Chapter 15 questions

A gas is allowed t expand (a) adiabatically and (b) isothermally. In each process, does the entropy increase, decrease, or stay the same? Explain.

(a) If a gas expands adiabatically, then Q = 0, so Δ = S 0 by Eq. 15-8, Δ= . S QT/ (b) If a gas expands isothermally, then there is no change in its internal energy, and the gas does work on its surroundings. Thus by the first law of thermodynamics, there must be heat flow into the gas, so Δ>⎯ S 0 the entropy of the gas increases.

(a) What happens if you remove the lid of a bottle containing chlorine gas? (b) Does the reverse process ever happen? Why or why not? (c) Can you think of two other examples of irreversibility?

(a) When the lid is removed, the chlorine gas mixes with the air in the room around the bottle so that eventually both the room and the bottle contain a mixture of air and chlorine. (b) The reverse process, in which the individual chlorine particles reorganize so that they are all in the bottle, violates the second law of thermodynamics and does not occur naturally. It would require a spontaneous decrease in entropy. (c) Adding a drop of food coloring to a glass of water is another example of an irreversible process; the food coloring will eventually disperse throughout the water but will not ever gather into a drop again. The toppling of buildings during an earthquake is another example. The toppled building will not ever become "reconstructed" by another earthquake.

What are the high-temperature and the low-temperature areas for (a) an internal combustion engine, and (b) a steam engine? Are they, strictly speaking, heat reservoirs?

8. (a) In an internal combustion engine, the high-temperature reservoir is the ignited gas-air mixture in the cylinder. The low-temperature reservoir is the "outside" air. The burned gases leave through the exhaust pipe. (b) In the steam engine, the high-temperature reservoir is the heated, high-pressure steam from the boiler. The low-temperature reservoir is the condensed water in the condenser. In the cases of both these engines, these areas are not technically heat "reservoirs," because each one is not at a constant temperature.

You are asked to test the machine that the inventor calls an "in-room air conditioner": a big box, standing in the middle of the room, with a cable that plugs into a power outlet. When the machine is switched on, you feel a stream of cold air coming out of it. How do you know that this machine cannot cool the room?

Any air conditioner-type heat engine will remove heat from the room L ( the Q ⎯ low-temperature input). Work ( ) W is input to the device to enable it to remove heat from the low-temperature region. By the second law of thermodynamics (conservation of energy), there must be a high-temperature exhaust heat QH which is larger than L Q . Perhaps the inventor has come up with some clever method of having that exhaust heat move into a well-insulated heat "sink," like a container of water. But eventually the addition of that heat to the device will cause the device to become warmer than the room itself, and then heat will be transferred to the room. One very simple device that could do what is described in the question would be a fan blowing over a large block of ice. Heat from the room will enter the ice; cool air from near the surface of the ice will be blown by the fan. But after the ice melts, the fan motor would again heat the air.

Explain why the temperature of a gas increases when it is compressed adiabatically.

If the gas is compressed adiabatically, then no heat enters or leaves from the gas. The compression means that work was done ON the gas. By the first law of thermodynamics, Δ=− U QW , since Q = 0, then Δ= . U W2 The change in internal energy is equal to the opposite of the work done by the gas or is equal to the work done on the gas. Since positive work was done on the gas, the internal energy of the gas increased, and that corresponds to an increase in temperature. This is conservation of energy—the work done on the gas becomes internal energy of the gas particles, and the temperature increases accordingly.

Entropy is often called "time's arrow" because it tells us in which direction natural processes occur. If a movie were run backward, name some processes that you might see that would tell you that time was "running backward."

In an action movie, you might see a building or car changing from an exploded state to an unexploded state, or a bullet that was fired going backward into the gun and the gunpowder "unexploding." In a movie with vehicle crashes, you might observe two collided vehicles separating from each other, becoming unwrecked as they separate, or someone "unwrite" something on a piece of paper—moving a pen over paper, taking away written marks as the pen moves.

Can you warm a kitchen in winter by leaving the oven door open? Can you cool the kitchen on a hot summer day by leaving the refrigerator door open? Explain.

It is possible to warm the kitchen in the winter by having the oven door open. The oven heating elements radiate heat energy into the oven cavity, and if the oven door is open, then the oven is just heating a bigger volume than usual. There is no thermodynamic cycle involved here. However, you cannot cool the kitchen by having the refrigerator door open. The refrigerator exhausts more heat than it removes from the refrigerated volume, so the room actually gets warmer with the refrigerator door open because of the work done by the refrigerator compressor. If you could have the refrigerator exhaust into some other room, then the refrigerator would be similar to an air conditioner, and it could cool the kitchen, while heating up some other space. Or you could unplug the refrigerator and open the door. That would cool the room somewhat, but would heat up the contents of the refrigerator, which is probably not a desired outcome!

In an isothermal process, 3700 J of work is done by an ideal gas. Is this enough information to tel how much heat has been added to the system? If so, how much? If not, why not?

Since the process is isothermal, there is no change in the internal energy of the gas. Thus Δ = − =→ = U QW Q W 0 , so the heat absorbed by the gas is equal to the work done by the gas. Thus 3700 J of heat was added to the gas.

Suppose a lot of papers are strewn all over the floor; then you stack them neatly. Does this violate the second law of thermodynamics? Explain.

While the state of the papers has changed from disorder to order, they did not do so spontaneously. An outside source (you) caused the increase in order. You had to provide energy to do this (through your metabolic processes), and in doing so, your entropy increased more than the entropy of the papers decreased. The overall effect is that the entropy of the universe increased, satisfying the second law of thermodynamics.