Entropy changes in the ideal gas

The relations derived in the preceding sections are applicable to any system. They have a particularly simple form when applied to the ideal gas, which is the result of the fact that in the ideal gas the energy and the temperature are equivalent variables 17 = C dT. Using this value of dU in Eq. (9.23), we obtain 

This derivative is always positive in an isothermal transformation, the entropy of the ideal gas increases with increase in volume. The rate of increase is less at large volumes, since V appears in the denominator. 

Note that in this example, as well as in the earlier ones, it is essential to express the temperature in kelvins. Note also that in the second part of the problem where only a pressure ratio is involved, we may use any unit of pressure so long as it is related to the pascal by a multiplicative constant. In forming the ratio, the conversion factor will disappear, and thus need not have been introduced in the first place. 

Suppose that we integrate this equation from p = I atm to any pressure p. Then 

To calculate a numerical value of the logarithm on the right-hand side of Eq. (9.52), it is essential that the pressure be expressed in atmospheres. Then the ratio (p/1 atm) will be a pure number, and the operation of taking the logarithm is possible. (Note that it is not possible to take the logarithm of five oranges.) It is customary to abbreviate Eq. (9.52) to the simple form 

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Figure 4-3 Path for the calculation of entropy changes in the ideal-gas state.

If we further write V /Vi = Pi/P (recall that both states are at the same temperature), we obtain the familiar result for isothermal entropy change in the ideal-gas state ... 

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