Suppose an ideal gas is used as the working substance in the Carnot cycle. (a) I
ID: 2212281 • Letter: S
Question
Suppose an ideal gas is used as the working substance in the Carnot cycle. (a) If the gas absorbs 3 kJ during the isothermal expansion, how much work is performed? (b) If the gas performs 1 kJ of work during the adiabatic expansion, what is the change in internal energy? (c) How much work must be performed on the gas during the isothermal compression for it to remove 2 kJ of heat? (d) How much work must be performed during the adiabatic compression? (e) What is the net amount of work performed by the gas during the cycle? (f) What is the efficiency of the cycle?Explanation / Answer
a) at A, PV = n RT >> n=1 gram mole R = 8.314 put the value of P & V calculate T(A) ----------------- during AB constant Pressure, >> PV(A) = R T(A) and PV(B) = R T(B) divide >> T(B) = [V(B)/V(A)]*T(A) = ?? T(A) calculate as T(A) known ------------------------ during CA constant volume, >> P(c) V = R T(c) and P(a) V = R T(a) divide >> T(c) = [P(c)/P(a)]*T(a) calculate as T(A) known =============== what you find relation in T(c) & T(b) if equal then B-C process is Isothermal compression >> dU = 0 --------------------- net work done = area of ABCA W = 0.5*AB*AC ---------------------- change in internal energy = 1 cycle = 0 because system started fro T(A) temp and reached back at T(A) U = f (T) and depends on end points not the path taken internal energy = 1 cycle = same dU = >> u = constant --------- head added/removed >> dQ = dU + dW = 0 +work done in cylce for this >> dW1 = P dV (constant pressure A-B process) its expansion, work done by gas >>> i take that +ve C-A process> dV=0 constant volume dw2 =0 B-C process >> if isothermal (Tb=Tc) w(bc) = RT(b) log [vc/vb] >>> will be negative add all works in proper sign dw = + or - ve will decide dQ = + or -ve ------------------------- before cycle >> Ta = ? heat content Q(initial) >>> S(initial) = entropy = Q(ini)/Ta after cycle Ta = ? same heat content Q2 = Q(initial) + or - dQ from above S(final) = Q2/Ta result & calculations will fix things precisly. Source(s): can not keep link with PDF (long file) and Qs so not able to fit in values