# Entropy Changes in the Surroundings Now that we have seen how to calculate D S system for reversible and irreversible - PowerPoint PPT Presentation  Download Presentation Entropy Changes in the Surroundings Now that we have seen how to calculate D S system for reversible and irreversible

Entropy Changes in the Surroundings Now that we have seen how to calculate D S system for reversible and irreversible
Download Presentation ## Entropy Changes in the Surroundings Now that we have seen how to calculate D S system for reversible and irreversible

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1. Entropy Changes in the Surroundings Now that we have seen how to calculate DSsystem for reversible and irreversible processes, the relation: DS total = DS system + DS surroundings can be used to calculate either DSsurroundingsor DS totalgiven values for the other entropy changes. Alpha iron and beta iron are in equilibrium at 1033 K and 1.00 atm and therefore beta iron converts reversibly into alpha iron when it is cooled at 1033 K and 1.00 atm:1033 K Fe (b) -------------> Fe (a) DH1033 K = - 879 J / mole 1.00 atm Which of these two phases is stable above 1033 K? The entropy change in the system, the iron, when 1.00 mole of iron undergoes this phase change is: DSFe= dqrev, p / T = qrev, p / T = DH1033 K / T = (- 879 J / mole) / 1033 K = - 0.851 J / mole K Why does the entropy of the iron decrease in this case? The entropy change in the surroundings is: DS surroundings = DS total - DS Fe = 0 - (- 0.851 J / mole K) = + 0.851 J / K Why is the total entropy change zero? 26.1

2. vacuum Note that in the exothermic condensation of beta iron to alpha iron heat was transferred to the isothermal surroundings and the entropy of the surroundings increased. While not explicitly stated we can infer that the surroundings were isothermal and at the same temperature as the system, since, if they weren’t, thermal contact between the system and the surroundings would have driven a temperature change in the system. In general when heat is transferred to an infinite thermal reservoir, the entropy change in the reservoir will be positive and vice versa. 1.000 mole of an ideal gas initially at 25.0 oC expands adiabatically into a vacuum doubling its volume.Expansions into a vacuum are known asfree expansions: For this adiabatic free expansion calculate the work, w, the heat, q, the internal energy change, DE, the temperature change, DT, the enthalpy change, DH, the entropy change in the system, DSsystem, the entropy change in the surroundings, DSsurroundings , and the total entropy change, DStotal . 26.2

3. 100.0 g of Al at 50.0 oC 50.0 g of Cu at 150.0 oC Al at Tf Cu at Tf + A 100.0 gram Al block initially at 50.0 oC is brought into thermal contact with a 75.0 gram Cu block initially at 150.0 oC: The specific heat capacities of Al and Cu are 0.216 cal / g oC and 0.0919 cal / g oC, respectively. Assume that heat transfer only occurs between the blocks. What is the final temperature, when the blocks have come to thermal equilibrium? The entropy change for the Al block is: DSAl=  dqrev, p / T = 323.2 K TfmAl cp, Al dT / T = mAl cp, Al ln (Tf / 323.2 K) Should you view the Al and Cu blocks as an isolated system or should you view one metal block as the system and the other as the surroundings? Does it matter which view you adopt? Calculate the entropy changes in the system and the surroundings. What do you expect the sign on DStotal to be? Calculate DStotal. Does the calculated sign on DStotal confirm your expectations? 26.3

4. A 10.0 gram ice cube is tossed into a swimming pool, which is at 25.0 oC: What is the entropy change for the ice cube as it comes to thermal equilibrium with the pool? DS ice cube melting = DH melting, 0.0 oC / 273.2 K = (10.0 g) (+79.7 cal / g) / 273.2 K = + 2.92 cal / K Why do we assume that the ice cube melts at 0.0 oC? What is the entropy change for the warming of the liquid water that resulted from the melted ice cube? DS warming water=  dqrev, p / T = 273.2 K 298.2KmH2O cp, H2O (l) dT / T = (10.0 g) (1.00 cal / g K) ln (298.2 K / 273.2 K) = + 0.876 cal / K If we choose the ice cube to be the system, the entropy change for the system is: DS ice cube = DS ice cube melting + DS warming water = (+2.92 cal / K) + (+0.876 cal / K) = + 3.80 cal / K 26.4

5. Since we have choosen the ice cube to be the system, the pool will be the surroundings. The entropy change for the pool is: DS pool =  dqrev, pool / Tpool = qrev, pool / Tpool Why does the temperature of the pool remain constant? Does the pool fit our definition of an infinite thermal reservoir? = - (q melting of ice cube + q warming of melted water) / Tpool Why is qrev, pool = - (q melting of ice cube + q warming of melted water) ? = - [ (10.0 g) (+ 79.7 cal / g) + 273.2 K 298.2KmH2O cp, H2O (l) dT ] / Tpool = - [(10.0 g) (+ 79.7 cal / g) + (10.0 g)(1.00 cal / g K) (298.2 K - 273.2 K)] / 298.2 K = - 3.51 cal / K The total entropy change is: DS total = DS ice cube + DS pool =(+ 3.80 cal / K) + (- 3.51 cal / K) = + 0.29 cal / K Is the positive sign on DStotal consistent with the spontaneity of the process? 26.5