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In other words, the temperature of any given physical system must be finite, that is, T > 0 . This produces a very interesting question in physics: Do we know how a system would behave if it were at the absolute zero temperature?

The reason a system is unable to reach 0 K is fundamental and requires quantum mechanics to fully understand its origin. But we can certainly ask what happens to the entropy of a system when we try to cool it down to 0 K. Because the amount of heat that can be removed from the system becomes vanishingly small, we expect that the change in entropy of the system along an isotherm approaches zero, that is,

lim T 0 ( Δ S ) T = 0 .

This can be viewed as another statement of the third law, with all the isotherms becoming isentropic    , or into a reversible ideal adiabat. We can put this expression in words: A system becomes perfectly ordered when its temperature approaches absolute zero and its entropy approaches its absolute minimum .

The third law of thermodynamics puts another limit on what can be done when we look for energy resources. If there could be a reservoir at the absolute zero temperature, we could have engines with efficiency of 100 % , which would, of course, violate the second law of thermodynamics.

Entropy change of an ideal gas in free expansion

An ideal gas occupies a partitioned volume V 1 inside a box whose walls are thermally insulating, as shown in [link] (a). When the partition is removed, the gas expands and fills the entire volume V 2 of the box, as shown in part (b). What is the entropy change of the universe (the system plus its environment)?

Part a of the figure shows a container which has gas of volume V subscript 1 on the left side and nothing on the right side. Part b shows a container which is completely filled with gas of volume V subscript 2.
The adiabatic free expansion of an ideal gas from volume V 1 to volume V 2 .

Strategy

The adiabatic free expansion of an ideal gas is an irreversible process. There is no change in the internal energy (and hence temperature) of the gas in such an expansion because no work or heat transfer has happened. Thus, a convenient reversible path connecting the same two equilibrium states is a slow, isothermal expansion from V 1 to V 2 . In this process, the gas could be expanding against a piston while in thermal contact with a heat reservoir, as in step 1 of the Carnot cycle.

Solution

Since the temperature is constant, the entropy change is given by Δ S = Q / T , where

Q = W = V 1 V 2 p d V

because Δ E int = 0 . Now, with the help of the ideal gas law, we have

Q = n R T V 1 V 2 d V V = n R T ln V 2 V 1 ,

so the change in entropy of the gas is

Δ S = Q T = n R ln V 2 V 1 .

Because V 2 > V 1 , Δ S is positive, and the entropy of the gas has gone up during the free expansion.

Significance

What about the environment? The walls of the container are thermally insulating, so no heat exchange takes place between the gas and its surroundings. The entropy of the environment is therefore constant during the expansion. The net entropy change of the universe is then simply the entropy change of the gas. Since this is positive, the entropy of the universe increases in the free expansion of the gas.

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Entropy change during heat transfer

Heat flows from a steel object of mass 4.00 kg whose temperature is 400 K to an identical object at 300 K. Assuming that the objects are thermally isolated from the environment, what is the net entropy change of the universe after thermal equilibrium has been reached?

Strategy

Since the objects are identical, their common temperature at equilibrium is 350 K. To calculate the entropy changes associated with their transitions, we substitute the irreversible process of the heat transfer by two isobaric, reversible processes, one for each of the two objects. The entropy change for each object is then given by Δ S = m c ln ( T B / T A ) .

Solution

Using c = 450 J/kg · K , the specific heat of steel, we have for the hotter object

Δ S h = T 1 T 2 m c d T T = m c ln T 2 T 1 = ( 4.00 kg ) ( 450 J/kg · K ) ln 350 K 400 K = −240 J/K .

Similarly, the entropy change of the cooler object is

Δ S c = ( 4.00 kg ) ( 450 J/kg · K ) ln 350 K 300 K = 277 J/K .

The net entropy change of the two objects during the heat transfer is then

Δ S h + Δ S c = 37 J/K .

Significance

The objects are thermally isolated from the environment, so its entropy must remain constant. Thus, the entropy of the universe also increases by 37 J/K.

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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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