Question

drive the change in entropy for an ideal gas

Answer 1

For an ideal gas, . Thus

Using the equation of state for an ideal gas ( ), we can write the entropy change as an expression with only exact differentials:

(5..2)

We can think of Equation (5.2) as relating the fractional change in temperature to the fractional change of volume, with scale factors and ; if the volume increases without a proportionate decrease in temperature (as in the case of an adiabatic free expansion), then increases. Integrating Equation (5.2) between two states ``1'' and ``2'':

For a perfect gas with constant specific heats

In non-dimensional form (using )

(5..3)

Equation 5.3 is in terms of specific quantities. For moles of gas,

This expression gives entropy change in terms of temperature and volume. We can develop an alternative form in terms of pressure and volume, which allows us to examine an assumption we have used. The ideal gas equation of state can be written as

Taking differentials of both sides yields

Using the above equation in Eq. (5.2), and making use of the relations ; , we find

or

Integrating between two states 1 and 2

(5..4)

Using both sides of (5.4) as exponents we obtain

(5..5)

Equation (5.5) describes a general process. For the specific situation in which , i.e., the entropy is constant, we recover the expression . It was stated that this expression applied to a reversible, adiabatic process. We now see, through use of the second law, a deeper meaning to the expression, and to the concept of a reversible adiabatic process, in that both are characteristics of a constant entropy, or isentropic, process.