Consider the reversible Carnot cycle discussed in class with 1 mol of an ideal g
ID: 3308591 • Letter: C
Question
Consider the reversible Carnot cycle discussed in class with 1 mol of an ideal gas with Cv-3/2R as the working substance. The initial isothermal expansion occurs at the hot reservoir temperature of Thot-600C from an initial volume of 3.50 L to a volume 10.0 L. The system then undergoes an adiabatic expansion until the temperature falls to Tcold-150 C. The system then undergoes an isothermal compression until the initial state Tho-600C and V = 3.51 is reached. Q5.1 (20 points). Calculate the work, q, au and H for each step in the cycle and for the total cycle. Q 5.2 (10 points). Calculate and the amount of heat that is extracted from the hot reservoir to do 1.0 kJ of work in the surroundings.Explanation / Answer
given carnot cycle
n = 1 mol
Cv = 3R/2
isothermal expansion 1-2
Th = 600 C = T1 = T2 = 873.16 K
V1 = 3.5 L
V2 = 10 L
isothermal work, W = nRT1*ln(V2/V1) = 7617.46692 J
chang ein internal energy = 0 J ( isothermal process)
heat supplied = W = 7617.46692 J
Adiabatic expansion, 2-3
T3 = Tc = T4 = 150 C = 423.16 K
heat supplied, Q = 0 ( adiabatic)
change in internal energy U = nCv(Tf - Ti) = -5609.25 J
work = -U = 5609.25 J
isothermal compression
isothermal work = nRT3*ln(V1/V2) = -3691.657 J
change in internal energy = 0
heat supplied = -3691.657 J
Adiabatic compression
Q = 0
U = 5609.25 J
W = - 5609.25 J
hence total work = 3925.80992 J
change in internal energy = 0 J
heat supplie d= 3925.80992 J
hence
for 1 kJ work to be done, heat extraceted = 7617.46692*1000/3925.80992 = 1940.355 J = 1.94035J