Question 4.5-2: An Alternative Way to Solve One Problem Sometimes it is poss...

The mass, energy, and entropy balances (on a molar basis) for this system are

N_{1}^{f}=N_{1}^{i}                                   (a) 

N_{1}^{f} \underline{U}_{1}^{f}=N_{1}^{i} \underline{U}_{1}^{i}-\int_{V_{1}^{i}}^{V_{1}^{f}} P d V                       (b) 

and

N_{1}^{f} \underline{S} _{1}^{f}=N_{1}^{i} \underline{S} _{1}^{i}                                      (c) 

Now the important observation is that by combining Eqs. a and c, we obtain

\underline{S}_{1}^{f}=\underline{S}_{1}^{i}                              (d) 

so the process is isentropic (i.e., occurs at constant entropy) for the system we have chosen. Using Eq. 4.4-3,

and Eq. d yields

This is precisely the result obtained in Eq. f of Illustration 3.4-5 using the energy balance on the open system consisting of the total contents of cylinder 1. The remainder of the problem can now be solved in exactly the same manner used in Illustration 3.4-5.
Although the system choice used in this illustration is an unusual one, it is one that leads quickly to a useful result. This demonstrates that sometimes a clever choice for the thermodynamic system can be the key to solving a thermodynamic problem with minimum effort.

However, one also has to be careful about the assumptions in unusual system choices. For example, consider the two cylinders connected as in Fig. 4.5-2, where the second cylinder is not initially evacuated. Here we have chosen to treat that part of the initial contents of cylinder 1 that
will be in that cylinder at the end of the process as one system and the total initial contents of cylinder 2 as a second system. The change that occurs in the first system, as we already discussed, is adiabatic and reversible, so that

\left(\frac{T_{1}^{f}}{T_{1}^{i}}\right)^{C_{ P }^{*} / R}=\left(\frac{P_{1}^{f}}{P_{1}^{i}}\right)                                           (e)

Figure 4.5-2 Incorrect system choice for gas contained in cylinder 2.

One might expect that a similar relation would hold for the system shown in cylinder 2. This is not the case, however, since the gas entering cylinder 2 is not necessarily at the same temperature as the gas already there (hydrodynamics will ensure that the pressures are the same). Therefore, temperature gradients will exist within the cylinder, and our system, the partial contents of cylinder 2, will not be adiabatic. Consequently, Eq. e will not apply.