MME6701 Solution to Assignment #1 1. 5 moles of an ideal gas of volume 2.5 m3 and temperature 25 °C are expanded isothermally to 3.4

m3. The change in entropy during the process is calculated as 5.0 J/mol-K. Calculate the change in internal energy of the system.

Given data: n = 5 mol V1 = 2.5 m3 V2 = 3.4 m3 T1 = 25 °C = 298 K T2 = T1 = 298 K ΔS = 5 J/mol-K For an isothermal process, the work done on the system W = –nRT ln (V2/V1) = –(5 mol).(8.314 J/mol-K).(298 K). ln (3.4 m3 / 2.5 m3) = –3809.08 J The amount of heat absorbed by the system Q = nTΔS = (5 mol).(298 K).(5 J/mol-K) = 7450 J Then, the change in internal energy of the system ΔU = Q + W = 7450 - 3809.08 J = 3640.92 J

2. One kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and 1

bar until its volume triples. Calculate W, Q, ΔU for the process. Assume air obeys the relation PV/T = 83.14 bar cm3/mol-K and that CP = 29 J/mol.K.

Given data: m = 1 kg PV/T = 83.14 bar-cm3/mol-K T1 = 300 K V2 = 3V1 P1 = 1 bar CP = 29 J/mol-K Molecular mass of air at STP = 28.97 g/mol So, number of air moles, n = (1000 g)/(28.97 g/mol) = 34.52 mol Using the given relation, V1 = (83.14 bar-cm3/mol-K) n(T/P) = (83.14 bar-cm3/mol-K).(34.52 mol).(300 K / 1 bar) = 860997.84 cm3 V2 = 3V1 = 3 (860997.84 cm3) = 2582993.52 cm3 For an isobaric process, the work done on the system W = –P(V2–V1) = – (1 bar).(2582993.52–860997.84 cm3) = –1721995.68 bar-cm3 = (–1721995.68 bar-cm3).(8.314 J/83.14 bar-cm3) = –172199.57 J = –172.20 kJ Now the final temperature of the system T2 = PV2/83.14 = (1 bar)( 2582993.52 cm3) / (83.14 bar-cm3/mol-K) = 31068 K/mol = 900 K The heat absorbed by the system Q = ∫ nCPdT = 29 n (T2–T1) J = (29 J/mol-K).(34.52 mol).(900-300 K) = 600648 J = 600.65 kJ Then, the change in internal energy of the system ΔU = Q + W = 600.65 – 172.20 kJ = 428.45 kJ

3. Two moles of monatomic ideal gas are contained at 1 atm pressure and 300 K temperature. 34166 joules of heat are transferred to the gas, as a result of which the gas expands and does 1216 joules of work against its surroundings. The process is reversible. Calculate the final temperature of the gas.

Given data: n = 2 mol Q = +34166 J P1 = 1 atm W = –1216 J T1 = 300 K cV = 3R/2 The change in internal energy of gas ΔU = Q + W = 34166 – 1216 J = 32950 J The change in internal energy of an ideal gas can also expressed by the relation ΔU = ncV ΔT = (3R) (T2–T1) T2 = T1 + ΔU/3R = 300 K + (32950 J) / 3(8.314 J/mol-K) = 1621.06 K

4. One mole of an ideal gas is compressed isothermally but irreversibly at 130 °C from 2.5 bar to 6.5

bar in a piston-cylinder device. The work required is 30% greater than the work of reversible, isothermal compression. The heat transferred from the gas during compression flows to a heat reservoir at 25 °C. Calculate the entropy changes of the gas, the heat reservoir, and DStotal.

Given data: n = 1 mol T1 = 130 °C = 403 K P1 = 2.5 bar P2 6.5 bar Work done to the system during the reversible isothermal process Wrev = –nRT ln (V2/V1) = nRT ln (P2/P1) = (1 mol).(8.314 J/mol-K).(403 K) ln (6.5/2.5) = 3201.48 J Using 30% more, the irreversible work done to the system Wirr = 1.3 (3201.4 J) = 4161.92 J Since for an isothermal process, ΔU = CV dT = 0 = Q+W, and the heat input into the system Qsys = –Wirr = –4161.92 J Then, the entropy change for the system (the gas) ΔSsys = Qirr/T1 = (–4161.92 J)/(403 K) = –10.33 J/mol Now, the heat input into the surroundings (i.e., the heat reservoir) at 25 °C (298 K) Qres = –Qsys = +4161.92 J Then, the entropy change for the heat reservoir ΔSres = Qirr/T1 = (+4161.92 J)/(298 K) = +13.97 J/mol Then the total entropy change for the process ΔStotal = ΔSsys + ΔSres = –10.33 J/mol + 13.97 J/mol = +3.64 J/mol