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Chapter 2: Q. 2.41 (page 83)

Describe a few of your favorite, and least favorite, irreversible processes. In each case, explain how you can tell that the entropy of the universe increases.

Short Answer

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The entropy of the universe increases an example of irreversible processes are

Allowing a hot cup of tea to cool on its own and an egg that has fallen on the floor and cracked open, spilling its contents and in general, anything that cannot occur spontaneously in a time-reversed form is irreversible. If you leave a cup of tea alone, it will never heat up. A broken egg, likewise, will never rejoin itself.

Step by step solution

01

Step: 1 Definition of irreversible process in entropy:

The overall energy of the system and its environment grows when an irreversible event occurs. To establish whether or not a hypothetical process is reversible, the second law of thermodynamics can be applied. When there's no dissipation, a process appears to be reversible.The entropy of the cosmos remains unaltered in a reversible process, whereas the entropy of the universe grows in an irreversible process. It also rises when a quantifiable non-spontaneous process occurs. Because energy continually goes downward, entropy increases.

02

Step: 2 Example of irreversible process:

The entropy of the universe increases an example of irreversible processes are:
1Allowing a hot cup of tea to cool on its own.

2An egg that has fallen on the floor and cracked open, spilling its contents.

3In general, anything that cannot occur spontaneously in a time-reversed form is irreversible. If you leave a cup of tea alone, it will never heat up. A broken egg, likewise, will never rejoin itself.

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Most popular questions from this chapter

For either a monatomic ideal gas or a high-temperature Einstein solid, the entropy is given by times some logarithm. The logarithm is never large, so if all you want is an order-of-magnitude estimate, you can neglect it and just say . That is, the entropy in fundamental units is of the order of the number of particles in the system. This conclusion turns out to be true for most systems (with some important exceptions at low temperatures where the particles are behaving in an orderly way). So just for fun, make a very rough estimate of the entropy of each of the following: this book (a kilogram of carbon compounds); a moose of water ; the sun of ionized hydrogen .

Fun with logarithms.
a Simplify the expressionealn⁡b. (That is, write it in a way that doesn't involve logarithms.)
b Assuming that b<<a, prove that ln⁡(a+b)≈(ln⁡a)+(b/a). (Hint: Factor out the afrom the argument of the logarithm, so that you can apply the approximation of part d of the previous problem.)

Consider again the system of two large, identical Einstein solids treated in Problem 2.22.

(a) For the case N=1023, compute the entropy of this system (in terms of Boltzmann's constant), assuming that all of the microstates are allowed. (This is the system's entropy over long time scales.)

(b) Compute the entropy again, assuming that the system is in its most likely macro state. (This is the system's entropy over short time scales, except when there is a large and unlikely fluctuation away from the most likely macro state.)

(c) Is the issue of time scales really relevant to the entropy of this system?

(d) Suppose that, at a moment when the system is near its most likely macro state, you suddenly insert a partition between the solids so that they can no longer exchange energy. Now, even over long time scales, the entropy is given by your answer to part (b). Since this number is less than your answer to part (a), you have, in a sense, caused a violation of the second law of thermodynamics. Is this violation significant? Should we lose any sleep over it?

Use a computer to plot formula 2.22directly, as follows. Define z=q_{A}/q, so that (1-z)=q_{B}/q. Then, aside from an overall constant that we'll ignore, the multiplicity function is [4z(1-z)]N, where zranges from 0to1and the factor of 4ensures that the height of the peak is equal to 1for any N. Plot this function forN=1,10,100,1000, and 10,000. Observe how the width of the peak decreases asNincreases.

According to the Sackur-Tetrode equation, the entropy of a monatomic ideal gas can become negative when its temperature (and hence its energy) is sufficiently low. Of course this is absurd, so the Sackur-Tetrode equation must be invalid at very low temperatures. Suppose you start with a sample of helium at room temperature and atmospheric pressure, then lower the temperature holding the density fixed. Pretend that the helium remains a gas and does not liquefy. Below what temperature would the Sackur-Tetrode equation predict that Sis negative? (The behavior of gases at very low temperatures is the main subject of Chapter 7.)
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