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  3. Unit 1. thermodynamics
  4. The second law of thermodynamics
  5. Entropy on a microscopic scale

4.7 Entropy on a microscopic scale  (Page 25/10)

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By the end of this section you will be able to:
  • Interpret the meaning of entropy at a microscopic scale
  • Calculate a change in entropy for an irreversible process of a system and contrast with the change in entropy of the universe
  • Explain the third law of thermodynamics

We have seen how entropy is related to heat exchange at a particular temperature. In this section, we consider entropy from a statistical viewpoint. Although the details of the argument are beyond the scope of this textbook, it turns out that entropy can be related to how disordered or randomized a system is—the more it is disordered, the higher is its entropy. For example, a new deck of cards is very ordered, as the cards are arranged numerically by suit. In shuffling this new deck, we randomize the arrangement of the cards and therefore increase its entropy ( [link] ). Thus, by picking one card off the top of the deck, there would be no indication of what the next selected card will be.

The photo shows a person’s hand shuffling a deck of cards.
The entropy of a new deck of cards goes up after the dealer shuffles them. (credit: “Rommel SK”/YouTube)

The second law of thermodynamics requires that the entropy of the universe increase in any irreversible process. Thus, in terms of order, the second law may be stated as follows:

In any irreversible process, the universe becomes more disordered . For example, the irreversible free expansion of an ideal gas, shown in [link] , results in a larger volume for the gas molecules to occupy. A larger volume means more possible arrangements for the same number of atoms, so disorder is also increased. As a result, the entropy of the gas has gone up. The gas in this case is a closed system, and the process is irreversible. Changes in phase also illustrate the connection between entropy and disorder    .

Entropy change of the universe

Suppose we place 50 g of ice at 0 °C in contact with a heat reservoir at 20 °C . Heat spontaneously flows from the reservoir to the ice, which melts and eventually reaches a temperature of 20 °C . Find the change in entropy of (a) the ice and (b) the universe.

Strategy

Because the entropy of a system is a function of its state, we can imagine two reversible processes for the ice: (1) ice is melted at 0 °C ( T A ) ; and (2) melted ice (water) is warmed up from 0 °C to 20 °C ( T B ) under constant pressure. Then, we add the change in entropy of the reservoir when we calculate the change in entropy of the universe.

Solution

  1. From [link] , the increase in entropy of the ice is
    Δ S ice = Δ S 1 + Δ S 2 = m L f T A + m c ∫ A B d T T = ( 50 × 335 273 + 50 × 4.19 × ln 293 273 ) J/K = 76.3 J/K .
  2. During this transition, the reservoir gives the ice an amount of heat equal to
    Q = m L f + m c ( T B − T A ) = 50 × ( 335 + 4.19 × 20 ) J = 2.10 × 10 4 J .

This leads to a change (decrease) in entropy of the reservoir:

Δ S reservoir = − Q T B = −71.7 J/K .

The increase in entropy of the universe is therefore

Δ S universe = 76.3 J/K − 71.7 J/K = 4.6 J/K > 0 .

Significance

The entropy of the universe therefore is greater than zero since the ice gains more entropy than the reservoir loses. If we considered only the phase change of the ice into water and not the temperature increase, the entropy change of the ice and reservoir would be the same, resulting in the universe gaining no entropy.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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