## Ql t T J Ql TT

Since temperatures are always positive, and since the initial restriction was T1 > T2, the above equation shows that Q2, must be positive. That is, the direction of heat flow is from the hot body to the cold body.

The above application of the Second Law may seem needlessly formal, but a more challenging analysis of the process depicted in Fig. 1.16 is the following. If the initial state of the isolated system is T1 > T2,

• what is the final common equilibrium temperature of the two systems?

• what is the entropy increase of the isolated system when the final (equilibrium) state is attained?

• does the final state represent the maximum possible entropy of the isolated system?

Answering these questions requires knowing how temperature affects internal energy (or enthalpy) and entropy, which are considered in Problem 3.15

Example: Entropy change resulting from a rapid (irreversible) expansion of an ideal gas

In Fig. 1.17, an ideal gas is initially partitioned into the left half of an isolated system by a membrane. The process consists of puncturing the membrane, thereby allowing the gas to occupy both halves equally.

isolated . system isolated . system

Fig. 1.17 An ideal gas contained on one side of an isolated system by a membrane that is subsoquently ruptured to allow the gas So redistribute

The system is isolated, so AU = 0, hence the gas temperature does not change. The gas, however, has doubled its volume. The entropy change occasioned by this expansion cannot be calculated for the irreversible process depicted in Fig. 1.17. However, since entropy is a state function, AS can be calculated by constructing a reversibly process (any convenient one) that brings the system between the same states. Such a reversible process is the reverse of the one shown in Fig. 1.12, in which the work done on the system is given by Eq (1.3a). Since AU = 0 for isothermal expansion of an ideal gas, the First Law requires that the heat absorbed be equal to the work performed. The entropy change for the reversible isothermal expansion of an ideal gas to twice the volume is:

The entropy change in the irreversible process of Fig. 1.17 is also nRln2, despite the absence of heat exchange with the surroundings. It is positive, in conformance with the principle of maximum entropy for an isolated system. From a microscopic perspective, doubling the volume increases the number of quantum states available to the gas, and so increases its state of disorder. That the uniform state on the right in Fig. 1.17 is the equilibrium state is intuitively obvious. It can be shown that any other distribution has a lower entropy than the uniform distribution, which leads to equal gas pressures in the two halves of the container.

Problem 3.17 shows that the entropy increase due to this irreversible expansion is not restricted to ideal gases as in the above example. Several examples of the 2nd law applied to reversible compression of an ideal gas are given below

Example # 1 What is the entropy change of an ideal gas that is compressed reversibly and isothermally from volume V1 to V2? (see Fig. 1.12). Explain the sign of the entropy change.

The internal energy of an ideal gas depends only on temperature; for an isothermal process, AU = 0. The First law (Eq(1.4) gives:

where the last equality comes from Eq (1.3a). Because the gas is compressed, V2/V1 < 1, and the heat is negative (heat is released during the process) and the work is negative (work is done on the system by the surroundings). n is the moles of gas in the cylinder.

Since the process is reversible, Eq (1.9) is the appropriate form of the 2nd law, and the entropy change is:

The entropy change is negative in the process, and can be explained in two ways. From a macroscopic viewpoint, the loss of heat from the system implies a reduction in entropy. From a microscopic viewpoint, the reduced volume of the final state 2 means that the molecules occupy fewer translational quantum states. Hence, order is increased, and entropy is decreased.

Example #2 (from Ref. 5) Consider the piston-spring device shown below. Initially, the piston rests on stops and the spring at its equilibrium length (no force on piston)

/rigid structure

/rigid structure piston, 300 kg area = 0.0314 m2

(a) Heat is added to the gas in the cylinder until the piston just leaves the stops. What is the gas temperature at this point?

Since the spring is still at its equilibrium length, the pressure at piston liftoff is due only to the gas pressure of the surroundings and the force due to the piston mass:

where g = 9.81 m/s is the acceleration of gravity. The temperature at piston liftoff is given by the ideal gas law:

(b) Heat is added reversibly to the gas until the final volume is V2 =0.000628 m . How much pV work is done by the system?

For a volume V greater than Vl, the pressure is greater than pL due to compression of the spring. The spring compression distance is x = (V - Vl)/A, and the pressure is:

where k is the spring constant. The work done by the expanding gas is:

## Getting Started With Solar

Do we really want the one thing that gives us its resources unconditionally to suffer even more than it is suffering now? Nature, is a part of our being from the earliest human days. We respect Nature and it gives us its bounty, but in the recent past greedy money hungry corporations have made us all so destructive, so wasteful.

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