In other words, the temperature of any given physical system must be finite, that is,
. 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,
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
, 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
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
of the box, as shown in part (b). What is the entropy change of the universe (the system plus its environment)?
The adiabatic free expansion of an ideal gas from volume
to volume
.
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
to
. 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
where
because
Now, with the help of the ideal gas law, we have
so the change in entropy of the gas is
Because
,
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.
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
Solution
Using
, the specific heat of steel, we have for the hotter object
Similarly, the entropy change of the cooler object is
The net entropy change of the two objects during the heat transfer is then
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.
