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Tutorial 2. By Miss Anis Atikah Ahmad. Question 1. A heat engine uses reservoirs at 800°C and 0°C. Calculate maximum possible efficiency I f q H is 1000 J, calculate the maximum - w and the minimum value of - q C. Question 1. Calculate maximum possible efficiency
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Tutorial 2 By Miss Anis Atikah Ahmad
Question 1 A heat engine uses reservoirs at 800°C and 0°C. • Calculate maximum possible efficiency • If qH is 1000 J, calculate the maximum -w and the minimum value of -qC
Question 1 • Calculate maximum possible efficiency TH= 800°C = 1073.15K, TC =0°C =273.15K Work output per cycle Energy input per cycle
Question 1 b) If qH is 1000 J, calculate the maximum -w and the minimum value of -qC Work output per cycle Energy input per cycle
Question 2 Calculate ΔS for each of the following changes in state of 2.50 mol of a perfect monoatomic gas with CV,m =1.5R for all temperatures: • (1.50 atm, 400K) (3.00 atm, 600K) • (2.50 atm, 20.0L) (2.00 atm, 30.0L) • (28.5L, 400K) (42.0L atm, 400K)
Question 2 • For perfect gas;
Question 2 • Since CV,mis constant (same for all temperatures) ; • Given, , thus:
Question 2 a) P1 = 1.5 atm, P2= 3 atm, T1 = 400 K, T2= 600K
Question 2 b) P1 = 2.5 atm, P2= 2 atm, V1 = 20L, V2= 30L
Question 2 c) V1 = 28.5L, V2= 42L, T1= T2 = 400K
Question 3 • After 200 g of gold [cP= 0.0313 cal/(g °C)] at 120°C is dropped into 25.0 g of water at 10°C, the system is allowed to reach equilibrium in an adiabatic container. Calculate: • The final temperature • ΔSAu • ΔSH2O • ΔSAu+ ΔSH2O
Question 3 (a) • At constant pressure; • At equilibrium;
Question 3 • Solving for T2;
Question 3 (b)
Question 3 (c)
Question 3 (d)
Question 4 A sample consisting of 2.00 mol of diatomic perfect gas molecules at 250 K is compressed reversibly and adiabatically until its temperature reaches 300 K. Given that CV,m= 27.5 JK-1mol-1, calculate: • q • w • ΔU • ΔH • ΔS.
Question 4 n = 2.00 mol CV,m= 27.5 JK-1mol-1 T1= 250 K T2 = 300 K Process: reversible adiabatic of a perfect gas • qrev =0 b) w = ? Recall first law: ΔU = q + w For perfect gas, Thus,
Question 4 • ΔU ΔU = q + w = 0 + w = w = 2750 J d) ΔH For a perfect gas; is not given. However, we know that, Thus;
Question 4 e) ΔS • Since qrev =0 (reversible adiabatic),
Question 5 A system consisting of 1.5 mol CO2 (g), initially at 15°C and 9 atm and confined to a cylinder of cross-section 100.0 cm2. It is allowed to expand adiabatically against an external pressure of 1.5 atm until the piston has moved outwards through 15 cm. Assume that carbon dioxide may be considered a perfect gas with CV,m= 28.8 JK-1mol-1, calculate: • q • w • ΔU • ΔT • ΔS
Question 5 A system consisting of 1.5 mol CO2 (g), initially at 15°C and 9 atm and confined to a cylinder of cross-section 100.0 cm2. It is allowed to expand adiabatically against an external pressure of 1.5 atm until the piston has moved outwards through 15 cm. Assume that carbon dioxide may be considered a perfect gas with CV,m= 28.8 JK-1mol-1, calculate: • q • w • ΔU • ΔT • ΔS Pext= 1.5 atm T1 = 15°C P1 = 9 atm A = 100 cm2 V1 V2 15 cm
Question 5 Pext= 1.5 atm T1 = 15°C • n = 1.5 mol CO2 • CV,m= 28.8 JK-1mol-1 • A perfect gas • Adiabatic • q = 0 (adiabatic) • w = ? P1 = 9 atm A = 100 cm2 V1 V2 15 cm
Question 5 Pext= 1.5 atm T1 = 15°C • n = 1.5 mol CO2 • CV,m= 28.8 JK-1mol-1 • A perfect gas • Adiabatic • ΔU = ? • ΔT = ? For perfect gas, Thus, P1 = 9 atm A = 100 cm2 V1 V2 15 cm
Question 5 Pext= 1.5 atm T1 = 15°C • n = 1.5 mol CO2 • CV,m= 28.8 JK-1mol-1 • A perfect gas • Adiabatic e) ΔS=? From part d) P1 = 9 atm A = 100 cm2 V1 V2 15 cm The gas undergoes constant-volume cooling followed by isothermal expansion
Question 5 Pext= 1.5 atm T1 = 15°C • n = 1.5 mol CO2 • CV,m= 28.8 JK-1mol-1 • A perfect gas • Adiabatic e) ΔS=? P1 = 9 atm A = 100 cm2 V1 V2 15 cm Constant volume cooling Find this first!
Question 5 Pext= 1.5 atm T1 = 15°C • n = 1.5 mol CO2 • CV,m= 28.8 JK-1mol-1 • A perfect gas • Adiabatic e) ΔS=? P1 = 9 atm A = 100 cm2 V1 V2 15 cm