1. An ideal gas undergoes a reversible isothermal expansion at 132.C. The entropy of the gas increases by 46.2 J/K. The heat absorbed is_ 18711J △S=→Q=△ST=46.2×(273+132)=18711(0) 2. In Fig. 1, suppose that the change in entropy of the system in passing from state a to state b along path 1 is +0.60 J/K. The entropy change in passing from state a to b along path 2 0.6 J/K, and the entropy change in passing from state b to a, along path 2 is-0.6 J/K Solution The entropy is a state variable of a thermodynamic system. T(K) Fig1 It is settled by the initial and final states 3. For the Carnot cycle shown in Fig. 2, the heat that enters 200F is_ J, the work done on the system is_ 22.5 J Solutio 0.6 According to the entropy change 4Ss O 1g2 We have O= AsT. So the heat enters is Q=Qn=(0.6-0.2)×400=160() According to the first law of thermodynamics, the work done on the system is 250 W=a=( 160=60J 4. A Carnot heat engine has an efficiency of 0.300 operating between a high-temperature reservoir at temperature TH and a low-temperature reservoir at 20C. To increase the efficiency of the Carnot heat engine to 0.400, the temperature of the hot reservoir be increased is 69.76K (in Kelvin) Solution The efficiency of the Carnot heat engine is e=l T273+20 =41857(K) 1-0.3 =48833(K) T E21-0.41. An ideal gas undergoes a reversible isothermal expansion at 132°C. The entropy of the gas increases by 46.2 J/K. The heat absorbed is 18711 J . Solution Q ST 46.2 (273 132) 18711 (J ) T Q ∆S = ⇒ = ∆ = × + = 2. In Fig. 1, suppose that the change in entropy of the system in passing from state a to state b along path 1 is +0.60 J/K. The entropy change in passing from state a to b along path 2 is 0.6 J/K , and the entropy change in passing from state b to a along path 2 is -0.6 J/K . Solution: The entropy is a state variable of a thermodynamic system. It is settled by the initial and final states. 3. For the Carnot cycle shown in Fig.2, the heat that enters is 60 J , the work done on the system is 22.5 J . Solution: According to the entropy change T Q ∆S = , We haveQ = ∆ST . So the heat enters is = = (0.6 − 0.2) × 400 = 160(J) Q QH According to the first law of thermodynamics, the work done on the system is ) 160 60J 400 250 W = εQ = (1− × = 4. A Carnot heat engine has an efficiency of 0.300 operating between a high-temperature reservoir at temperature TH and a low-temperature reservoir at 20 °C. To increase the efficiency of the Carnot heat engine to 0.400, the temperature of the hot reservoir be increased is 69.76 K (in Kelvin). Solution: The efficiency of the Carnot heat engine is H C T T ε =1− 418.57(K) 1 0.3 273 20 1 1 1 1 1 1 = − + = − = − ⇒ = ε ε C H H C T T T T 488.33(K) 1 0.4 273 20 1 1 2 2 2 2 = − + = − = − ⇒ = ε ε C H H C T T T T 1 P a b Fig.1 2 0 S(J/K) T(K) Fig.2 0.2 0.4 0.6 100 200 300 400