Thermodynamics

dQ T 0 (7–1)Chapter 7 | 333It appears that the combined system is exchanging heat with a single thermalenergy reservoir while involving (producing or consuming) work W Cduring a cycle. On the basis of the Kelvin–Planck statement of the secondlaw, which states that no system can produce a net amount of work whileoperating in a cycle and exchanging heat with a single thermal energyreservoir, we reason that W C cannot be a work output, and thus it cannot bea positive quantity. Considering that T R is the thermodynamic temperatureand thus a positive quantity, we must havewhich is the Clausius inequality. This inequality is valid for all thermodynamiccycles, reversible or irreversible, including the refrigeration cycles.If no irreversibilities occur within the system as well as the reversiblecyclic device, then the cycle undergone by the combined system is internallyreversible. As such, it can be reversed. In the reversed cycle case, allthe quantities have the same magnitude but the opposite sign. Therefore, thework W C , which could not be a positive quantity in the regular case, cannotbe a negative quantity in the reversed case. Then it follows that W C,int rev 0since it cannot be a positive or negative quantity, and therefore a dQ T b int rev 0(7–2)for internally reversible cycles. Thus, we conclude that the equality in theClausius inequality holds for totally or just internally reversible cycles andthe inequality for the irreversible ones.To develop a relation for the definition of entropy, let us examine Eq. 7–2more closely. Here we have a quantity whose cyclic integral is zero. Letus think for a moment what kind of quantities can have this characteristic.We know that the cyclic integral of work is not zero. (It is a good thingthat it is not. Otherwise, heat engines that work on a cycle such as steampower plants would produce zero net work.) Neither is the cyclic integral ofheat.Now consider the volume occupied by a gas in a piston–cylinder deviceundergoing a cycle, as shown in Fig. 7–2. When the piston returns to its initialposition at the end of a cycle, the volume of the gas also returns to itsinitial value. Thus the net change in volume during a cycle is zero. This isalso expressed as dV 0(7–3)That is, the cyclic integral of volume (or any other property) is zero. Conversely,a quantity whose cyclic integral is zero depends on the state onlyand not the process path, and thus it is a property. Therefore, the quantity(dQ/T ) int rev must represent a property in the differential form.Clausius realized in 1865 that he had discovered a new thermodynamicproperty, and he chose to name this property entropy. It is designated S andis defined as1 m 3 3 m 31 m 3∫ dV = ∆V cycle = 0FIGURE 7–2The net change in volume (a property)during a cycle is always zero.dS a dQ T b int rev1kJ>K2(7–4)