Question

P.5.2) Consider the Carnot cycle of Fig 5.2 for 1 mol of gas ideal (Cym=3/2R). A isothermal expansion 4.--= nRT In(V/V) 40 B PI adiabatic compression 40 - T adiabatic CNN on YO T 4.- W RT InV/V) C isothermal compression If Thot = 600 °C(873K) and if Tcola = 150°C(423K), and VA = 3.5L, V. = 10.0L. Then, the system closes the cycle. Calculate:

a) Vc and Vd

b) W for each step and the overall cycle

c) Efficiency and qhot to do work = -1kJ

Answer 1

16- B Ť Thoto Thot = 873K to Teodd = 423K. & Ter va = 3.5L sig v = 10.0L.' Step-I : Isothermal expansion' Step-I : Adiabetie. eupension Step-m : Isothermal compression at als Slep - TV : Adliabetic compression.

ce How lon stepa I and one pi Po vb P = Pe Va me That You = me Tole vee a. That (8-1) = Teold ve (8-1) (1) Similarly, That Ta (P-1) - Teold Ved (2-1) Deciding aan (1) by equ (2) che have (3) For step -I . WE - MR Then ( vo n =1 mole Q = %R =-1 moke x 8.314 7 Kimata *8731%dn( o ) =-7.6 kJ. Fon step I : W = ner (Te-TH) = 1 yol * ☆ *8-314 Ik mod (423-873)* =-5.6 kJ.

Fon step II : Cm = MRT In =-1 molle * 8.315 3 molt *423% 1 t= 3.7kJ. sk For step : come = nevo (TH-Te) = 1 motex - *8.314 J K imot " (873-423)K = 5.6 kJ. Bil ... The cool done for overall cycle ! clay =(-766-5-6+ 3-7 +5-6) kJ. = *-3.9kJ.

Teold hot 423 8 +3 c7 E officiency n = 1- Teclado = 1- the - 6 0.51. How to do -1 kJ coook we have to find We know, n=1 Thot that La cold tot

Phot Nowo, :: Hold = i-n = 1-0.51. > 9 cold not 10:49 way = 9 hot - 9 cold on not - Yhet *0.49 = -10 > 9not (1–0.48) = -2.. 7 qner Thot I-o.49. = -1.96 kJ. =-1.96 kJ Thot