the system. Since the energy of the system is the heat plus the work, and the work is negative and the energy is positive, then the heat added must also be (c) positive (and greater than the absolute value of the work).

For B, the work done on the gas is (a) zero, since there is no change in volume. By equipartition of energy for an ideal gas, U = \frac{3}{2}PV, the energy increase is (b) positive. You can also get this from observing that increased pressure at the same volume means more collisions with the container per unit time, which means that the molecules of the gas must be moving faster, and so must have more energy. Since there is more energy and no work has been done, the heat added is (c) positive.

For C, the work done on the gas must be (a) positive, because the volume decreases. The energy added to the gas is (b) negative, because even though the volume is decreasing, the pressure is going down. Since the work done on the gas is positive and the overall energy change is negative, the heat added to the gas is also (c) negative.

For the whole cycle, the net work must be (a) positive, because the average pressure is higher during step C than during step A. Since the pressure and volume at the end of the cycle are the same as at the start, the net energy change must be (b) zero. And since the work is positive and the energy change is zero, the heat added must be (c) negative. So this cycle takes in energy as work and emits the energy as heat.

There’s another problem here, but it’s basically the same, except it’s a four-step rectangular cycle that goes clockwise, and turns heat added into work done by the gas. It’s late, so I’m not going to go through the details.