An Alternative Way to Solve One Problem Sometimes it is possible to solve a thermodynamic problem several ways, based on different choices of the system. To see this, we consider Illustration 3.4-5, which was concerned with the partial evacuation of a compressed gas cylinder into an evacuated cyli

Question

An Alternative Way to Solve One Problem

Sometimes it is possible to solve a thermodynamic problem several ways, based on different choices of the system. To see this, we consider Illustration 3.4-5, which was concerned with the partial evacuation of a compressed gas cylinder into an evacuated cylinder of equal volume. Suppose we now choose for the system of interest only that portion of the contents of the first cylinder that remains in the cylinder when the pressures have equalized (see Fig. 4.5-1, where the thermodynamic system of interest is within the dashed lines). Note that with this choice the system is closed, but of changing volume. Furthermore, since the gas on one side of the imaginary boundary has precisely the same temperature as the gas at the other side, we can assume there is no heat transfer across the boundary, so that the system is adiabatic. Also, with the exception of the region near the valve (which is outside what we have taken to be the system), the gas in the cylinder is undergoing a uniform expansion so there will be no pressure, velocity, or temperature gradients in the cylinder. Therefore, we can assume that the changes taking place in the system occur reversibly.

Figure 4.5-1 The dashed lines enclose a system consisting of gas initially in the first cylinder that remains in that cylinder at the end of the process.

Accepted Answer

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

[latex]N_{1}^{f}=N_{1}^{i}[/latex] (a)

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

and

[latex]N_{1}^{f} \underline{S} _{1}^{f}=N_{1}^{i} \underline{S} _{1}^{i}[/latex] (c)

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

[latex]\underline{S}_{1}^{f}=\underline{S}_{1}^{i}[/latex] (d)

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

[latex]\underline{S}_{1}\left(P^{f}, T^{f}\right)-\underline{S}_{1}\left(P^{i}, T^{i}\right)=C_{ P }^{*} \ln \left(\frac{T_{1}^{f}}{T_{1}^{i}}\right)-R \ln \left(\frac{P_{1}^{f}}{P_{1}^{i}}\right)[/latex]

and Eq. d yields

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

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

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

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

This Question has been Answered.

Already have an account?

Login

Related Answered Questions