S = Si + S2 + ••• U = Ui + U2 + ••• H = Hi + H2 + •••
Example 4.12: a column of air We can think of a column of air as a composite system. Its pressure and temperature might be varying with altitude z but we can add up these contributions from individual slabs:
-lower where p(z) is mass density, u(z) is the specific internal energy, etc. □
In general, if two systems are brought into "contact" we can say the sum of the internal energies of the system will remain the same. Bringing two systems into contact amounts to removing or relaxing a constraint. For example, if two gases A and B at the same pressure are in a chamber, but separated by a partition, the removal of the partition will allow the gases to mix. No change in internal energy will occur if the chamber containing both subsystems is insulated from its environment. On the other hand, the change in the total entropy must be zero or positive. Processes (either reversible or not) in which the sum of the entropies of a system and its surroundings yield a negative change in total entropy do not occur in nature. This principle can be of great utility in determining which way a process will proceed when a constraint is relaxed or removed. Recall that removal of a constraint means that in order to restore the system to its original state additional work must be performed by some external agent. This last is really the essence of the Second Law of Thermodynamics. Solutions to problems of this type may not always be facilitated by use of the internal energy alone, since it may remain fixed when the constraint is removed. But restoration of the original conditions does require work, and hence changes in the internal energy alone are insufficient to describe what has happened. It turns out that the entropy change provides this additional information.
Example 4.13 How could we restore the gas in Example 4.11 to its original condition and how much work would be required?
Answer: First we would have to bring the ideal gas into contact with a reservoir of temperature T, then we would perform an isothermal compression of the gas from volume 2V0 to V0. As we perform the compression of the gas, infinitesimal temperature differences will develop between the system and the reservoir: heat will transfer from one system to the other to maintain the fixed temperature. The work performed (by the system) in the isothermal compression is just -MRT ln 2.
Note that in the case of mixing two gases A and B mentioned above we would need to recompress each to its original volume by an isothermal route in order to restore them to their original states. We might in this case use a membrane which allows molecules A to pass into a new adjacent chamber but not B. But suppose A and B are the same gas. Does the entropy increase when the partition is removed? No. □
It is possible to derive an analytical expression for the entropy of an ideal gas. To proceed consider the expression for an ideal gas dry air parcel. Recall that the entropy S is given by
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